On the Degree of Univariate Polynomials Over the Integers Gil Cohen∗ Amir Shpilka † Avishay Tal‡ Abstract We study the following problem raised by von zur Gathen and Roche [GR97]: What is the minimal degree of a nonconstant polynomial f : {0, . . . , n} → {0, . . . , m}? Clearly, when m = n the function f (x) = x has degree 1. We prove that when m = n − 1 (i.e. the point {n} is not in the range), it must be the case that deg(f ) = n − o(n). This shows an interesting threshold phenomenon. In fact, the same bound on the degree holds even when the image of the polynomial is any (strict) subset of {0, . . . , n}. Going back to the case m = n, as we noted the function f (x) = x is possible, however, we show that if one excludes all degree 1 polynomials then it must be the case that deg(f ) = n − o(n). Moreover, the same conclusion holds even if m = O(n1.475− ). In other words, there are no polynomials of intermediate degrees that map {0, . . . , n} to {0, . . . , m}. Furthermore, we give a meaningful answer when m is a large polynomial, or even exponential, in n. Roughly, we show that if m < n/c , for some constant c, then either deg(f ) ≤ d − 1 (e.g. d f (x) = x−n/2 is possible) or deg(f ) ≥ n/3−O(d log n). So, again, no polynomial of intermediate d−1 degree exists for such m. We achieve this result by studying a discrete version of the problem of giving a lower bound on the minimal L∞ norm that a monic polynomial of degree d obtains on the interval [−1, 1]. √ We complement these results by showing that for every d = o( n/ log n) there exists a polyd+0.5 nomial f : {0, . . . , n} → {0, . . . , O(n )} of degree n/3 − O(d log n) ≤ deg(f ) ≤ n − O(d log(n)). Our proofs use a variety of techniques that we believe will find other applications as well. One technique shows how to handle a certain set of diophantine equations by working modulo a well chosen set of primes (i.e. a Boolean cube of primes). Another technique shows how to use lattice theory and Minkowski’s theorem to prove the existence of a polynomial with certain properties. ∗ Department of Computer Science and Applied Mathematics, The Weizmann Institute of Science, Rehovot, Israel. Email: gil.cohen@weizmann.ac.il. † Faculty of Computer Science, Technion-Israel Institute of Technology, Haifa, Israel and Microsoft Research, Cambridge MA. Email: shpilka@cs.technion.ac.il. This research was partially supported by the Israel Science Foundation (grant number 339/10). ‡ Faculty of Computer Science, Technion-Israel Institute of Technology, Haifa, Israel. Email: avishay.tal@gmail.com. Contents 1 Introduction 1.1 Our results . 1.2 Related work 1.3 Techniques . 1.4 Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Preliminaries 2.1 Stirling’s formula . . . . . . . . . . . 2.2 Newton basis . . . . . . . . . . . . . 2.3 Lucas’ theorem . . . . . . . . . . . . 2.4 The gap between consecutive primes 2.5 Linear recurrence relations . . . . . . 3 Warm up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 3 3 4 . . . . . 4 5 5 6 6 7 8 4 Proof of Theorem 1 11 4.1 A cube of primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.2 Missing proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 5 The range of a degree d polynomial 20 5.1 A possible route for improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.2 The case of small degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 6 Proof of Theorem 4 25 6.1 Basic properties of lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 6.2 Proof of Theorem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 7 Back to the Boolean case 30 8 Discussion 30 A Newton polynomials 34 B Lucas’ theorem 35 1 Introduction In this paper we study the following problem that was raised by von zur Gathen and Roche [GR97] What is the minimal degree of a nonconstant polynomial f : {0, . . . , n} → {0, . . . , m}? As f is defined over n + 1 points, its degree is at most n, so the question basically asks whether the degree can be much smaller than n. The answer must of course depend on the choice of m. For example, when m = n we have the polynomial f (x) = x whereas when m = 1 the degree of f is at least n − o(n) [GR97]. Von zur Gathen and Roche observed an obvious lower bound on the degree of nonconstant polynomials f : {0, . . . , n} → {0, . . . , m}, that follows from the pigeonhole principle, namely, deg(f ) ≥ (n + 1)/(m + 1). They also noted that their techniques for the case m = 1 cannot yield bounds better than n − Ω(n) for larger values of m. Thus, prior to this work no lower bounds of the form n − o(n) were known on the degree of polynomials f : {0, ..., n} → {0, . . . , m}, when m > 1. We note that von zur Gathen and Roche were mainly interested in the case that m is independent of n, but the problem is also relevant when m = n − 1 and in fact even for m ≥ n. In such cases, one should omit other ‘trivial’ examples besides the constant functions. The reason that a meaningful answer can be obtained is that the requirement that f takes values in the domain {0, . . . , m} restricts the freedom that the coefficients of f a priori had and puts a severe limitation on their structure. In this paper we focus in the case of large m, although our results clearly hold for small values of m as well. The goal to better understand the degree of polynomials is well motivated by the important role that polynomials (both multivariate and univariate) play in theoretical computer science. For example, polynomials are prominent in areas such as circuit complexity [Raz87, Smo87, Bei93], learning theory [LMN93, MOS04], decision tree complexity and quantum query complexity [BdW02], Fourier analysis of Boolean functions [KLM+ 09, ST10], explicit constructions (see e.g. [Gop06]) and more. Understanding the complexity of univariate polynomials is one of the most important problems in algebraic complexity as it is closely related to the question of hardness of integer factorization (see e.g. Section B.3 in [Gol08]). The degree of polynomials is probably the most simple and natural complexity measure that is associated with them. Indeed, a basic question in the study of polynomials that attracted a lot of interest concerns the minimal degree that a polynomial, belonging to some predetermined family of polynomials, can have. This fundamental question was studied before in the context of multivariate real polynomial approximation of Boolean functions (see the survey [BdW02]), in the study of representations of symmetric Boolean functions as univariate polynomials [GR97] (where the problem that we study here was raised) and in relation to learning symmetric juntas [MOS04, KLM+ 09, ST10]. In [ST10] it was showed that in order to better understand the Fourier spectrum of symmetric functions one needs to study polynomials f : {0, ..., n} → {0, 1, 2} and prove lower bounds on their degree, which is exactly the question that we study here for the case m = 2. Besides its connection to complexity theory, the question of understanding univariate polynomials is important from an approximation theory point of view. A different angle to look at our problem is asking, for a given degree d how small can the range of a degree d polynomial mapping {0, . . . , n} to N be. This question is a discrete version of a fundamental question in approximation theory concerning the minimal L∞ norm of monic polynomials1 over the real interval [−1, 1]. That is, the question is what is minf maxx∈[−1,1] |f (x)|, where f ranges over all monic polynomials of degree d. It is well known that Chebyshev polynomials are the only extremal example. The problem that we study in this paper basically asks for the minimum L∞ norm that a monic polynomial of degree d attains at the points {−1, −1 + n2 , . . . , 1}. Namely, what is minf maxx∈{−1,−1+ 2 ,...,1} |f (x)|, where n 1 A polynomial is monic if its leading coefficient is 1. 1 f ranges over all monic polynomials of degree d. There is a significant difference from the original question as we allow the polynomial to take arbitrarily high values on other points in the interval. √ While for d < n one can get a good estimate using the classical theory of Chebyshev polynomials, this is not the case for larger values of d. We discuss this connection in more detail in Section 5.1. 1.1 Our results We prove two main results concerning the degree of polynomials over the integers. Both results present a dichotomy behavior. That is, given a function f : {0, . . . , n} → {0, . . . , m}, either deg(f ) is very small (we consider those cases as ‘trivial’) or deg(f ) is very high. The first result gives a strong lower bound when m is not too large (but still larger than n). Theorem 1. For every > 0 there exists n such that for every n > n and f : {0, 1, . . . , n} → {0, 1, . . . , n1.475− }, either deg(f ) ≤ 1 or deg(f ) ≥ n − 4n/ log log n. As an immediate corollary we get that if a polynomial tries to “compress” the domain even by one value, then it must have a nearly full degree. Corollary 1.1. Let S ( {0, . . . , n} and f : {0, . . . , n} → S be a nonconstant polynomial. Then, deg(f ) ≥ n − 4n/ log log n. Note that such a strong result cannot hold for m ≥ n as, for example, the function f (x) = x maps {0, . . . , n} to itself. Our second main result concerns larger values of m at the price of a slightly weaker dichotomy. 2 Theorem 2. There exists a constant nn integers satisfying d ≤ 15 n and n > n0 0 such that j if d, n are ko d n−d 1 then the following holds. If f : [0, n] → 0, . . . , √7d · 2d is a polynomial then deg(f ) ≤ d − 1 1 n or deg(f ) ≥ 13 n − 1.2555 · [d ln( n−d 2d ) − 2 ln( d )]. In other words, besides the (“trivial”) case where deg(f ) ≤ d − 1, the only other option is that f has a relatively high degree. The proof of Theorem 2 relies on the following theorem that gives a lower bound on the maximum value that any monic polynomial must obtain on the points {0, . . . , n}. d Theorem 3. Let f : R → R be a degree d monic polynomial. Then, maxi=0,1,...,n |f (i)| > n−d . In 2e particular, if f : Z → Z is a degree d polynomial (not necessarily monic) then 1 n−d d 1 n−d d max |f (i)| > · ≥√ · . i=0,1,...,n d! 2e 2d 7d As mentioned before, this question is a discrete analog of a question from approximation theory asking for the minimal L∞ norm of a monic polynomial of degree d over the real interval [−1, 1]. √ Our next result gives an upper bound on the degree when the range is of size at most exp(o( n)). √ Theorem 4. For every large enough integer n > 0 and integer d = o( n/ log n) there exists f : {0, . . . , n} → {0, . . . , O(nd+0.5 )} of degree 2d < deg(f ) ≤ n − d log(n). In particular, by Theorem 2, it holds that n/3 − O(d log n) ≤ deg(f ) ≤ n − d log(n). We note that in [GR97] von zur Gathen and Roche conjectured that any such nonconstant polynomial to {0, 1} must be of degree n − O(1). While this conjecture is still open, Theorem 4 shows that one can get polynomials of lower degree when the range is larger, even after excluding the obvious examples. Finally, we consider polynomials f : {0, . . . , n} → {0, 1}, where n = p2 − 1 and p is a prime √ number. We are able to show that in this case deg(f ) ≥ p2 − p > n − n. This improves the result of [GR97] for this special case. Theorem 5. Let p be a prime number, n = p2 − 1 and f : {0, . . . , n} → {0, 1} be nonconstant. Then √ deg(f ) ≥ p2 − p > n − n. 2 We summarize our results in Table 1 that can be found on page 33. 1.2 Related work The most relevant result is the aforementioned work of von zur Gathen and Roche [GR97] that raised and studied the question of bounding (from below) the minimal degree that a real polynomial representing a nonconstant symmetric Boolean function can have. As any symmetric function f : {0, 1}n → {0, 1} is actually a function of the number of ones in x, it can be represented by a unique polynomial f : {0, ..., n} → {0, 1} (we abuse notations here and think of f both as a univariate polynomial and as a symmetric function). Thus, von zur Gathen and Roche basically studied the question of giving a lower bound on the minimal degree of nonconstant polynomials f : {0, ..., n} → {0, 1}. They showed that when n = p − 1, p prime, it must be the case that deg(f ) = n (when f is not constant). Using the density of prime numbers (see Theorem 2.6) they concluded that deg(f ) ≥ n−o(n) for every n (in the notations of Theorem 2.6, deg(f ) ≥ n−Γ(n)). For the case of polynomials taking values in {0, . . . , m}, von zur Gathen and Roche observed that deg(f ) ≥ (n + 1)/(m + 1) and mentioned that their techniques cannot give any result of the form deg(f ) = n − o(n). However, they suggested that “...for each m there is a constant Cm such that deg(f ) ≥ n − Cm for all n.” In particular, when m = O(1), this amounts to having deg(f ) ≥ n − O(1). This conjecture is still open, even for the case m = 1. Another line of work concerning symmetric Boolean functions f : {0, 1}n → {0, 1} has focused on bounding from above the minimal size of a nonempty set S such that fˆ(S) 6= 0, where fˆ(S) is the Fourier coefficient of f at S. We do not want to delve into the definition of the Fourier transform, so we only mention that when f is balanced, i.e. takes the values 0 and 1 equally often, this is the same as bounding from below the degree of f ⊕ PARITY, see [KLM+ 09] for details. As symmetric Boolean functions can be represented by univariate polynomials from {0, . . . , n} to {0, 1}, this problem is closely related to the questions studied here. A strong motivation for studying the case m > 1 was given in [ST10] where it was shown that bounding from below the degree of univariate polynomials to {0, 1, 2}, will give an upper bound on the size of such a set S (for which fˆ(S) 6= 0), even when f is not balanced. Thus, an advance in understanding the degree of polynomials over the integers, that obtain more than two values, may shed new light on a well studied problem concerning the Fourier spectrum of symmetric Boolean functions. 1.3 Techniques The proofs of Theorems 1, 3 and 4 use a completely different set of techniques. In the proof of Theorem 1 we rely on solving systems of diophantine equations by working modulo a well chosen set of primes. The proof of Theorem 3 is more elementary and follows from some averaging argument. For the proof of Theorem 4 we use lattice theory and Minkowski’s theorem to prove the existence of a polynomial with the required properties. We shall now extend more on each of the proofs. We give a very rough sketch of the idea of the proof of Theorem 1. Our goal is to show that every nonlinear polynomial f : {0, . . . , n} → {0, . . . , m}, for m ∼ n1.475 , must have high degree. As the coefficients of f are determined by the set of values {f (0), f (1), . . . , f (n)}, and in fact are linear combinations of them, a natural approach is to look at these dependencies and prove that one of the highest coefficients cannot be zero. Specifically, representing f in the basis of the Newton polynomials (see Definition 2.2) we get an explicit and nice formula for each coefficient. If f is not of high degree, many of those coefficients vanish and this gives a set of linear equations that the values {f (0), f (1), . . . , f (n)} must satisfy. In fact, we manage to get many linear equations from every zero 3 coefficient. The idea is that if the degree of f is smaller than a prime number p, then the values f (r) and f (r + p) must be strongly correlated for r ∈ {0, . . . , n − p}. Using such correlations for many different primes, we obtain a set of special linear equations (which we call linear recurrence relations) on the values of f . A similar approach was taken in [KLM+ 09] (and arguably also in [GR97]) where the authors used different primes to obtain information for the case m = 1. It is not clear, however, how to exploit the information from the different primes. We manage to do so by considering prime numbers that form a ‘nice’ and ‘rigid’ structure that we call a cube of primes. An r-dimensional cube of primes is a set P = Pp;δ1 ,...,δr ⊆ [n] of the form ( ) r X P = p+ ai δi | a1 , . . . , ar ∈ {0, 1} , i=1 such that all the elements of P are prime numbers. The idea is that we can partition P , in many different ways, to pairs of primes such that the differences, between the primes in each pair, are the same. This enables us to combine the different linear recurrences obtained from each prime in a way that reveals more information on the values that f takes. Theorem 2 is an immediate corollary of Theorem 3 whose proof goes along completely different lines than the proof of Theorem 1. The idea is to observe that since f has at most d roots in the interval {0, . . . , n} then some point in that interval is relatively far from all roots of f . This immediately implies that f obtains a large value at this point. To prove Theorem 4 we note that the polynomials that we study belong to a certain lattice. Proving the existence of a polynomial with a (relatively) small range amounts to proving that a lattice point exists outside the linear span of the lattice points corresponding to low degree polynomials. To show this fact we use Minkowski’s theorem that gives an upper bound on the minimal product of the lengths of linearly independent vectors from the lattice. Using this upper bound we can prove the existence of a lattice point that is linearly independent from points corresponding to low degree polynomials and whose length is not too large. By the construction of the lattice, the fact that the length of the point is small implies that the polynomial corresponding to that point has a relatively small range, as needed. 1.4 Organization The paper is organized as follows. In Section 2 we give the basic definitions and discuss mathematical tools that we shall later use. In Section 3 we demonstrate our general technique by considering the case of 2-dimensional cube of primes. In Section 4 we prove Theorem 1 and conclude Corollary 1.1. In Section 5 we prove Theorems 2 and 3 and discuss their tightness. We then present the connection to Chebyshev polynomials in Section 5.1 and conclude Theorem 6 that improves Theorem 3 for √ d ≤ n/2. We prove the existence of a polynomial that has degree n − d log n in Section 6. Finally, in section 7 we consider the case m = 1 and n = p2 − 1 for a prime p. We note that the results in Sections 4, 5 and 6 are independent of each other so it is not required to read the paper in a linear order. 2 Preliminaries For two integers a, b we denote with [a, b] the set of all integers between a and b. Namely, [a, b] , {c ∈ Z | a ≤ c ≤ b} = {a, a + 1, . . . , b}. We also denote [m] , [1, m]. We sometimes abuse notation 4 and speak of the real interval [a, b] (in this case [a, b] = {a ≤ x ≤ b | x ∈ R}). We will always mention the words ‘real interval’ whenever we speak of the real interval. For a prime number pPand integers a, b we denote a ≡p b when a and b are equal modulo p. For a polynomial f (x) = ni=0 ai xi we denote with monom(f ) the number of monomials in f . I.e. the number of nonzero ai ’s. We denote the family of all polynomials from [0, n] to [0, m] by Fm (n). Namely, Fm (n) = {f ∈ Q[x] | deg(f ) ≤ n, f : {0, 1, . . . , n} → {0, 1, . . . , m}}. Throughout the paper we avoid the use of floor and ceiling in order not to make the equations even more cumbersome. This does not affect our results and only makes the reading easier. We denote by log(·) and ln(·) the logarithms to the base 2 and to the base e (that is, the natural logarithm) respectively. In the next subsections we present some well known technical tools that we require for our proofs. 2.1 Stirling’s formula We shall make use of the well known Stirling approximation for the factorial function. Theorem 2.1 (Stirling’s formula). For every natural number n ∈ N it holds that n n √ n! = 2πn · · eλn e with 1 1 < λn < . 12n + 1 12n A proof of this theorem can be found, e.g., in [Rob55] (see also pages 50-53 of [Fel68]). 2.2 Newton basis Definition 2.2. For every k ∈ N, define the polynomial xk as follows x(x − 1) · · · (x − k + 1) x = . k k! The set of polynomials xk : k ∈ N is called the Newton basis. It is easy to see that xk : k = 0, 1, . . . , d forms a basis to the vector space of polynomials of degree at most d. An interesting property of the Newton basis is given in the next theorem whose simple proof can be found in Appendix A. Theorem 2.3. Let f ∈ Q[x] be a polynomial of degree ≤ n. Then f can be represented as f (x) = n X d=0 x γd · d where γd = d X d−j (−1) j=0 d · · f (j) . j As noted in [GR97], Theorem 2.3 implies that a polynomial f is of degree smaller than d iff for all d ≤ s ≤ n it holds that s X s (−1)j f (j) = (−1)s γs = 0 . j j=0 As an immediate corollary we get the following useful lemma. 5 Lemma 2.4. Let f : [0, n] → Z be such that deg(f ) < d. Then for all r ∈ [0, n − d] we have that d X d j=0 j · (−1)j · f (j + r) = 0 . Proof. For r ∈ [0, n − d] set gr (x) = f (x + r). We think of gr as a function gr : [0, n − r] → Z. As deg(gr ) = deg(f ) < d, Theorem 2.3 implies that d X (−1)j j=0 2.3 d X d d f (j + r) = (−1)j gr (j) = 0 . j j j=0 Lucas’ theorem The following theorem of Lucas [Luc78] allows one to compute a binomial coefficient modulo a prime number. In Appendix B we give one of the many known proofs. Theorem 2.5 (Lucas’ theorem). Let a, b ∈ N \ {0} and let p be a prime number. Denote with a = a0 + a1 p + a2 p2 + · · · + ak pk and b = b0 + b1 p + b2 p2 + · · · + bk pk their base p expansion. Then k Y a ai ≡p , b bi i=0 where 2.4 ai bi = 0 if ai < bi . The gap between consecutive primes Denote with pn the n-th prime number. Understanding the asymptotic behavior of pn+1 − pn is a long standing open question in number theory. Cramér conjectured that pn+1 − pn = O((log pn )2 ) √ and, assuming the correctness of Riemann hypothesis, he proved that pn+1 − pn = O( pn log pn ) [Cra36]. The strongest unconditional result is due to Baker et al. [BHP01].2 Denote with π(n) the number of primes numbers less than or equal to n. Theorem 2.6 ([BHP01]). For any large enough integer n and any y ≥ n0.525 we have that π(n) − π(n − y) ≥ 9 y · . 100 log n For convenience, we denote Γ(n) , n0.525 . We will usually apply the theorem above to claim, for some integer n, that there exists a prime number p ∈ [n − Γ(n), n]. 2 The main theorem of [BHP01] only claims that there exists a prime number in the interval [n − n0.525 , n], however they actually prove the stronger claim that is stated here. 6 2.5 Linear recurrence relations P Definition 2.7. Let Φ(t) = si=0 αi ti be a polynomial with rational coefficients.3 For f ∈ Q[x] we define the action of Φ on f as s X (Φ ◦ f )(x) , αi · f (x + i) . i=0 When we consider Φ as an operator acting on other polynomials, we call Φ a linear recurrence polynomial. From now on we will always denote linear recurrence polynomials with capital Greek letters: Φ, Ψ, Υ. Following is a list of properties of linear recurrence polynomials. Lemma 2.8. For polynomials f, g and linear recurrences Φ, Φ0 the following claims hold. 1. Φ ◦ f ∈ Q[x]. 2. deg(Φ ◦ f ) ≤ deg(f ). 3. (Φ + Φ0 ) ◦ f = Φ ◦ f + Φ0 ◦ f . 4. Φ ◦ (f + g) = Φ ◦ f + Φ ◦ g. 5. (Φ · Φ0 ) ◦ f = Φ ◦ (Φ0 ◦ f ). Proof. Properties 1-4 follow from the definition. Property 5 follows by a simple calculation. Pe Pd trivially j i 0 Denote, w.l.o.g., Φ(t) = i=0 αi x and Φ (t) = j=0 βj x . We have that 0 Φ · Φ ◦ f (x) = d X e X αi βj xi+j ◦ f (x) = i=0 j=0 d X e X αi βj f (x + i + j) i=0 j=0 d e X X = αi βj f (x + i + j) = Φ ◦ (Φ0 ◦ f ) (x) . i=0 j=0 | {z (Φ0 ◦f )(x+i) } While property 2 of Lemma 2.8 states the obvious fact that applying a linear recurrence cannot increase the degree, the following lemma assures that the degree can decrease by (roughly) at most the number of monomials in the linear recurrence polynomial. P Lemma 2.9. Let f ∈ Q[x] be a nonconstant polynomial and let Φ(t) = si=1 αi · tdi be some linear recurrence, Φ 6= 0. Then, for g = Φ ◦ f we have that ( s−2 g≡0 deg(f ) ≤ s + deg(g) − 1 otherwise Proof. As Φ 6= 0 we can assume w.l.o.g. that the exponents d1 , . . . , ds are distinct (indeed if they are not distinct then we can rewrite Φ as a polynomial with s0 < s monomials and obtain stronger results). Similarly, if deg(f ) ≤ s−2 then we are done. So, we may assume w.l.o.g. that deg(f ) ≥ s−1. 3 There is nothing special about Q and the only reason that we use it is that in our proofs we encounter rational coefficients. 7 P i Let f (x) = D `=0 bi x , where bD 6= 0. Let U be a (D + 1) × (D + 1) lower triangular matrix whose (i, j) entry (for i, j = 0, . . . , D) is Ui,j , bD+j−i · D+j−i (where bD+j−i = 0 if j > i). This is j clearly a lower triangular matrix with a nonzero diagonal. Let V be a (D + 1) × s Vandermonde matrix defined as (for i = 0, . . . , D and j = 1, . . . , s) Vi,j , (dj )i . It is now easy to verify that the coefficients of the polynomial g = Φ◦f the result of the matrix-vector multiplication U ·V ·~ α where Pare D i ~ . I.e. cD−r = (U · V · α ~ )r . α ~ = (α1 , . . . , αs ). Namely, if g(x) = i=0 ci x then (cD , . . . , c0 ) = U · V · α Indeed, (Φ ◦ f )(x) = s X αi f (x + di ) = i=1 s X i=1 αi D X j bj (x + di ) = j=0 s X αi i=1 D X j=0 j X j j−k k d x = bj k i k=0 X s D D s D D−k X X X X j `+k X j−k k k x bj αi di = x b`+k αi d`i . k k k=0 j=k i=1 k=0 `=0 i=1 Hence, the coefficient of xD−r is: r X `=0 s r D X X `+D−r X b`+D−r αi d`i = Ur,` (V · α ~ )` = Ur,` (V · α ~ )` = (U · V · α ~ )r . D−r i=1 `=0 `=0 As the first s rows (recall that D + 1 = deg(f ) + 1 ≥ s) of U · V form an invertible matrix (as a product of a Vandermonde matrix with a lower triangular matrix that has a nonzero diagonal), we see that the top s coefficients of g are zero iff α ~ = 0 (which is a contradiction to the assumption). Hence, the degree of g is at least D − s + 1 = deg(f ) − s + 1. 3 Warm up In this section we prove some preliminary results that give good intuition to the proofs of Theorem 1 (and also to the proof of Theorem 7). Similarly to other works that studied the degree of polynomials over the integers [GR97, KLM+ 09], we shall consider properties of the polynomial modulo different prime numbers. As a first step we show that if f ∈ Fn−1 (n) is of low degree then it is actually a constant function. The proof of the lemma already contains some of the ingredients that we will later use in a more sophisticated manner. Lemma 3.1. Let f ∈ Fn−1 (n) be such that deg(f ) < n/6 − Γ(n), then f is a constant. Proof. Let p ∈ [n/2, n/2 + Γ(n)] be a prime number, guaranteed to exist by Theorem 2.6. Since deg(f ) < p, Lemma 2.4 implies that for all r ∈ [0, n/2 − Γ(n)] ⊆ [0, n − p] we have that 0= p X k=0 (−1)k p f (k + r) ≡p f (r) − f (p + r) . k In particular, if we define g by g(r) = f (r)−fp (p+r) , then we have that g : [0, n/2 − Γ(n)] → [−1, 1] (indeed, f (r) − f (p + r) ∈ [−n + 1, n − 1]). Clearly, g + 1 ∈ F2 (n). Note that if g is not constant then its degree must be at least (n/2 − Γ(n))/3 as one of the values in its range is obtained at least that many times. Since in this case n/6 − Γ(n) < deg(g) ≤ deg(f ) we get a contradiction. Therefore, g must be constant. However, in this case we get by Lemma 2.9 that deg(f ) ≤ deg(g) + 2 − 1 = 1. Indeed, for Φ(t) = p1 − p1 tp , it holds that g = Φ ◦ f . Hence, deg(f ) ≤ 1. Since the range of f is smaller than its domain (and f takes integer values), f is constant. 8 Clearly, for m ≥ n, we cannot expect such a strong behavior (that is, degree 0 as opposed to degree Ω(n)). However, the following lemma, which relies on Lemma 3.1, shows that a slightly weaker dichotomy behavior exists for m which is roughly quadratic in n. We later strengthen this result (Corollary 5.4). Lemma 3.2. Let m < then deg(f ) ≤ 1. n2 −4Γ(n)2 8 be an integer and f ∈ Fm (n) be such that deg(f ) < n/12 − Γ(n), Proof. Let p ∈ [ n2 − Γ(n), n2 ] be a prime number, guaranteed to exist by Theorem 2.6. As before, Lemma 2.4 implies that for all r ∈ [0, n − p] we have that 0= p X k=0 p f (k + r) ≡p f (r) − f (p + r) . (−1) k In particular, if we define g by g(r) = Clearly, g + m p ∈ F 2m (n − p), and k f (r)−f (p+r) , p then we have that g : [0, n − p] → [−m/p, m/p]. p 2 Hence, g + m p ( n − Γ(n))( n2 + Γ(n)) m < 2 ≤n−p. p p is actually in Fn−p−1 (n − p), and deg(g + m n n−p ) ≤ deg(f ) ≤ − Γ(n) ≤ − Γ(n − p) . p 12 6 Now we can apply Lemma 3.1 to conclude that g + deg(f ) ≤ 1 which completes the proof. m p is constant. From Lemma 2.9 it follows that n2 −4Γ(n)2 is very close to being tight. Indeed, 8 n−1 2 [0, n] → [0, n 8−1 ] defined as f (x) = x− 2 2 . We note that the choice m < assume that n is odd and consider the function f : An important ingredient in the proof of Theorem 1 is the use of prime numbers that form a structure analogous to a cube. To illustrate our approach, consider four prime numbers of the form p < p + δ1 < p + δ2 < p + δ1 + δ2 . Using Theorem 2.6 one can show that such primes exist and that we can even choose them so that they all lie in an interval of the form [n/3 − o(n), n/3]. Lemma 3.3. Let n be a large enough integer. Then, there exist four prime numbers n n − Γ(n) ≤ p < p + δ1 < p + δ2 < p + δ1 + δ2 ≤ . 3 3 Proof. The lemma follows from the more general Lemma 4.1 that is proved in Section 4.1, however, for clarity we prove this special case here. Theorem 2.6 guarantees that for a large enough n there are at least4 Γ(n)/12 log(n) prime numbers in the interval [n/3 − Γ(n), n/3]. Consider all possible differences between two primes in this set. There are at least, say, 13 (Γ(n)/12 log(n))2 such differences. As all the differences are smaller than 1 (Γ(n)/12 log(n))2 many Γ(n) it follows that one of the differences is obtained for at least 3 ≥ 500Γ(n) Γ(n) log2 (n) pairs of primes. Denote the i-th pair with (pi,1 , pi,2 ) where pi,1 < pi,2 . Consider any two distinct pairs in the set, (p1,1 , p1,2 ) and (p2,1 , p2,2 ). Denote δ1 = p1,2 − p1,1 = p2,2 − p2,1 and δ2 = |p1,1 − p2,1 | > 0. We have that 0 < δ1 + δ2 < Γ(n). In particular, {p1,1 , . . . , p2,2 } is the required cube.5 4 5 There is nothing special about 12, it is just a large enough constant. We can of course make sure that p2,1 6= p1,2 , and hence δ1 6= δ2 , by ‘throwing’ away one pair. 9 As a warmup for our main result and to demonstrate our proof technique we shall prove here the following easier theorem. Theorem 3.4. If f ∈ Fm (n), where m < n/7, is nonconstant then deg(f ) ≥ 2n/3 − 2Γ(n). Although the theorem is much weaker than Theorem 1, its proof demonstrates our general technique and, hopefully, will make the proof of Theorem 1 easier to follow. Proof. Let p, δ1 , δ2 be as guaranteed in Lemma 3.3. Assume for a contradiction that f ∈ Fm (n) is such that deg(f ) < 2n/3 − 2Γ(n) ≤ 2p. Consider the identity guaranteed by Lemma 2.4 modulo each of the four primes. For example, taking s = 2p (in the notations of Lemma 2.4), we get that for all r = 0, . . . , n − 2p 0= 2p X k=0 2p f (k + r) ≡p f (r) − 2f (p + r) + f (2p + r) . (−1) k k (1) Since |f (r) − 2f (p + r) + f (2p + r)| < 2n/7 < p, Equation (1) is actually satisfied over the integers. Namely, f (r) − 2f (p + r) + f (2p + r) = 0. In the same manner we get, for all r ∈ [0, n − 2(p + δ1 + δ2 )] f0,0 (r) , f (r) − 2f (p + r) + f (2p + r) = 0 , f1,0 (r) , f (r) − 2f (p + δ1 + r) + f (2p + 2δ1 + r) = 0 , (2) f0,1 (r) , f (r) − 2f (p + δ2 + r) + f (2p + 2δ2 + r) = 0 , f1,1 (r) , f (r) − 2f (p + δ1 + δ2 + r) + f (2p + 2δ1 + 2δ2 + r) = 0 . We now show how to combine these equations in a way that will give information not only for small values of r (i.e. r ≤ n − 2(p + δ1 + δ2 )) but also for larger values of r. By considering the following linear combinations of the equalities f0,0 , . . . , f1,1 we get that for r ∈ [0, n − 2(p + δ2 + 2δ1 )] it holds that 0 = f0,0 (r + 2δ1 ) − f1,0 (r) = f (r + 2δ1 ) − f (r) − 2f (p + r + 2δ1 ) + 2f (p + r + δ1 ) , 0 = f0,1 (r + 2δ1 ) − f1,1 (r) = f (r + 2δ1 ) − f (r) − 2f (p + r + 2δ1 + δ2 ) + 2f (p + r + δ1 + δ2 ) . Therefore, 0 = (f0,0 (r + 2δ1 + δ2 ) − f1,0 (r + δ2 )) − (f0,1 (r + 2δ1 ) − f1,1 (r)) = f (r + 2δ1 + δ2 ) − f (r + δ2 ) − f (r + 2δ1 ) + f (r) . (3) Similarly, 1 0 = − · ((f0,0 (r + 2δ1 ) − f1,0 (r)) − (f0,1 (r + 2δ1 ) − f1,1 (r))) 2 = f (p + r + 2δ1 ) − f (p + r + δ1 ) − f (p + r + 2δ1 + δ2 ) + f (p + r + δ1 + δ2 ) (4) and 0 = f0,0 (r + δ1 ) − f1,0 (r) − f0,1 (r + δ1 ) + f1,1 (r) = f (2p + r + δ1 ) − f (2p + r + 2δ1 ) − f (2p + r + δ1 + 2δ2 ) + f (2p + r + 2δ1 + 2δ2 ) . (5) We thus get the following equations for every 0 ≤ r ≤ n − 2(p + δ1 + δ2 ) 0 = f (r + 2δ1 + δ2 ) − f (r + δ2 ) − f (r + 2δ1 ) + f (r) (6) 0 = f (p + r + 2δ1 ) − f (p + r + δ1 ) − f (p + r + 2δ1 + δ2 ) + f (p + r + δ1 + δ2 ) (7) 0 = f (2p + r + δ1 ) − f (2p + r + 2δ1 ) − f (2p + r + δ1 + 2δ2 ) + f (2p + r + 2δ1 + 2δ2 ) (8) 10 These equations give linear recurrence relations on the values of f on the intervals [0, n − 2p − 2(δ1 + δ2 )], [p, n − p − 2(δ1 + δ2 )] and [2p, n − 2(δ1 + δ2 )]. Indeed, Equations 7 and 8 are equivalent to 0 = f (r + 2δ1 ) − f (r + δ1 ) − f (r + 2δ1 + δ2 ) + f (r + δ1 + δ2 ) 0 = f (r + δ1 ) − f (r + 2δ1 ) − f (r + δ1 + 2δ2 ) + f (r + 2δ1 + 2δ2 ) (9) for r ∈ [p, n − p − 2(δ1 + δ2 )] and r ∈ [2p, n − 2(δ1 + δ2 )], respectively. Let Φ(t) = (t2δ1 +δ2 − tδ2 − t2δ1 + 1) · (t2δ1 − tδ1 − t2δ1 +δ2 + tδ1 +δ2 ) · (tδ1 − t2δ1 − tδ1 +2δ2 + t2δ1 +2δ2 ) (10) It follows that (Φ◦f )(r) = 0 for all r ∈ [0, n−2p−6(δ1 +δ2 )]∪[p, n−p−6(δ1 +δ2 )]∪[2p, n−6(δ1 +δ2 )] (see Property 5 in Lemma 2.8).6 We have two cases: • The three ranges are distinct. In this case, Φ◦f has at least 3·(n−2p−6(δ1 +δ2 )) ≥ n−18(δ1 +δ2 ) many roots. • The three ranges overlap. In this case, Φ ◦ f has at least n − 6(δ1 + δ2 ) many roots. Either way, Φ ◦ f has at least n − 18(δ1 + δ2 ) many roots. We conclude that either Φ ◦ f ≡ 0 or deg(Φ ◦ f ) ≥ n − 18(δ1 + δ2 ). As deg(Φ ◦ f ) ≤ deg(f ) < 32 n < n − 18(δ1 + δ2 ) it must be the case that Φ ◦ f ≡ 0. Hence, by Lemma 2.9 it follows that deg(f ) = O(1). However, at this point we can apply Lemma 3.1 and conclude that f is constant. In the general case, we will not be able to deduce that in (the analogous equation to) Equation (2) the sum is equal to 0, but rather we will only bound it from above. Furthermore, we will work with 2Ω(log log n) many prime numbers that form a structure of an Ω(log log n)-dimensional cube (in the sense that {p, p + δ1 , p + δ2 , p + δ1 + δ2 } is a 2-dimensional cube). This will make the construction of the relevant Φ more complicated, but the high level ideas will be similar. 4 Proof of Theorem 1 In this section we prove Theorem 1. We repeat it for the sake of readability. Theorem (Theorem 1). For every > 0 there exists n such that for every n > n and f : {0, 1, . . . , n} → {0, 1, . . . , n1.475− }, either deg(f ) ≤ 1 or deg(f ) ≥ n − 4n/ log log n. Proof of Theorem 1. For convenience, set η = log log(n)/2 and m = n1.475− . Let f ∈ Fm (n) be a function such that 2 4n deg(f ) < n · (1 − ) = n − . η log log n As was demonstrated in Section 3, we will consider the behavior of f modulo various prime numbers that form a high dimensional cube of primes. The existence (and properties) of this structure is guaranteed by the next lemma. Lemma 4.1. Let 0 < < 1/2, there exists n0 () such that for any n > n0 () and η = log log(n)/2, there exists a set ( ) η X n n Pp;δ0 ,δ1 ,δ2 ,...,δη = p + ai · δi | ∀i ai ∈ {0, 1} ⊆ − 4Γ(n), − Γ(n) η+1 η+1 i=0 with the following properties: 6 The change in the range of r occurs since we want all the evaluations points of Φ ◦ f to be inside the interval [0, n]. 11 1. Every q ∈ Pp;δ0 ,δ1 ,δ2 ,...,δη is a prime number. 2. δi > 0 for all i = 1, . . . , η. P 3. ∆ , ηi=1 δi ≤ n . 4. δ0 ∈ [Γ(n), 3Γ(n)]. We defer the proof of the lemma to Section 4.1 and continue with the proof of Theorem 1. We shall consider two subcubes of Pp;δ0 ,δ1 ,δ2 ,...,δη . Denote B , Pp;δ1 ,δ2 ,...,δη and B0 , Pp+δ0 ;δ1 ,δ2 ,...,δη . Note that in both B, B0 we do not consider shifts by δ0 . Let q ∈ Pp;δ0 ,δ1 ,δ2 ,...,δη = B ∪B0 be a prime number. From the construction of Pp;δ0 ,δ1 ,δ2 ,...,δη it follows that (for a large enough n) 2 n deg(f ) < n · (1 − ) < · η < qη . η η+2 (11) Combining Lemma 2.4 and Lucas’ theorem (Theorem 2.5) we get that for every r ∈ [0, n − qη] it holds that qη η X X qη η j 0= · (−1) · f (j + r) ≡q · (−1)j · f (qj + r) . (12) j j j=0 j=0 Notice that this equality is analogous to Equation (1) from the proof of Theorem 3.4. Since f ∈ Fm (n) we can rewrite Equation (12) as η X η j=0 where |Kq,r (f )| < j · (−1)j · f (qj + r) = Kq,r (f ) · q, (13) 2η · m m η m 2η 2η · m < = · 2 · (η + 2) < · 2 = n0.475− · 22η . q n/(η + 2) n n (14) Thus, instead of summing to 0 as was the case in Equation (2), we get that the sum equals a relatively small (i.e., at most poly log(n) · n0.475− ) multiple of q. In the language of linear recurrence, when applying the linear recurrence η X η Ψq (t) = · (−1)j · tqj (15) j j=0 to f we get (Ψq ◦ f )(r) = Kq,r (f ) · q (16) for every r ∈ [0, n − qη]. We now combine all the different Ψq ’s to obtain a linear recurrence in an analogous way to the way that we combined the different equalities in (2) to create the linear recurrences given by (6),(7) and (8). Let p̃ be either p or p + δ0 . We will cancel out all the monomials of the linear recurrence except those whose exponents lie in a small range: [p̃k, p̃k + η∆]. Consider the following linear recurrence for k ∈ [0, η] Φ0p̃,k (t) = X Pη (−1) i=1 ai Pk · Ψ(p̃+Pηi=1 ai ·δi ) (t) · t i=1 (1−ai )·(i−1)·δi + Pη i=k+1 (1−ai )·i·δi . (17) ~a∈{0,1}η The reason for this complicated looking expression will become clear soon when we showP that this linear recurrence give information about f (r) for r ∈ [p̃k, p̃k+n−η(p̃+∆)] (recall that ∆ = ηi=1 δi ≤ 12 n ). The following claim shows that indeed Φ0p̃,k has the required property. To simplify the statement of the claim let7 if ai = 1 k i − 1 if ai = 0 and i ≤ k (18) c~a,k,k (i) , i if ai = 0 and i ≥ k + 1 Pη Pη P Claim 4.2. Φ0p̃,k (t) = tkp̃ · (−1)k · kη · ~a∈{0,1}η (−1) i=1 ai · t i=1 c~a,k,k (i)·δi . To ease the reading we postpone the proof of the claim to Section 4.2 and proceed with the proof of Theorem 1. Claim 4.2 has two interesting consequences. The first is that p̃ only appears in the term tkp̃ . The second is that Φ0p̃,k is actually divisible by tkp̃ . In particular if we set Φp̃,k (t) , Φ0p̃,k (t)/tkp̃ (19) then we get that Φp̃,k gives a recurrence relation for every r ∈ p̃k + [0, n − η(p̃ + ∆)] = [p̃k, p̃k + n − η(p̃ + ∆)]. This is similar to the way that we obtained Equation (9) from Equations (6),(7) and (8). Furthermore, since we factored out the term tkp̃ , it follows that Φp,k = Φp+δ0 ,k . (20) We now wish to better understand the value of Φp̃,k ◦ f . Equations (16),(17) and (19) imply that for X Lp̃,r (f ) , Pη (−1) i=1 ai · K(p̃+Pηi=1 ai ·δi ),r0 (f ) ~ a (21) ~a∈{0,1}η and X Lp̃,j,r (f ) , Pη (−1) i=1 ai · K(p̃+Pηi=1 ai ·δi ),r0 (f ) ~ a (22) ~a∈{0,1}η :aj =1 where r~a0 , r − k p̃ + k X (1 − ai ) · (i − 1) · δi + i=1 η X (1 − ai ) · i · δi , i=k+1 we have that (Φp̃,k ◦ f )(r) = Lp̃,r (f ) · p̃ + η X Lp̃,i,r (f ) · δi . (23) i=1 Notice that r~a0 ∈ [0, n − η(p̃ + η X ai · δi )] . i=1 From the bound in Equation (14) it follows that |Lp̃,r (f )| < 23η · n0.475− |Lp̃,i,r (f )| < 23η−1 · n0.475− . and (24) The following claim shows that we actually have Lp,r (f ) = Lp+δ0 ,r (f ) = 0, so, in fact, (Φp̃,k ◦ f )(r) = η X Lp̃,i,r (f ) · δi . (25) i=1 Therefore, |(Φp̃,k ◦ f )(r)| ≤ 23η−1 · n0.475− · ∆ ≤ 23η−1 · n0.475 . 7 In the proof of Claim 4.2 we use the more general notation c~a,j,k (i). 13 (26) Claim 4.3. Lp,r (f ) = Lp+δ0 ,r (f ) = 0. We defer the proof of the claim to Section 4.2 and proceed with the proof of the theorem. The good thing about Equation (26) is that it will allow us to reduce to the case of a polynomial with a bounded range. This somewhat resembles the way that we concluded the proof of Theorem 3.4, although it is done in a slightly more involved manner. Let Υ(t) = η Y Φp,i (t) and Υk (t) = i=0 Υ(t) . Φp,k We now bound the value of g(r) , (Υ ◦ f )(r) for r ∈ [kp, Q kp + n − η(p + ∆) − deg(Υk )]. Notice that g(r) = (Υk ◦ (Φp,k ◦ f ))(r). Furthermore, Υk (t) = i6=k Φp,i (t). Claim 4.2 implies that each Φp,i (t) contains 2η monomials 8 , and that its coefficients are upper bounded (in absolute value) by 2η . Therefore, since Υk (t) is a product of 2 η such Φp,i ’s, it follows that Υk (t) is a sum of 2η monomials with coefficients upper bounded (in 2 absolute value) by 2η . Moreover, as a polynomial, the degree of each Φp,i (t) is at most η · ∆ (this follows as c~a,k,k ≤ η). Hence, the degree of Υk (t) is at most η 2 · ∆. Thus, we have that η2 Υk (t) = 2 X αi · tdi where 0 ≤ di ≤ η 2 · ∆ , and 2 |αi | ≤ 2η . i=1 This implies that for every k ∈ [0, η] and every9 r ∈ Ik , [kp, kp + n − η(p + ∆) − deg(Υk )] we have that 2 η X 2 |g(r)| = |(Υk ◦ (Φp,k ◦ f ))(r)| = αi · (Φp,k ◦ f )(r + di ) i=1 η2 ≤ 2 X 2 2 |αi | · |(Φp,k ◦ f )(r + di )| ≤ 2η · 2η · 23η−1 · n0.475 ≤ n0.475+o(1) , (27) i=1 where we also used the bound on |Φp,k ◦ f | given in (26). Notice that the size of the interval Ik satisfies n n |Ik | = n − η(p + ∆) − deg(Υk ) + 1 > n − η( − n0.525 ) − deg(Υk ) + 1 > >p η+1 η+1 and therefore every two consecutive intervals Ik and Ik+1 have a nonzero intersection. Hence, we conclude that for every r ∈ [0, n − η∆ − deg(Υη )] (note that n − η∆ − deg(Υη ) is the endpoint of Iη ) it holds, by (27), that |g(r)| ≤ n0.475+o(1) < n0.5 . We thus have that g : [0, n − η∆ − deg(Υη )] → [−n0.5 , n0.5 ] . (28) In addition we have (by Lemma 2.8) that deg(g) ≤ deg(f ) < ηp . (29) We now would like to show that deg(g) is small and then use Lemmas 2.9 and 3.2 to conclude that f is constant. Before applying Lemma 2.9, we must ensure that Φp,k (t) 6= 0. 8 9 Note that here we allow different monomials with the same exponent The drop by deg(Υk ) in the range of relevant r’s is so that r + di will be in the range [kp, kp + n − η(p + ∆)]. 14 Claim 4.4. For every k ∈ [0, η] it holds that Φp,k (t) 6= 0. We defer the proof of Claim 4.4 and continue with the proof of the Theorem. Assume first that g is not a constant. The point is that now we can repeat the whole proof for g instead of f , with n0 = n − η∆ − deg(Υη ) instead of n. Note that due to the bound on the range of g we get that Equation (14), applied to g instead of f , gives |Kq,r (g)| < 2η · n0.5 2η · n0.5 < < 1. q n/(η + 2) I.e. Kq,r (g) = 0. Continuing, we see that (Φp̃,k ◦ g)(r) = 0 for r ∈ [p̃k, p̃k + n0 − η(p̃ + ∆)]. Therefore, if we define h = Υ ◦ g then for every k ∈ [0, η] and r ∈ Ik0 , [kp, kp + n0 − η(p + ∆) − deg(Υk )] we 0 have that h(r) = 0. As before, we see that any two consecutive intervals Ik0 and Ik+1 have a nonzero intersection. Indeed |Ik0 | = >(∗) n0 − η(p + ∆) − deg(Υk ) + 1 = n − ηp − 2η∆ − deg(Υk ) − deg(Υη ) + 1 n n − n0.525 ) − 2(η∆ + η 2 ∆) > >p n − η( η+1 η+1 where inequality (∗) follows from the properties of the construction in Lemma 4.1. It therefore follows that h(r) is zero for all r ∈ [0, n0 − η∆ − deg(Υη )]. Thus, if h 6≡ 0 then it must be the case that deg(h) ≥ n0 − η∆ − deg(Υη ) + 1 > (η + 1)p. Since deg(g) ≥ deg(h) we get that deg(g) ≥ (η + 1)p . (30) Combining Equations (29) and (30) yields a contradiction. On the other hand, if h ≡ 0 then by Lemma 2.9, deg(g) ≤ monom(Υ) − 2 . However, if this is the case then applying Lemma 2.9 again yields that deg(f ) ≤ deg(g) + monom(Υ) − 1 ≤ 2 · monom(Υ) − 3 ≤ 2η 2 +η+1 − 3 = o(n) . Lemma 3.2 now implies that f is constant. On the other hand, if g is constant (in which case there was no point in trying to run the argument again for g) then applying Lemmas 2.9 and 3.2 again we conclude that in this case too f is a constant. This completes the proof of the theorem (the missing proofs are given in Sections 4.1 and 4.2). Corollary 1.1 follows immediately from Theorem 1. Indeed, as S is contained in and not equal to the domain [0, n], any function with degree at most 1 is in fact a constant function. 4.1 A cube of primes We shall now prove Lemma 4.1. To ease the reading we repeat the statement of the lemma. Lemma (Lemma 4.1). Let 0 < < 1/2, there exists n0 () such that for any n > n0 () and η = log log(n)/2, there exists a set ( ) η X n n Pp;δ0 ,δ1 ,δ2 ,...,δη = p + ai · δi | ∀i ai ∈ {0, 1} ⊆ − 4Γ(n), − Γ(n) η+1 η+1 i=0 with the following properties: 15 1. Every q ∈ Pp;δ0 ,δ1 ,δ2 ,...,δη is a prime number. 2. δi > 0 for all i = 1, . . . , η. P 3. ∆ , ηi=1 δi ≤ n . 4. δ0 ∈ [Γ(n), 3Γ(n)]. As in the proof of Lemma 3.3, the proof of Lemma 4.1 is by the pigeonhole principle and relies on Theorem 2.6. Proof of Lemma 4.1. The high level idea is the same as in the proof of Lemma 3.3. However, since we are looking for η-dimensional ‘cubes’ it will be convenient to first prove the following combinatorial lemma. Note that the lemma does not necessarily concern prime numbers. Lemma 4.5. Let A ⊆ [a1 , a2 ] and let ` = a2 − a1 , α = |A|/` . Then, if r ≤ log log(`) − log log( α4 ), there is an r-dimensional ‘cube’ which is a subset of A ( ) r X ai · δi | ∀i ai ∈ {0, 1} ⊆ A Px;δ1 ,...,δr , x + i=1 where δi > 0 for i = 1, 2, . . . , r. Note that we do not require that the δi ’s are distinct. Proof. We shall prove, by induction on r that for every r ∈ [0, log log(`) − log log( α4 )], there exist r δ1 , . . . , δr such that there are at least A. `·α2 42r −1 r-dimensional cubes Px;δ1 ,...,δr (with different x’s) inside The case r = 0: This case is trivial as there are exactly ` · α = |A| elements in A, each is a 0-dimensional ‘cube’. The induction step: Assume that we already proved the claim for r and we wish to prove it for r + 1. Consider the smallest number in each r-dimensional cube that was found in the r-th r `·α2 step. By the induction hypothesis we have 42r −1 such different numbers, all of which in A ⊆ [a1 , a2 ]. 2r Looking at all the differences between those numbers, we get that if 4`·α 2r −1 ≥ 2 then there are at least 2r 2r 2 `·α r ≥ 41 4`·α many such differences, all between 1 and `. Using the pigeonhole principle, we 42 −1 2r −1 2 2r 2 conclude that there is a ‘popular’ difference, δr+1 , with at least 1` · 14 · 4`·α many occurrences. 2r −1 For such a ‘popular’ difference δr+1 and every pair of cubes at distance δr+1 we have that Px;δ1 ,δ2 ,...,δr ∪ Px+δr+1 ;δ1 ,δ2 ,...,δr = Px;δ1 ,δ2 ,...,δr ,δr+1 . This gives the required 1 · 4` r ` · α2 42r −1 2 (r + 1)-dimensional cubes. 16 r+1 ` · α2 = 2r+1 −1 4 To conclude the proof of Lemma 4.5 we need to show that for r ≤ log log(`) − log log( α4 ), it holds 2r r r 2 −1 · ( 1 )2 . It is clearly enough to that 4`·α 2r −1 ≥ 2, which is equivalent to showing that ` ≥ 2 · 4 α 4 2r show that ` ≥ ( α ) , which follows since r ≤ log log(`) − log log( α4 ). This completes the proof of the lemma. We now proceed with the proof of Lemma 4.1. Recall that we have to find δ0 that will be much larger than the other δi ’s (in fact, it has to be much larger than their sum, as we consider which is relatively small). We therefore start by first choosing δ0 and only then apply Lemma 4.5. Let p, q be prime numbers such that: q ∈ Iq , [ n n − 2Γ(n), − Γ(n)] η+1 η+1 q ∈ Ip , [ n n − 4Γ(n), − 3Γ(n)] η+1 η+1 Clearly, |Ip | = |Iq | = Γ(n) and Γ(n) ≤ q − p ≤ 3Γ(n) for any such p and q. Theorem 2.6 implies that Γ(n) 9 · log each of the intervals Iq , Ip contains at least 100 n different prime numbers. By the pigeonhole 1 n principle, each of the intervals Ip , Iq has a sub-interval of length n that contains at least 12 · log n many prime numbers. Denote these sub-intervals as Ip0 , Iq0 respectively: Ip0 = [rp , rp + n ] Iq0 = [rq , rq + n ]. n 2 Looking at all the differences between pairs of primes in Iq0 ×Ip0 we get that there are at least ( 12·log n) many differences, each of which is between rq − rp − n and rq − rp + n . Hence, one of the differences n n 2 occurs at least ( 12·log n ) /2n = 2(12·log n)2 many times. Let δ0 be that popular difference. Clearly, property 4 holds from this choice of δ0 . Consider the following set A , x ∈ Ip0 | x + δ0 ∈ Iq0 and both x and x + δ0 are prime numbers . Obviously, A ⊆ Ip0 , and by the choice of δ0 we are guaranteed that |A| ≥ |A|/|Ip0 | ≥ 1 . 2(12·log n)2 n . 2(12·log n)2 Let α = Note that 4 log log n log log(n ) − log log( ) ≥ log log(n) − log log log(n) − log(1/) − O(1) > =η. α 2 We now apply Lemma 4.5 with parameters ` = |Ip0 | = n α = |A|/|Ip0 | ≥ and 1 2(12 · log n)2 and obtain that there exists an η-dimensional cube B = Px;δ1 ,...,δη ⊆ A. By the definition of A it follows that all the elements in B + δ0 , {b + δ0 | b ∈ B} are prime numbers. Our final (r + 1)dimensional cube is therefore, ( ) η X Px;δ0 ,δ1 ,...,δη = x + ai · δi | ∀i ai ∈ {0, 1} . i=0 We note that Lemma 4.5 also guarantees that all the δi ’s are positive and that ∆, n X δi ≤ |Ip0 | = n . i=1 17 4.2 Missing proofs We now give the proofs of Claims 4.2, 4.3 and 4.4. For the sake of readability we repeat the statement of each of the claims before proving it. Claim (Claim 4.2). Pη Pη X η 0 kp̃ k (−1) i=1 ai · t i=1 c~a,k,k (i)·δi . Φp̃,k (t) = t · (−1) · · k η ~a∈{0,1} Proof of Claim 4.2. Recall that Pk Pη Pη X (−1) i=1 ai · Ψ(p̃+Pηi=1 ai ·δi ) (t) · t i=1 (1−ai )·(i−1)·δi + i=k+1 (1−ai )·i·δi . Φ0p̃,k (t) = (31) ~a∈{0,1}η Denote if ai = 1 j i − 1 if ai = 0 and i ≤ k c~a,j,k (i) , i if ai = 0 and i ≥ k + 1 This is consistent with the previous definition of c~a,k,k (see Equation (18)). By expanding Ψ (recall Equation (15)) and using the c~a,j,k ’s we get that Φ0p̃,k (t) X = Pη (−1) i=1 ai · ~a∈{0,1}η η X j=0 Pη η (−1) · · tj p̃+ i=1 c~a,j,k (i)·δi . j j Considering the coefficients for different j’s we have the following cases. ~ Case 1 : j < k. For every ~a = (a1 , . . . ,P aj , 0, aj+2 , . . . , aP η ), let b = (a1 , . . . , aj , 1, aj+2 , . . . , aη ). η η It is easy to verify that c~a,j,k = c~b,j,k . As (−1) i=1 ai = −(−1) i=1 bi we get that ~a and ~b cancel each other. Case 2 : j > k. Quite similarly, for every ~a = (a1 , . . . , aj−1 , 0, aj+1 , . . . , aη ), let ~b = (a1 , . . . , aj−1 , 1, aj+1 , . . . , aη ). Again, ~a and ~b cancel each other. Case 3 : j = k. get that This is the only case where coefficients do not get canceled out. We therefore Φ0p̃,k = X ~a∈{0,1}η Pη (−1) i=1 ai Pη η · (−1) · · tkp̃+ i=1 c~a,k,k (i)·δi k k as claimed. We now proceed to proving Claim 4.3. The specific properties of the cube (that may have seemed somewhat arbitrary) play a major role in this proof. Claim (Claim 4.3). Lp,r (f ) = Lp+δ0 ,r (f ) = 0. 18 Proof of Claim 4.3. Recall that Φp,k = Φp+δ0 ,k (Equation (20)). Therefore, Lp,r (f ) · p + η X Lp,i,r (f ) · δi = Φp,k (r) (32) i=1 = Φp+δ0 ,k (r) = Lp+δ0 ,r (f ) · (p + δ0 ) + η X Lp+δ0 ,i,r (f ) · δi . i=1 Rearranging (32) gives (Lp,r (f ) − Lp+δ0 ,r (f )) · p = Lp+δ0 ,r (f ) · δ0 + η X (Lp+δ0 ,i,r (f ) − Lp,i,r (f )) · δi . i=1 Recall that |Lp,r (f )|, |Lp+δ0 ,r (f )| < 23η · n0.475− and |Lp,i,r (f )|, |Lp+δ0 ,i,r (f )| < 23η−1 · n0.475− (Equation (24)). By our choice of parameters we have that |Lp+δ0 ,r (f ) · δ0 + η X (Lp+δ0 ,i,r (f ) − Lp,i,r (f )) · δi | ≤ 23η · n0.475− · (δ0 + i=1 n 0.475− η X δi ) = i=1 · Γ(n) · poly log(n) = n 1− · poly log(n) < p . As (Lp,r (f )−Lp+δ0 ,r (f ))·p is an integer multiple of p, it must be the case that Lp,r (f )−Lp+δ0 ,r (f ) = 0. We now show that Lp+δ0 ,r (f ) = 0 which will conclude the proof. As we just proved that Lp,r (f ) − Lp+δ0 ,r (f ) = 0 we can rewrite (32) as Lp+δ0 ,r (f ) · δ0 = − η X (Lp+δ0 ,i,r (f ) − Lp,i,r (f )) · δi . i=1 Similarly to the previous argument we note that Lp+δ0 ,r (f ) · δ0 is an integer multiple of δ0 and that, by our choice of parameters (Lemma 4.1) | η X (Lp+δ0 ,i,r (f ) − Lp,i,r (f )) · δi | < 2 · 23η−1 · n0.475− · i=1 η X δi ≤ 23η · n0.475 < Γ(n) ≤ δ0 . i=1 Hence, Lp+δ0 ,r (f ) = 0. This completes the proof of the claim. Claim (Claim 4.4). For every k ∈ [0, η] it holds that Φp,k (t) 6= 0. Proof of Claim 4.4. By claim 4.2, Φp,k (t) is the sum of 2η (not necessarily different) monomials. To prove that the different monomials do not cancel each other we will show that there is a unique monomial of maximal degree. Note that for every ~a ∈ {0, 1}η we have a monomial of degree Pη i=1 c~a,k,k (i) · δi in Φp,k (t). Let ~a , (1, 1, . . . , 1, 0, 0, . . . , 0). | {z } | {z } k η−k η Then, for every other binary vector ~a 6= ~b ∈ {0, 1} we have For i ≤ k, c~b,k,k (i) ≤ k = c~a,k,k (i) and the inequality is strong if bi = 0. For i ≥ k + 1, c~b,k,k (i) ≤ i = c~a,k,k (i) and the inequality is strong if bi = 1 . As ~a 6= ~b, it follows that c~b,k,k < c~a,k,k . Namely, ∀i ∈ [1, η] : c~b,k,k (i) ≤ c~a,k,k (i) and ∃i ∈ [1, η] : c~b,k,k (i) < c~a,k,k (i) . P P Since all the δi ’s are positive, we get that ηi=1 c~b,k,k (i) · δi < ηi=1 c~a,k,k (i) · δi , and the monomial that corresponds to ~a is the unique monomial of maximal degree. 19 5 The range of a degree d polynomial In this section we prove Theorem 2. It will be an easy corollary of Theorem 3 which we first prove. The proof is quite elementary and basically follows from averaging arguments. At the end of the section we present a possible approach for improving our results using the Chebyshev polynomials, however at this stage we get more general results using our simple argument. To ease the reading we repeat the statement of Theorem 3. Theorem (Theorem 3). Let f : R → R be a degree d monic polynomial. Then, maxi∈[0,n] |f (i)| > n−d d . In particular, if f : Z → Z is a degree d polynomial (not necessarily monic) then 2e 1 max |f (i)| > · d! i∈[0,n] n−d 2e d 1 ≥√ · 7d n−d 2d d . Proof of Theorem 3. For d = 1 the theorem holds. So we can assume w.l.o.g that d ≥ 2. Consider the factorization of f over C, d Y f (x) = (x − αi ) . (33) i=1 Recall that if αi ∈ C is a root of f then its conjugate ᾱi is also a root of f . As we are interested in bounding the range of f from above, we can assume w.l.o.g. that all the roots of f are real. Indeed, for any complex α and real x it holds that (x − α) · (x − ᾱ) ≥ (x − R(α))2 , where R(α) is the real part of α. We would Q like to give a lower bound on the maximum (absolute) value of f by showing that the product ni=0 f (i) is large. However, since some of the i’s can be roots of f , or very close to roots of f , we need to remove them from the product first. Call an element i ∈ [0, n] an approximate root of f if there is a root of f , αj (in the notations of Equation (33)), such that 10 round(αj ) = i. Clearly, there are at most d approximate roots in the set [0, n]. Denote with S ⊆ [0, n] the set of all non-approximate roots. Clearly |S| ≥ n + 1 − d. Note that " #1 |S| Y max |f (i)| ≥ |f (i)| . (34) i∈[0,n] As Y i∈S |f (i)| = i∈S d Y Y | i − αj | , (35) j=1 i∈S Q it will suffice for our needs to bound from below the value of each product i∈S | i − αj | and then apply it in Equation 34. Fix some j ∈ [d]. Notice that the closest element to αj in S has distances at least 1/2 from it. The next element has distance at least 1 from it. The next has distance at least 3/2 from it, etc. In other words, if we sort the elements in S according to their distances from αj , S = {i1 , . . . , i|S| }, then the k element, ik will be at distance at least k/2. Hence, |S| Y k |S|! ∗ |S| |S| p | i − αj | ≥ | ik − αj | ≥ = |S| ≥ · 2π|S| 2 2e 2 i∈S k=1 k=1 Y 10 |S| Y round(x) is the integer closest to x, if x = i + 1/2 then round(x) = i. In other words, round(x) = dx − 1/2e 20 (36) where inequality (∗) follows from Stirling’s formula (Theorem 2.1). Plugging Equation (36) back to Equations (35) and (34) we get " max |f (i)| ≥ i∈[0,n] |S| 2e |S| 1 #d |S| d d p d |S| n − d · 2π|S| = (2π|S|) 2|S| · > . 2e 2e This proves the first statement of the theorem. For the second statement we note that if f is a polynomial over the integers, then by Theorem 2.3 the coefficient of xd in f is an integer multiple of 1/d!. In particular there is an integer c 6= 0 such that (d!/c) · f (x) is monic. Therefore, d d c d! 1 n − d 1 n − d max |f (i)| = · max · f (i) > · ≥√ · , d! i∈[0,n] c d! 2e 2d i∈[0,n] 7d where we used Stirling’s formula (and the assumption that d ≥ 2) in the last inequality. We believe that Theorem 3 can be improved. Nevertheless, the next example shows that the theorem is not far from being tight. x− n−d+1 2 Example 5.1. For an odd integer n and an even integer d ≤ n, the polynomial f (x) = is d d a degree d polynomial mapping [0, n] to [0, 2nd ·d! ]. Proof. It is not difficult to see that since d is even, f (x) = f (n − x). In particular, f (x) ≥ 0 for all x ∈ [0, n]. Furthermore, for all r ∈ [0, n] n+d−1 f (r) ≤ f (n) = 2 d < 1 · d! n2 − 1 4 d/2 < nd . · d! 2d This upper bound is larger by a factor of (roughly) ed from the lower bound on the range that is stated in Theorem 3. It is an interesting question to understand the ‘correct’ bound. To derive Theorem 2 we will need the following easy property of the function 1 n−x x Dn (x) , √ · . 2x 7x Lemma 5.2. In the real interval [1, n] the function Dn (x) is first strictly increasing and then strictly decreasing. Furthermore, it attains its maximum at some 0.135 · n < x < 0.136 · n (for n ≥ 450). Proof. It is clearly sufficient to prove that the function n−x x 1 1 1 ln(Dn (x)) = ln √ · = x ln(n − x) − x ln(x) − x ln(2) − ln(x) − ln(7) 2x 2 2 7x has the claimed property. This will follow from the observation that the second derivative of ln(Dn (x)) is negative. Indeed, (ln(Dn (x)))0 = ln(n − x) − x 1 − ln(x) − 1 − ln(2) − n−x 2x 21 and (ln(Dn (x)))00 = − 1 n 1 1 − − + <0 n − x (n − x)2 x 2x2 where the last inequality holds since x ≥ 1. To see the ‘furthermore’ part we note that (ln(Dn ))0 (0.135 · n) > 0 for n ≥ 450 and that (ln(Dn ))0 (0.136 · n) < 0 for every n. Hence, by the intermediate value theorem, (ln(Dn (x)))0 = 0 for some 0.135 · n < x < 0.136 · n (when n ≥ 450). We denote the unique maximum point of Dn as xDn . We can now derive Theorem 2. We first repeat its statement. 2 Theorem (Theorem 2). There exists a constant n0 such satisfying d ≤ 15 n n thatjif d, n are integers ko d and n > n0 then the following holds. If f : [0, n] → 0, . . . , √17d · n−d is a polynomial then 2d 1 n deg(f ) ≤ d − 1 or deg(f ) ≥ 31 n − 1.2555 · [d ln( n−d 2d ) − 2 ln( d )]. Proof. If deg(f ) ≤ d − 1 we are done. We may therefore assume that deg(f ) ≥ d. If deg(f ) ≤ xDn then by Theorem 3 and Lemma 5.2, we get that the maximal value that f attains on [0, n] is larger d than Dn (deg(f )) ≥ Dn (d) > √17d · n−d , in contradiction to the assumption of the theorem. Since 2d Dn (x) is decreasing for x > xDn we observe, by substituting x = 13 n − 1.2555 · [d ln( n−d 2d ) − into Dn , that Dn 1 2 ln( nd )] 1 n 1 n−d d n−d 1 n − 1.2555 · d ln − ln >√ · . 3 2d 2 d 2d 7d Indeed, it is not hard to see that for any c such that c < n/3 − 0.136 · n (which in particular means that xDn < n/3 − c) it holds that Dn (n/3 − c) = ≥(∗) n/3−c n − (n/3 − c) n/3−c 1 3c/2 p · =p · 1+ 2n/3 − 2c n/3 − c 7(n/3 − c) 7(n/3 − c) √ 1 1 1 p · e0.531·3c/2 = 3 · √ · e0.7965·c− 2 ln(n/d) , 7d 7n/3 1 where to prove inequality (∗) we used the simple fact that (1 + x) ≥ e0.531·x for x ≤ 2.1765, together 2 with the bound on c. In our case, since d ≤ 15 n, it is not hard to verify that c , 1.2555 · [d ln( n−d 2d ) + 1 n ln( )] satisfies c < n/3 − 0.136 · n (for n large enough) as required. 2 d We therefore obtain that √ 1 1 n−d 1 n 1 Dn n − 1.2555 · d ln − ln ≥ 3 · √ · e0.7965·c− 2 ln(n/d) 3 2d 2 d 7d n−d 1 1 n−d d > √ · ed ln( 2d ) = √ · 2d 7d 7d 1 as claimed. By Lemma 5.2, deg(f ) ≥ 13 n − 1.2555 · d ln n−d − 2 ln nd . 2d To summarize, Theorem 2 uses the fact that Dn has a unique maximum, xDn , and aims to find, for a given degree d < xDn , another degree d0 > xDn such that Dn (d0 ) ≥ Dn (d). In the theorem we gave a relatively simple way to derive d0 from d. With more work one can push this result for d’s closer to xDn . 22 We note that Theorem 2 implies that when Ω(n) ≤ deg(f ) < (1 − )n/3 then the range of f is exponentialin n. As a corollary of Example 5.1 one can show that if we allow the range to be as √ n n+d−1 1+ 5 2 large as O then f can have any degree. Indeed, taking the maximum over , when 2 d d + n is odd, we get an upper bound on that range that is smaller than the n-th Fibonacci number, FIBn . n+d−1 2 Lemma 5.3. For integers d, n such that n + d is odd, let Rn,d , , and set d Rn , max{Rn,d | d ∈ [0, n], d + n is odd } . Then, Rn ≤ Rn−1 + Rn−2 for n > 2. Proof. Since n > 2, we can assume that maximum of Rn,d is achieved for some d > 0. We use them−1 m−1 to conclude that: + = the combinatorial identity m k−1 k k n+d−1 Rn,d = 2 n+d−1 2 = d = d n+d−1 d (n−2)+d−1 2 −1 + −1 = d−1 2 (n−1)+(d−1)−1 + 2 d−1 = Rn−2,d + Rn−1,d−1 maximizing over d in both sides we conclude that Rn ≤ Rn−2 + Rn−1 . As an immediate corollary, using the fact that R1 = R2 = 1, we deduce that √ !n 1 1+ 5 Rn ≤ FIBn ≤ √ · 2 5 which completes our argument. 5.1 A possible route for improvements √ In this section we present a possible approach towards improving Theorem 2, when d ≤ n/2, based on Chebyshev polynomials. We will only give a sketch of the approach and we will not cover all necessary background on Chebyshev polynomials. The interested reader is referred to [MH03]. A natural approach to proving that a polynomial must take large values is by comparing it to the Chebyshev polynomial of the same degree. Roughly, the Chebyshev polynomial of degree d is defined on the real interval [−1, 1] in the following way: Td (x) = cos(d arccos(x)) . It is not hard to prove that Td is a degree d polynomial, having exactly d roots in the interval [−1, 1], that its leading coefficient is 2d−1 and that it has d + 1 extremal values in the same interval, on which it is equal, in absolute value, to 1. Specifically, its roots lie on the points cos( π(2k−1) ) and its extremal 2d πk points are cos( d ), on which it alternates between 1 and −1. A well known fact of the Chebyshev polynomials is that among the degree d monic polynomials the polynomial fd (x) = 21−d Td (x) whose maximum on the real interval [−1, 1] is the smallest and equals 21−d . The problem in using this fact is that we are interested in the maximum of a function on a relatively small set of points. Consider a polynomial f : [0, n] → [0, m]. Let g(x) = f ( n2 x + n2 ). I.e. g : [−1, 1] → [0, m], (where [−1, 1] is the real interval) and we are interested in the value of g on the points {−1, −1 + n2 , −1 + n4 , . . . , 1}. Denote for simplicity xk = 2k/n − 1, k = 0, . . . , n. We would like 23 to say that as Td obtains the smallest maximum on [−1, 1] then (after we normalize g by its leading coefficient) it must obtain a value larger than 21−d on one of the xk ’s. However, all that we know is that the maximum of g on the whole interval [−1, 1] is large and not necessarily on one of the xk ’s. To tackle this problem one has to prove that the values that Td obtains on the xk ’s is relatively large (close to its overall maximum). A possible way for proving this is by observing that we can find a point xk near any extremal point and then, since we have a reasonable bound on the derivative of Td , conclude that Td obtains a relatively large value there as well. This approach in fact works; √ Since the derivative of Td is bounded by d2 it follows that when d < n/2 there are d + 1 points among the xk ’s on which Td alternates in sign and obtains absolute value larger than, say, 1/2. Now, let g̃ = g/gd , where gd is the leading coefficient of g. Assume that |g̃(xk )| < 21 · 21−d , for every k. Then the polynomial 21−d Td − g̃ has degree at most d − 1 (it is the difference of two degree d monic polynomials) and it changes sign d times (between the xk ’s on which Td obtains large value), which is a contradiction. It therefore follows that maxk∈[0,n] |g(xk )| ≥ 21 |gd | · 21−d . 1 As gd equals fd · (n/2)d , where fd is the leading coefficient of f , and since |fd | ≥ d! , we get that d n −d d maxk∈[0,n] |f (k)| = maxk∈[0,n] |g(xk )| ≥ 2 · (n/2) /d! = 22d d! . We summarize this in the next theorem. Theorem 6. There exists a constant n0 such that for every two integers d, n such that n > n0 √ and d ≤ n/2 it holds that if f : Z → Z is a degree d polynomial (not necessarily monic) then nd maxi∈[0,n] |f (i)| ≥ 22d . d! d 1 This result is slightly better than the bound maxi∈[0,n] |f (i)| ≥ d! · n−d that was obtained in 2e √ the proof of Theorem 3, but it holds only for d ≤ n/2. We note, however, that this approach √ cannot work for d = ω( n) as for such large d many roots of Td are very close to each other. Indeed, the distances among the first roots (and among the last roots) are smaller than 1/n while the xk ’s are separated from one other. For that reason we cannot use Theorem 6 instead of Theorem 3; In order to show that the degree must be larger than Ω(n) we must claim something about the range of polynomials of degree, say, n/ log(n) and Theorem 6 does not give any information in this case. 5.2 The case of small degrees In this section we give two small improvements for the case of polynomials of degrees 1 or 2. The first improvement concerns polynomials whose range is (roughly) [0, n2.475 ]. Theorem 7. For every 0 < there exists n0 such that for every integer n0 < n the following holds: Every f : [0, n] → 0, n2.475− must satisfy deg(f ) ≤ 2 or deg(f ) ≥ n/2 − 2n/ log log n. Notice that Theorem 2 implies that if the range of f is, say, [0, n3 /1000] then either deg(f ) ≤ 2 or deg(f ) ≥ n/3 − O(log n). Thus, the improvement that Theorem 7 gives is that if the range is [0, n2.475− ] then either deg(f ) ≤ 2 (as before) or it is at least n/2 − 2n/ log log n (compared to roughly n/3). The proof is quite similar to the proof of Lemma 3.1. Proof. We first explain how n0 is defined. A corollary of Theorem 1 is that there exists n1 such that for every n > n1 and f : [0, n] → [0, 17n1.475− ], either deg(f ) ≤ 1 or deg(f ) > n−4n/ log log n. Define n2 (guaranteed to exist from Theorem 2.6) such that for every n > n2 it holds that there is a prime number in the range [ n2 − Γ(n), n2 ] and such that Γ(n) = n0.525 < n2 − n3 . We set n0 = max(2n1 , n2 ). The proof is by a reduction to Theorem 1. Let p ∈ [ n2 − Γ(n), n2 ] be a prime number If deg(f ) ≥ p 24 then we are done, as in this case deg(f ) ≥ p ≥ n n − Γ(n) ≥ − 2n/ log log n . 2 2 Therefore, we may assume that deg(f ) < p. By Lemma 2.4, working modulo p, we get that f (r) ≡p f (p + r) for every r ∈ [0, n − p]. As in the proof of Lemma 3.1, we consider the polynomial g(r) = f (r)−f (r+p) which is defined over r ∈ [0, n − p]. It follows that p −n2.475− n2.475− ⊆ −3 · n1.475− , 3 · n1.475− . g : [0, n − p] → , p p In particular, g+3·n1.475− maps [0, n/2] to 0, 6 · n1.475− ⊆ 0, 17(n/2)1.475− . Since n > n0 ≥ 2n1 Theorem 1 implies that either deg(g) ≤ 1 or deg(g) > n/2 − 2n/ log log n. By Lemma 2.9 we get that deg(f ) ≤ deg(g) + 1 and so the case deg(g) ≤ 1 translates to deg(f ) ≤ 2. In the second case where deg(g) > n/2 − 2n/ log log n we get the same conclusion for f as deg(g) ≤ deg(f ). As an immediate corollary we get our second improvement that provides a strengthening of Lemma 3.2. n j 2 ko 2 Corollary 5.4. There exists a constant n0 such that if n > n0 and f : [0, n] → 0, . . . , n −4Γ(n) 8 is a polynomial then deg(f ) ≤ 1 or deg(f ) ≥ n/2 − 2n/ log log n. Proof. Lemma 3.2 implies that if deg(f ) > 1 then it is at least n/12 − Γ(n). However, by Theorem 7 we get that actually deg(f ) ≥ n/2 − 2n/ log log n. n−1 The example given after Lemma 3.2, f (x) = x− 2 2 , gives a degree 2 polynomial mapping [0, n] h i 2 to 0, n 8−1 . Thus, up to an additive O(n1.05 ) term, the range in Corollary 5.4 is tight. 6 Proof of Theorem 4 In this section we prove Theorem 4. For convenience we repeat its statement. √ Theorem (Theorem 4). For every large enough integer n > 0 and integer d = o( n/ log n), there exists f : {0, . . . , n} → {0, . . . , O(nd+0.5 )} of degree 2d < deg(f ) ≤ n − d log(n). The proof of Theorem 4 is based on a reduction to the shortest vector problem (SVP) in lattice theory. In section 6.1 we introduce basic definitions and tools from lattice theory. We then turn to prove Theorem 4 in section 6.2. 6.1 Basic properties of lattices Definition 6.1. Let b1 , b2 , . . . , bn be linearly independent vectors in Rm . We define the lattice generated by them as ( n ) X Λ(b1 , b2 , . . . , bn ) = xi bi : xi ∈ Z . i=1 We refer to b1 , b2 , . . . , bn as a basis of the lattice. More compactly, if B is the m × n matrix whose columns are b1 , b2 , . . . , bn , then we define Λ(B) = Λ(b1 , b2 , . . . , bn ) = {Bx : x ∈ Zn } . 25 We say that the rank of the lattice is n and its dimension is m. p The lattice is called full-rank lattice if n = m. The determinant of Λ(B) is defined as det (Λ(B)) = det (B T B). Although a basis of a T T T lattice is not unique, e.g. both (0, 1) , (1, 0) and (1, 1) , (2, 1)T span Z2 , it can be shown that the determinant of a lattice is independent of the choice of basis. Definition 6.2. let Λ be a lattice of rank n. For i ∈ [n], the i-th successive minimum is defined as λi (Λ) = inf {r : dim (span (Λ ∩ B(r))) > i} where B(r) = {x ∈ Rn : kxk < r} is the open-ball of radius r. We shall need the following theorem, due to Minkowski. A proof can be found in, e.g., [MG02]. Theorem 6.3. For any full-rank lattice Λ of rank n, n Y λi (Λ) ≤ nn/2 det Λ. i=1 6.2 Proof of Theorem 4 The idea behind the proof of Theorem 4 is roughly as follows. We identify each function f : [0, n] → Z with its set of values (f (0), f (1), . . . , f (n)). That is, we think of functions as vectors in Zn+1 . We shall construct a lattice in Rn+1 which is not full-rank, and contains only points representing polynomials with degree deg(f ) ≤ n − d log(n). We then prove that this lattice has many (at least 2d + 2) linearly independent short vectors with `∞ -norm smaller than O(nd+0.5 ), i.e. many linearly independent polynomials whose image is (somewhat) bounded. One of these polynomials must be of degree at least 2d + 1. From technical reasons we will not work with the lattice described above but rather we shall consider a full rank lattice obtained by adding ‘long’ orthogonal vectors to the basis of our initial lattice. Proof of Theorem 4. Set D = n − d log n and let m = O(nd+0.5 ).11 We now describe the basis for j n+1 the lattice. For i ∈ [0, D] define the vector bi ∈ R as follows: (bi )j = i , for j = 0, . . . , n. Notice that bi corresponds to the polynomial fi (x) = xi . Let bD+1 , . . . , bn ∈ Rn+1 be arbitrary vectors of length ||bi ||2 = M , m/2 + 1 such that for every i ∈ [D + 1, n], bi is orthogonal to bk for all k 6= i (we can find the bi ’s by, say, the Gram-Schmidt procedure). Denote by B the matrix whose columns are b0 , . . . , bn and let Λn,D = Λ(B). Lemma 6.4. det (Λn,D ) ≤ 2(n+D+1)(n−D)/2 · M n−D . We defer the proof of the lemma and continue with the proof. By a theorem of Minkowski (see Theorem 6.3) we get n+1 Y λi (Λn,D ) ≤ (n + 1)(n+1)/2 · det Λn,D . (37) i=1 Note that for i ≥ D + 1, λi (Λn,D ) ≥ M . Indeed, if u is a point in Λn,D with a non-zero coefficient for some bi , i ≥ D + 1, then by orthogonality and the fact that the length of such a bi is M , we have that u is a vector of length at least M . Combining this observation with Equation (37) and Lemma 6.4, we get D+1 Y λi (Λn,D ) ≤ (n + 1)(n+1)/2 · 2(n+D+1)(n−D)/2 . (38) i=1 11 The exact value of m will be determined later. 26 Estimating the LHS from below gives D+1 Y λi (Λn,D ) ≥ i=1 D+1 Y λi (Λn,D ) ≥ λ2d+2 (Λn,D )D−2d . (39) i=2d+2 Combining Equations (38) and (39), we get n+1 λ2d+2 (Λn,D ) ≤ (n + 1) 2(D−2d) 2 (n+D+1)(n−D) 2D−4d ≤ β · nd+0.5 , where the last inequality holds for a large enough n and some constant12 β (recall that D = n−d log n √ and d = o( n/ log n)). Letting m = 2βnd+0.5 , we get that λ2d+2 (Λn,D ) ≤ m/2. Hence, by the definition of λ2d+2 , there are 2d + 2 linearly independent vectors, in Λn,D whose length is not greater n+1 . than m/2. In particular, those vectors are contained Pn in Λn,D ∩ [−m/2, m/2] Let v be any such vector. Denote with v = i=0 αi bi its representation according to the basis B. Recall that all the coefficients αi are integers. Since kvk ≤ m/2 and for every j > D, kbj k = M = m/2 + 1 > m/2, we get, by orthogonality, = αD+2 = · · · = αn = 0. Hence, for ` ∈ [0, n], Pthat αD+1 ` α the `-th coordinate of v is equal to v` = D the polynomial corresponding to v, i=0 i i . Therefore, PD PD ` fv , satisfies, fv (`) = i=0 αi i . In other words, fv (x) = i=0 αi xi . As v ∈ [−m/2, m/2]n+1 we get that fv (x) : [0, n] → [−m/2, m/2] is a polynomial of degree at most D. To end the proof we need to show that we can pick v such that deg(fv ) ≥ 2d + 1. Indeed, since there are 2d + 2 linearly independent vectors in Λn,D ∩ [−m/2, m/2]n+1 , we get 2d + 2 linearly independent polynomials fv . Consequently, there must exist v ∈ Λn,D ∩ [−m/2, m/2]n+1 such that deg(fv ) ≥ 2d + 1. The polynomial we were looking for is therefore, f (x) = fv (x) + m/2. This completes the proof of Theorem 4. We now prove Lemma 6.4. Proof of Lemma 6.4. By the orthogonality of bD+1 , . . . , bn det Λn,D = det (b0 , . . . , bn ) = det (b0 , . . . , bD ) · n Y ||bi ||2 = det (b0 , . . . , bD ) · M n−D , i=D+1 and so it is enough to show that det (b0 , . . . , bD ) ≤ 2(n+D+1)(n−D)/2 q. Let Bn,D be the (n+1)×(D +1) T B det(Bn,D n,D ). Using basic rows Q −2 D T T and columns operations on B, one can show that det(Bn,D Bn,D ) = det (An,D An,D ) · , i=0 i! matrix with columns b0 , . . . bD . By definition, det (b0 , . . . , bD ) = where An,D is a (n + 1) × (D + 1) matrix with entries (An,D )i,j = ij .13 The matrix Cn,D , ATn,D An,D P has the form (Cn,D )i,j = n`=0 `i+j for 0 ≤ i, j ≤ D. In [VD99], the determinant of Cn,D , which is a Vandermondian matrix, was computed. Theorem 6.5 ([VD99] subsection 6.10.4.). ∆n,D , det(Cn,D ) , X (V (k0 , k1 , . . . , kD ))2 , 0≤k0 <k1 <···<kD ≤n where V (k0 , k1 , . . . , kD ) is the determinantQ of the usual Vandermonde matrix with parameters k0 , k1 , . . . , kD . That is, V (k0 , k1 , . . . , kD ) = 0≤i<j≤D (kj − ki ). 12 13 The choice of β does not depend on the exact value of m. It is easy to prove this by, say, induction on j. 27 To get a more explicit upper bound on the determinant of Cn,D , ∆n,D , we prove the following lemma. Lemma 6.6. For any integer ` > 0, ∆D+`,D ≤ ∆D+`−1,D · 4D+` . We postpone the proof of Lemma 6.6 and continue with the proof. We note that 2 ∆D,D = Y (j − i) = D Y !2 i! , i=1 0≤i<j≤D and so, applying Lemma 6.6 multiple times, we get ∆n,D ≤ ∆n−1,D · 4n ≤ ∆n−2,D · 4n+(n−1) ≤ · · · D Y · · · ≤ ∆D,D · 4n+(n−1)+···+(D+1) = !2 i! · 2(D+n+1)(n−D) . i=1 Therefore, 2 (det (b0 , . . . , bD )) = T det (Bn,D Bn,D ) = det(Cn,D ) · D Y !−2 i! i=1 = ∆n,D · D Y !−2 i! ≤ 2(D+n+1)(n−D) . i=1 Taking the square root of both sides we obtain Lemma 6.4. We now prove Lemma 6.6. Proof of Lemma 6.6. We shall map each of the sequences 0 ≤ k0 < k1 < k2 < . . . < kD ≤ D + ` to a 0 ≤ D + ` − 1 as follows sequence 0 ≤ k00 < k10 < k20 < . . . < kD 1. If kD ≤ D + ` − 1, then ∀i ∈ [0, D] : ki0 = ki . 2. If 1 ≤ k0 , then ∀i ∈ [0, D] : ki0 = ki − 1. 3. Otherwise, let t be the first index satisfying kt < kt+1 − 1. Note that there is such an index since k0 = 0, kD = D + ` and ` > 0. We set ki if i ≤ t 0 ki := ki − 1 otherwise 0 ≤ D + ` − 1, and that at most D + 2 sequences 0 ≤ k < k < Note that 0 ≤ k00 < k10 < k20 < . . . < kD 0 1 0 ≤ D + ` − 1. k2 < . . . < kD ≤ D + ` were mapped to the same sequence 0 ≤ k00 < k10 < k20 < . . . < kD We now wish to give an upper bound on Q V (k0 , k1 , . . . , kD ) i<j kj − ki (40) =Q 0 0 . 0 0 0 V (k0 , k1 , . . . , kD ) i<j kj − ki In Cases 1,2 Equation (40) equals 1 since the mapping does not affect the differences between the ki ’s. In Case 3 we have 28 Q i<j (40) = Q i<j kj − ki kj0 − ki0 Y kj − ki Y kj − ki Y kj − ki · · kj0 − ki0 kj0 − ki0 kj0 − ki0 = i<j≤t t<i<j i≤t<j Y Y kj − ki Y kj − ki kj − ki · · = kj − ki kj − 1 − ki (kj − 1) − (ki − 1) = t Y D Y i=0 j=t+1 = = ≤ i=0 t Y i=0 t Y i=0 kj − ki kj − 1 − ki QD t Y t<i<j i≤t<j i<j≤t j=t+1 kj QD j=t+1 kj − ki − 1 − ki QD−1 kD − ki j=t+1 kj − ki · QD kt+1 − 1 − ki j=t+2 kj − 1 − ki kD − ki . kt+1 − 1 − ki Note, that by definition of t it must be the case that k0 = 0, k1 = 1,. . . , kt = t and kt+2 ≥ t + 2. Therefore, t t+1 Y Y (kt+1 − 1 − ki ) ≥ i. i=0 i=1 and, t Y (kD − ki ) ≤ i=0 t Y (D + ` − i), i=0 it follows that (40) ≤ t Y i=0 kD − ki ≤ kt+1 − 1 − ki Qt i=0 D + ` Qt+1 i=1 i −i D+` D+` 2D+` = ≤ <p , t+1 (D + `)/2 1.5 · (D + `) where the last inequality follows from Stirling’s approximation for a large enough D. Hence X ∆D+`,D = (V (k0 , k1 , . . . , kD ))2 0≤k0 <k1 <···<kD ≤D+` ≤ 0≤k0 <k1 ,...<kD ≤D+` = ≤ ≤ 4D+` · 1.5 · (D + `) 4D+` 1.5 · (D + `) !2 2D+` X p 1.5 · (D + `) X 0 · V (k00 , k10 , . . . , kD ) 0 V (k00 , k10 , . . . , kD ) 2 2 0≤k0 <k1 ,...<kD ≤D+` X · (D + 2) · 0≤k00 <k10 ,...<kD ≤D+`−1 4D+` · (D + 2) · ∆D+`−1,D 1.5 · (D + `) ≤ 4D+` · ∆D+`−1,D . This completes the proof of the lemma. 29 0 V (k00 , k10 , . . . , kD ) 2 7 Back to the Boolean case In this section we consider the Boolean case. Specifically, let m = 1 and n = p2 − 1 for some prime √ p. We prove that in this case the degree must be at least n − n. For completeness, we also give a proof for the case n = p − 1, that was previously proved in [GR97]. To ease the reading we restate Theorem 5. Theorem (Theorem 5). Let p be a prime number, n = p2 − 1 and f : [0, n] → {0, 1} be nonconstant. √ Then deg(f ) ≥ p2 − p > n − n. Proof. Let f be as in the statement of the theorem and assume that deg(f ) < p2 − p. By Lemma 2.4 we get that for all r ∈ [0, p − 1] 2 −p pX k=0 2 p −p f (k + r) = 0 . (−1) k k (41) Since p2 − p = (p − 1) · p + 0, it follows, by Lucas’ theorem, that if k = k1 · p + k0 , is the base p 2 2 representation of k, then p k−p ≡p 0 when k0 6= 0 and p k−p ≡p (−1)k1 when k0 = 0. Therefore, (41) is equivalent to 0= 2 −p pX k=0 2 p−1 X p −p (−1) f (k + r) ≡p f (k1 p + r) . k k k1 =0 Note that the RHS contains exactly p summands. As they are all in {0, 1} they must all be equal in order for their sum to be 0 modulo p. We thus get that for every r ∈ [0, p−1], f (r) = f (p+r) = . . . = f ((p − 1)p + r). In other words, if we set g(x) , f (x + p) − f (x) then g(x) = 0 for x ∈ [0, p2 − p − 1]. If g is identically zero, then Lemma 2.9 implies that deg(f ) = 0. I.e. that f is constant, as claimed. Otherwise, since g has p2 − p zeroes, it follows that deg(g) ≥ p2 − p. This is a contradiction as deg(f ) ≥ deg(g) (in fact, deg(f ) = deg(g) + 1). For completeness we also prove the following result of [GR97]. Theorem 7.1 ([GR97]). Let p be a prime number, n = p − 1 and f : [0, n] → {0, 1} be nonconstant. Then deg(f ) = p − 1 = n. Proof. Assume that deg(f ) < n. As in the proof of Theorem 5, we apply Lemma 2.4 and Lucas’ theorem to obtain p−1 p−1 X X k p−1 0= (−1) f (k + r) ≡p f (k) . k k=0 k=0 Again, it must be the case that f (0) = f (1) = . . . = f (p − 1). I.e., f is constant. 8 Discussion We proved that it is ‘hard’ for polynomials to ‘compress’ the interval [0, n]. Namely, that any such nonconstant polynomial to a strict subset of [0, n] must have degree n − o(n). We also proved that if d 1 · n−d then f can of course have degree < d, but all other polynomials mapping we allow m = d! 2e [0, n] to [0, m] must have degree ≥ n/3 − o(n). We are not able to prove however that our results are tight. In particular we believe that they can be improved both for the case m < n and for the case 30 of large m. We note that the following question, posed by von zur Gathen and Roche, is still open: “... for each m there is a constant Cm such that deg(f ) ≥ n − Cm ”. Furthermore, when m = 1 they raise the possibility that C1 = 3. As an intermediate goal it will be interesting to manage to break √ the n − Γ(n) upper bound. Specifically, show that when f ∈ F1 (n) is nonconstant, deg(f ) ≥ n − n. It seems that new techniques are required in order to prove this claim as all current proofs are based √ on modular calculations and we cannot guarantee the existence of a prime p in the range [n − n, n]. For the special case that n = p2 −1 we managed to obtain such a result, and of course when n = p−1 a stronger result is known, but the general case is still open. Another intriguing question is to understand what is the minimal range that a polynomial over the integers of degree exactly d can have. We note that in Example 5.1 the degree is d and the d 1 range is (roughly) of size d! · n2 . Theorem 2 asserts that if the degree is d then the range must be d d √ 1 1 (Theorem 6 actually improves it to d! larger than (roughly) d! · n−d · n4 for d ≤ n/2). It is 2e an interesting question to understand the ‘correct’ bound. Finally, we think that it will be interesting to find examples that are significantly better than those obtained in Theorem 4 and Example 5.1. Acknowledgements Gil Cohen would like to thank Orit Ashtamker-Cohen for lots of support. He also thanks Malte Beecken (Bonn), Johannes Mittmann (Bonn) and Pablo Azar (MIT) for helpful discussions on the problem. Avishay Tal would like to thank Benjamin Eliot Klein for lots of support and helpful discussion on the subject - especially in proving Theorem 2.3 and Lemma 6.4. Avishay also thanks Nathan Keller for helpful discussions and for pointing out the possible use of Chebyshev polynomials. The authors wish to thank Joachim von zur Gathen for interesting discussions on the problems studied here. References [BdW02] H. 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Range of f “Trivial” case [GR97] {0, 1} f is constant [GR97] {0, 1} f is constant Thm. 5 {0, 1} f is constant Cor. 1.1 S ( {0, . . . , n} f is constant deg(f ) ≥ n − 4n/ log log n Thm. 1 {0, 1, . . . , n1.475− } n j 2 ko 2 0, . . . , n −4Γ(n) 8 deg(f ) ≤ 1 deg(f ) ≥ n − 4n/ log log n deg(f ) ≤ 1 deg(f ) ≥ n/2 − 2n/ log log n {0, 1, . . . , n2.475− } n j ko n−d d 1 √ 0, . . . , 7d · 2d deg(f ) ≤ 2 deg(f ) ≥ n/2 − 2n/ log log n Cor. 5.4 Thm. 7 Thm. 2 d≤ deg(f ) ≤ d − 1 2 15 n Excluding “Trivial” case deg(f ) = n when n = p − 1, p is prime deg(f ) ≥ n − n0.525 √ deg(f ) ≥ n − n when n = p2 − 1, p is prime deg(f ) ≥ 1 1 n−d − 2 ln 3 n − 1.2555 · d ln 2d n d Upper Bounds on Degree Ex. 5.1 Thm. 4 0, . . . , n+d−1 2 d ≈ e(n+d) 2d 0, . . . , O nd+0.5 d f= x− n−d+1 2 d deg(f ) ≤ n − d log n deg(f ) ≤ d √ d = o( n/ log n) (and n/3 − O(d log n) ≤ deg(f )) Table 1: Summary of Results 33 A Newton polynomials TheoremP (Theorem 2.3). Let f ∈ of degree ≤ n. Then f can be represented PQ[x] be a polynomial as f (z) = nd=0 γd · dz and γd = dj=0 (−1)d−j · dj · f (j). Proof. The set of polynomials { z0 , z1 , . . . , nz } is a basis of the vector space of real polynomials with degree at most n. Thus, f as a linear combination of them. We prove by P we can express induction on d that γd = dj=0 (−1)d−j · dj · f (j). The basis for the induction is d = 0. Clearly, P f (0) = ni=0 γi · 0i = γ0 . We now prove the induction step. The value f (d) is given by f (d) = n X i=0 X d d−1 X d d d γi · = γi · = γd + γi · . i i i i=0 i=0 Rearranging the equation (isolating γd ) we get γd = f (d) − d−1 X γi · i=0 By the induction assumption we have that γi = Plugging this to Equation (42) we obtain d . i Pi i−j j=0 (−1) (42) · i j · f (j), for ` = 0, 1, . . . , d − 1. d−1 X i X d i i−j γd = f (d) − · (−1) · · f (j) i j = f (d) − i=0 j=0 d−1 X d−1 X f (j) · j=0 From the identity d i · i j = d j · d−j i−j γd = f (d) − (−1) i=j d i · i j it follows that d−1 X f (j) · j=0 = f (d) − i−j d−1 X j=0 X d−1 d d−j · (−1)i−j · j i−j i=j d−1−j X d d−j r f (j) · · (−1) · j r P P d−j r r Since d−j = (1 + (−1))d−j = 0, we conclude that d−j−1 r=0 (−1) · r=0 (−1) · r Rewriting Equation (43) we obtain γd = f (d) + d−1 X f (j) · j=0 = d X j=0 (43) r=0 f (j) · d · (−1)d−j j d · (−1)d−j , j as required. 34 d−j r = −(−1)d−j . B Lucas’ theorem Proof of Theorem 2.5. Expanding (1 + x)a we get (1 + x)a = (1 + x) Pk i=0 ai pi = k Y i (1 + x)ai p ≡p i=0 k Y i=0 a i (1 + xp ) ai = ai k X Y ai jpi x . j i=0 j=0 The coefficient of xb on the LHS is b . Since there is a unique way to represent b in base p we have Q that the coefficient of xb on the RHS is ki=0 abii . 35

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