Congruent number problem

Congruent number problem
THE CONGRUENT NUMBER PROBLEM
KEITH CONRAD
1. Introduction
A right triangle is called rational when its legs and hypotenuse are all rational numbers.
Examples of rational right triangles include Pythagorean triples like (3, 4, 5). We can scale
such triples to get other rational right triangles, like (3/2, 2, 5/2). Of√course, usually when
two sides are rational the third side is not rational, such as the (1, 1, 2) right triangle.
Any rational right triangle has a rational area, but not all (positive) rational numbers
can occur as the area of a rational right triangle. For instance, no rational right triangle has
area 1. This was proved by Fermat. The question we will examine here is: which rational
numbers occur as the area of a rational right triangle?
Definition 1.1. A positive rational number n is called a congruent number if there is a
rational right triangle with area n: there are rational a, b, c > 0 such that a2 + b2 = c2 and
(1/2)ab = n.
In Figure 1 are rational right triangles with respective areas 5, 6, and 7, so these three
numbers are congruent numbers.
Figure 1. Rational right triangles with respective areas 5, 6, and 7.
This use of the word congruent has nothing to do (directly) with congruences in modular
arithmetic. The etymology will be explained in Section 3. The history of congruent numbers
can be found in [3, Chap. XVI], where it is indicated that an Arab manuscript called the
search for congruent numbers the “principal object of the theory of rational right triangles.”
The congruent number problem asks for a description of all congruent numbers. Since
scaling a triangle changes its area by a square factor, and every rational number can be
multiplied by a suitable rational square to become a squarefree integer (e.g., 18/7 = 32 · 2/7,
1
2
KEITH CONRAD
so multiplying by (7/3)2 produces the squarefree integer 14), we can focus our attention in
the congruent number problem on squarefree positive integers. For instance, to say 1 is not
a congruent number means no rational square is a congruent number.
When n is squarefree in Z+ , to show n is a congruent number we just need to find an
integral right triangle whose area has squarefree part n. Then writing the area as m2 n
shows scaling the sides by m produces a rational right triangle with area n.
In Section 2, the parametrization of Pythagorean triples will be used to construct a lousy
algorithm generating all congruent numbers. The equivalence of the congruent number
problem with a problem about rational squares in arithmetic progressions is in Section
3. Section 4 gives an equivalence between the congruent number problem and the search
for rational points on y 2 = x3 − n2 x where y 6= 0, which ultimately leads to a solution
of the congruent number problem (depending in part on the Birch and Swinnerton-Dyer
conjecture, a famous open problem in mathematics). In the appendices we explain some
algebraically mysterious formulas from our treatment using projective geometry and give a
relation between the congruent number problem and other Diophantine equations.
Acknowledgments. I thank Lucas David-Roesler for generating the pictures.
2. A bad algorithm
There is a parametric formula for primitive Pythagorean triples, and using it we will
make a small list of squarefree congruent numbers. Any primitive triple (with even second
leg) is (k 2 − `2 , 2k`, k 2 + `2 ) where k > ` > 0, (k, `) = 1, and k 6≡ ` mod 2. In Table 1 we
list such primitive triples where k + ` ≤ 9. The squarefree part of the area is listed in the
last column. Each number in the fourth column is a congruent number and each number
in the fifth column is also a congruent number. The final row of the table explains how a
rational right triangle with area 5 can be found.
k
2
4
3
6
5
4
8
7
5
`
1
1
2
1
2
3
1
2
4
(a, b, c)
(1/2)ab Squarefree part
(3, 4, 5)
6
6
(15, 8, 17)
60
15
(5, 12, 13)
30
30
(35, 12, 37)
210
210
(21, 20, 29)
210
210
(7, 24, 25)
84
21
(63, 16, 65)
504
126
(45, 28, 53)
630
70
(9, 40, 41)
180
5
Table 1. Congruent Numbers.
Notice 210 shows up twice in Table 1. Do other numbers which occur once also occur
again? We will return to this question later.
Table 1 can be extended according to increasing values of k + `, and any squarefree
congruent number eventually shows up in the last column, e.g., the triangle (175, 288, 337)
with area 25200 = 7 · 602 occurs at k = 16 and ` = 9. Alas, the table is not systematic in
the appearance of the last column: we can’t tell by building the table when any particular
number should occur, if at all, in the last column, so this method of generating (squarefree)
THE CONGRUENT NUMBER PROBLEM
3
congruent numbers is not a good algorithm. For instance, 53 is a congruent number, but it
shows up for the first time when k = 1873180325 and ` = 1158313156. (The corresponding
right triangle has area 53 · 2978556542849787902 .)
Tabulations of congruent numbers can be found in Arab manuscripts from the 10th
century, and 5 and 6 appear there. Fibonacci discovered in the 13th century that 7 is
congruent and he stated that 1 is not congruent (that is, no rational right triangle has area
equal to a perfect square). The first accepted proof is due to Fermat, who also showed 2
and 3 are not congruent numbers.
Theorem 2.1 (Fermat, 1640). The number 1 is not congruent.
Proof. We will use the method of descent, which was discovered by Fermat on this very
problem. Our argument is adapted from [2, pp. 658–659].
Assume there is a rational right triangle with area 1. Calling the sides a/d, b/d, and c/d,
where a, b, c, and d are positive integers, we have a2 + b2 = c2 and (1/2)ab = d2 . (In other
words, if there is a rational right triangle with area 1 then there is a Pythagorean triangle
whose area is a perfect square. The converse is true as well.) Clearing the denominator in
the second equation,
(2.1)
a2 + b2 = c2 ,
ab = 2d2 .
We will show (2.1) has no positive integer solutions.
Assume there is a solution to (2.1) in positive integers. Let’s show there is then a solution
where a and b are relatively prime. Set g = (a, b), so g|a and g|b. Then g 2 |c2 and g 2 |2d2 ,
so g|c and g|d (why?). Divide a, b, c, and d by g to get another 4-tuple of positive integers
satisfying (2.1) with (a, b) = 1. So we may now focus on showing (2.1) has no solution in
positive integers with the extra condition that (a, b) = 1.
We will do this using Fermat’s method of descent: construct a new 4-tuple of positive
integers a0 , b0 , c0 , d0 satisfying (2.1) with (a0 , b0 ) = 1 and 0 < c0 < c. Repeating this enough
times, we reach a contradiction. Several times in the descent process we will use the following
(or minor variations on it): two positive relatively prime integers whose product is a perfect
square must each be perfect squares.
Now we start the descent. Since ab = 2d2 and a and b are relatively prime, a or b is even
but not both. Then c2 = a2 + b2 is odd, so c is odd. Since ab is twice a square, (a, b) = 1,
and a and b are positive, one is a square and the other is twice a square. The roles of a and
b are symmetric, so without loss of generality a is even and b is odd. Then
a = 2k 2 ,
b = `2
for some positive integers k and `, with ` odd (because b is odd). The first equation in (2.1)
c−b
4
now looks like 4k 4 + b2 = c2 , so c+b
2
2 = k . Because b and c are both odd and relatively
prime, (c + b)/2 and (c − b)/2 are relatively prime. Therefore
c+b
c−b
= r4 ,
= s4
2
2
for some relatively prime positive integers r and s. Solve for b and c by adding and subtracting these equations:
b = r4 − s4 , c = r4 + s4 ,
so `2 = b = (r2 + s2 )(r2 − s2 ). The factors r2 + s2 and r2 − s2 are relatively prime: any
common factor would be odd (since ` is odd) and divides the sum 2r2 and the difference
4
KEITH CONRAD
2s2 , so is a factor of (r2 , s2 ) = 1. Since the product of r2 + s2 and r2 − s2 is an odd square
and one of these is positive, the other is positive and
(2.2)
r2 + s2 = t2 ,
r2 − s2 = u2
for odd positive integers t and u which are relatively prime. Since u2 ≡ 1 mod 4, r2 − s2 ≡
1 mod 4, which forces r to be odd and s to be even. Solving for r2 in (2.2),
t2 + u2
t+u 2
t−u 2
2
(2.3)
r =
=
+
,
2
2
2
where (t ± u)/2 ∈ Z since t and u are odd.
Equation (2.3) will give us a “smaller” version of (2.1). Setting
t−u
t+u
, b0 =
, c0 = r,
a0 =
2
2
we have a02 + b02 = c02 . From (t, u) = 1 we get (a0 , b0 ) = 1. Moreover, using (2.2), a0 b0 =
(t2 − u2 )/4 = 2s2 /4 = 2(s/2)2 . Let d0 = s/2 ∈ Z, so we have a new solution (a0 , b0 , c0 , d0 ) to
(2.1). Since 0 < c0 = r ≤ r4 < r4 + s4 = c, by descent we get a contradiction.
√
√
√
√ Theorem 2.1 leads to a weird proof that 2 is irrational. If 2 were rational then 2,
2, and 2 would be the sides of a rational right triangle with area 1. This is a contradiction
of 1 not being a congruent number!
3. Relation to Arithmetic Progressions of Three Squares
The three squares 1, 25, 49 form an arithmetic progression with common difference 24.
The squarefree part of 24 is 6. This is related to 6 being a congruent number, by the
following theorem.
Theorem 3.1. Let n > 0. There is a one-to-one correspondence between right triangles
with area n and 3-term arithmetic progressions of squares with common difference n: the
sets
{(a, b, c) : a2 + b2 = c2 , (1/2)ab = n}, {(r, s, t) : s2 − r2 = n, t2 − s2 = n}
are in one-to-one correspondence by
(a, b, c) 7→ ((b − a)/2, c/2, (b + a)/2),
(r, s, t) 7→ (t − r, t + r, 2s).
Proof. It is left to the reader to check the indicated functions take values in the indicated
sets, and that the correspondences are inverses of one another: if you start with an (a, b, c)
and make an (r, s, t) from it, and then form an (a0 , b0 , c0 ) from this (r, s, t), you get back the
original (a, b, c). Similarly, starting with an (r, s, t), producing an (a, b, c) from it and then
producing an (r0 , s0 , t0 ) from that returns the same (r, s, t) you started with.
How could the correspondence in Theorem 3.1 be discovered? When s2 − r2 = n and
− s2 = n, adding gives t2 − r2 = 2n, so (t − r)(t + r) = 2n. This suggests using a = t − r
and b = t + r. Then a2 + b2 = 2(t2 + r2 ) = 2(2s2 ) = (2s)2 , so use c = 2s.
When n > 0 is rational, the correspondence in Theorem 3.1 preserves rationality and
positivity/monotonicity: (a, b, c) is a rational triple if and only if (r, s, t) is a rational triple,
and 0 < a < b < c if and only if 0 < r < s < t. Therefore n is congruent if and only if there
is a rational square s2 such that s2 − n and s2 + n are also squares. Note the correspondence
in Theorem 3.1 involves not the squares in arithmetic progression but their square roots r,
s, and t.
t2
THE CONGRUENT NUMBER PROBLEM
5
Example 3.2. For n = 6, using (a, b, c) = (3, 4, 5) in Theorem 3.1 produces (r, s, t) =
(1/2, 5/2, 7/2), whose termwise squares are the arithmetic progression 1/4, 25/4, 49/4 with
common difference 6.
Example 3.3. Taking n = 5 and (a, b, c) = (3/2, 20/3, 41/6), the correspondence in Theorem 3.1 yields (r, s, t) = (31/12, 41/12, 49/12): the rational squares (31/12)2 , (41/12)2 ,
(49/12)2 are an arithmetic progression with common difference 5.
Example 3.4. Since Fermat showed 1 and 2 are not congruent numbers, there is no arithmetic progression of 3 rational squares with common difference 1 or 2 (or, more generally,
common difference a nonzero square or twice a nonzero square).
We now can explain the origin of the peculiar name “congruent number.” Fibonacci, in
his book Liber Quadratorum (Book of Squares) from 1225, called an integer n a congruum
if there is an integer x such that x2 ± n are both squares. This means x2 − n, x2 , x2 + n
is a 3-term arithmetic progression of squares. Fibonacci’s motivation for writing his book
was the study of 3-term arithmetic progressions of integral (rather than rational) squares.
Both words congruum and congruence come from the Latin congruere, which means “to
meet together” (to congregate!). A congruum is a number related to three integer squares
in a kind of agreement (having a common difference). Considering a congruum multiplied
by rational squares (e.g., 24 · (1/2)2 = 6) gives the congruent numbers.
4. The Curve y 2 = x3 − n2 x
Whether or not n is congruent is related to solvability of pairs of equations: first, by
definition we need to solve a2 + b2 = c2 and (1/2)ab = n in positive rational numbers a,
b, and c. In Section 3 we saw this is equivalent to solving a second pair of equations in
positive rational numbers: s2 − r2 = n and t2 − s2 = n. It turns out that the congruent
number property is also equivalent to (nontrivial) rational solvability of the single equation
y 2 = x3 − n2 x.
This equation has three obvious rational solutions: (0, 0), (n, 0), and (−n, 0). These are
the solutions with y = 0.
Theorem 4.1. For n > 0, there is a one-to-one correspondence between the following two
sets:
{(a, b, c) : a2 + b2 = c2 , (1/2)ab = n}, {(x, y) : y 2 = x3 − n2 x, y 6= 0}.
Mutually inverse correspondences between these sets are
2
2n2
x − n2 2nx x2 + n2
nb
,
, (x, y) 7→
,
,
.
(a, b, c) 7→
c−a c−a
y
y
y
Proof. This is a direct calculation left to the reader. We divide by c − a in the first formula,
and c 6= a automatically since if c = a then b = 0, but (1/2)ab = n is nonzero. Restricting
y to a nonzero value is necessary since we divide by y in the second formula.
Remark 4.2. It is of course natural to wonder how the correspondence in Theorem 4.1
could be discovered in the first place. See the appendix.
The correspondence in Theorem 4.1 preserves positivity: if a, b, and c are positive then
(c−a)(c+a) = b2 > 0, so c−a is positive and thus x = nb/(c−a) > 0 and y = 2n2 /(c−a) > 0.
In the other direction, if x and y are positive then from y 2 = x3 − n2 x = x(x2 − n2 ) we see
x2 − n2 has to be positive, so a, b, and c are all positive. Also, for rational n > 0, (a, b, c)
6
KEITH CONRAD
is rational if and only if (x, y) is rational. Any solution to a2 + b2 = c2 and (1/2)ab = n
needs a and b to have the same sign (since ab = 2n > 0), and by a sign adjustment there
is a rational solution with a, b, and c all positive if there is any rational solution at all.
Therefore a rational number n > 0 is congruent if and only if the equation y 2 = x3 − n2 x
has a rational solution (x, y) with y 6= 0; we don’t have to pay attention to whether or not
x and y are positive.
A positive rational number n is not congruent if and only if the only rational solutions to
y 2 = x3 − n2 x have y = 0: (0, 0), (n, 0), and (−n, 0). For example, since 1 is not congruent
(Theorem 2.1), the only rational solutions to y 2 = x3 − x have y = 0.
Example 4.3. Since 6 is the area of a (3, 4, 5) right triangle, the equation y 2 = x3 − 36x
has a rational solution with y 6= 0. The solution corresponding to the (3, 4, 5) right triangle
by Theorem 4.1 is (x, y) = (12, 36). See Figure 2.
r(12, 36)
Figure 2. The rational point (12, 36) on y 2 = x3 − 36x.
Example 4.4. From the rational right triangle (3/2, 20/3, 41/6) with area 5, Theorem 4.1
gives us a rational solution to y 2 = x3 − 25x: (x, y) = (25/4, 75/8). If we allow sign changes
on the coordinates of (3/2, 20/3, 41/6), Theorem 4.1 will give us new rational solutions to
y 2 = x3 − 25x. Using the triples of the form (±3/2, ±20/3, ±41/6) where the first two
coordinates have the same sign, the new solutions we get to y 2 = x3 − 25x are collected in
Table 2 and they are plotted on y 2 = x3 − 25x in Figure 3.
Example 4.5. A rational solution to y 2 = x3 − 49x is (25, 120). Theorem 4.1 produces
from this solution the rational right triangle (24/5, 35/12, 337/60) with area 7, which we
met already in Figure 1.
Example 4.6. In Table 1 we found two rational right triangles with area 210: (35, 12, 37)
and (21, 20, 29). Using Theorem 4.1, these triangles lead to two rational solutions to y 2 =
x3 − 2102 x: (1260, 44100) and (525, 11025), respectively. In Figure 4, the line through
(1260, 44100) and (525, 11025) meets the curve y 2 = x3 −2102 x in a third point (240, −1800).
THE CONGRUENT NUMBER PROBLEM
7
(x, y)
Signs on (3/2, 20/3, 41/6)
(+, +, +)
(25/4, 75/8)
(+, +, −)
(−4, −6)
(−, −, +)
(−4, 6)
(25/4, −75/8)
(−, −, −)
Table 2. Solutions to y 2 = x3 − 25x.
(−4, 6) r
(−4, −6) r
r
25 75
4 , 8
r
25
75
4 ,− 8
Figure 3. Some rational points on y 2 = x3 − 25x.
Its second coordinate is negative, but the point (240, 1800) is also on that curve, and it leads
by Theorem 4.1 to the new rational right triangle (15/2, 56, 113/2) with area 210.
Example 4.7. Suppose (a, b, c) satisfies a2 + b2 = c2 and (1/2)ab = n. Such a solution
gives rise to seven additional ones: (−a, −b, −c) and
(a, b, −c), (−a, −b, c), (b, a, c), (b, a, −c), (−b, −a, c), (−b, −a, −c).
These algebraic modifications have a geometric interpretation in terms of constructing new
points from old ones on the curve y 2 = x3 − n2 x using secant lines. Say (a, b, c) corresponds
to (x, y) by Theorem 4.1, so y 6= 0. From the point (x, y) on the curve we get automatically
a second point: (x, −y). This corresponds by Theorem 4.1 to (−a, −b, −c). What points
on the curve correspond to the six remaining algebraic modifications above?
Well, there are three obvious points on the curve which have nothing to do with our
particular (x, y), namely (0, 0), (n, 0), and (−n, 0). The line through (x, y) and (0, 0) meets
the curve in the point (−n2 /x, −n2 y/x2 ), which corresponds by Theorem 4.1 to (a, b, −c).
More generally, the three lines through (x, y) and each of (0, 0), (n, 0), and (−n, 0) meet
the curve in three additional points, and their reflections across the x-axis are an additional
three points (which are where the lines through (x, −y) and each of (0, 0), (n, 0), and (−n, 0)
meet the curve). See Table 3 and Figure 5. The corresponding triples from Theorem 4.1
are collected in Table 4 and are exactly what we were looking for.
8
KEITH CONRAD
(1260,44100) r
r(525,11025)
r
r (240, −1800)
Figure 4. New rational point on y 2 = x3 − 2102 x from a secant line. (Not
drawn to scale.)
(−n, 0)
P5 rPr 3
r
r
r
r r (0,0)
P1 P7
P0 r
P2 r
(n, 0)
P4
r
P6
r
Figure 5. Intersecting y 2 = x3 − n2 x with lines through P0 and (0, 0),
(n, 0), (−n, 0), and reflected points.
We have seen that the following properties of a positive rational number n are equivalent:
• there is a rational right triangle with area n,
• there is a 3-term arithmetic progression of rational squares with common difference
n,
• there is a rational solution to y 2 = x3 − n2 x with y 6= 0.
THE CONGRUENT NUMBER PROBLEM
9
Third Point
First Point Second Point
(x, y)
(0, 0)
(−n2 /x, −n2 y/x2 )
(x, −y)
(0, 0)
(−n2 /x, n2 y/x2 )
(x, y)
(n, 0)
(n(x + n)/(x − n), 2n2 y/(x − n)2 )
(n, 0)
(n(x + n)/(x − n), −2n2 y/(x − n)2 )
(x, −y)
(x, y)
(−n, 0)
(−n(x − n)/(x + n), 2n2 y/(x + n)2 )
(x, −y)
(−n, 0)
(−n(x − n)/(x + n), −2n2 y/(x + n)2 )
Table 3. Third Intersection Point of a Line with y 2 = x3 − n2 x.
Triple
Pair
(x, y)
(a, b, c)
(x, −y)
(−a, −b, −c)
(−n2 /x, −n2 y/x2 )
(a, b, −c)
(−a, −b, c)
(−n2 /x, n2 y/x2 )
(n(x + n)/(x − n), 2n2 y/(x − n)2 )
(b, a, c)
(n(x + n)/(x − n), −2n2 y/(x − n)2 ) (−b, −a, −c)
(−b, −a, c)
(−n(x − n)/(x + n), 2n2 y/(x + n)2 )
(−n(x − n)/(x + n), −2n2 y/(x + n)2 )
(b, a, −c)
Table 4. Theorem 4.1 and Sign Changes.
The viewpoint of the equation y 2 = x3 − n2 x lets us use the geometry of the curve to
do something striking: produce a new rational right triangle with area n from two known
triangles. We saw an instance of this in Example 4.6. Notice there is nothing in the definition
of a congruent number which suggests it is possible to produce a new rational right triangle
with area n from two known ones. We can even find a new rational right triangle with area
n from just one such triangle, by using a tangent line in place of a secant line. Given a
rational point (x0 , y0 ) on y 2 = x3 − n2 x with y0 6= 0, draw the tangent line to this curve
at the point (x0 , y0 ). This line will meet the curve in a second rational point, and that
can be converted into a new rational right triangle with area n using the correspondence of
Theorem 4.1 (and removing any signs on a, b, c if they turn out negative.)
Example 4.8. In Example 4.6, we found a third rational right triangle from two known
ones by intersecting the line through the points (1260, 44100) and (525, 11025) with y 2 =
x3 − 2102 x. We can find a new rational right triangle with area 210 from the single point
(1260, 44100) by using the tangent line to y 2 = x3 − 2102 x at (1260, 44100). The tangent is
107
x − 23310
2
and it meets the curve in the second point (1369/4, −39997/8). See Figure 6. By Theorem
4.1, this point corresponds to (a, b, c) = (−1081/74, −31080/1081, −2579761/79994), which
after removing signs is the rational right triangle (1081/74, 31080/1081, 2579761/79994),
whose area is 210.
y=
Example 4.9. The (3, 4, 5) right triangle with area 6 corresponds to the point (12, 36)
on the curve y 2 = x3 − 36x, as we saw already in Example 4.3. The tangent line to this
curve at the point (12, 36) is y = (11/2)x − 30, which meets the curve in the second point
10
KEITH CONRAD
(1260,44100) r
r
1369 −39997
4 ,
8
Figure 6. New rational point on y 2 = x3 − 2102 x from a tangent line. (Not
drawn to scale.)
(25/4, 35/8) = (6.25, 4.375). Let’s repeat the tangent process on this new point. The
tangent line to the curve at (25/4, 35/8) has equation
1299
6005
x−
,
y=
140
112
which meets the curve in the new point
1442401 1726556399
(4.1)
,
≈ (73.59, 629.21).
19600
2744000
This is illustrated in Figure 7, where the second tangent line meets the curve outside the
range of the picture.1 A larger view, showing where the second tangent line meets the
curve, is in Figure 8. (The axes in Figures 7 and 8 are not given equal scales, which is
why the same tangent line in the two figures appears to have different slopes.)
Using
Theorem 4.1, (25/4, 35/8) corresponds to the rational right triangle with area 6 having
sides (7/10, 120/7, 1201/70). The rational right triangle with area 6 corresponding to the
point in (4.1) has sides
1437599 2017680 2094350404801
(4.2)
,
,
.
168140 1437599 241717895860
Armed with 3 rational right triangles with area 6, we can find 3 arithmetic progressions
of rational squares using Theorem 3.1. The (3, 4, 5) triangle, as we saw in Example 3.2,
yields the arithmetic progression 1/4, 25/4, 49/4. The (7/10, 120/7, 1201/70) right triangle
yields the arithmetic progression
1151 2
1201 2
1249 2
,
,
.
140
140
140
1The inflection points on the curve in Figure 7, for x > 0, occur where x =
q
√
12(3 + 2 3) ≈ 8.8.
THE CONGRUENT NUMBER PROBLEM
r(12,36)
r
25 35
4 , 8
Figure 7. Close view of successive tangents to y 2 = x3 − 36x starting from (12, 36).
r
rr
Figure 8. Far view of successive tangents to y 2 = x3 − 36x starting from
(12, 36). (Not drawn to scale.)
The right triangle with sides in (4.2) yields the arithmetic progression
1727438169601 2
2094350404801 2
77611083871 2
,
,
.
483435791720
483435791720
483435791720
All of these arithmetic progressions of squares have common difference 6.
11
12
KEITH CONRAD
Remark 4.10. The secant method is a way to “add” points and the tangent method is
essentially the special case of “doubling” a point. These tangent and secant constructions
can be used to give the rational points on y 2 = x3 − n2 x the structure of an abelian group
in which, for rational n > 0, any rational point (x, y) with y 6= 0 has infinite order. (This is
not at all obvious.) Therefore the curve y 2 = x3 − n2 x has infinitely many rational points as
soon as it has just one rational point with y 6= 0, so there are infinitely many rational right
triangles with area n provided there is one example and there are infinitely many 3-term
arithmetic progressions of rational squares with common difference n provided there is one
example. In terms of Table 1, this means any area arising in the table at least once will
arise in the table infinitely often.2
The importance of thinking about congruent numbers in terms of the curves y 2 = x3 −n2 x
goes far beyond this interesting construction of new rational right triangles with area n from
old ones: this viewpoint in fact leads to a tentative solution of the whole congruent number
problem! In 1983, Tunnell [8] used arithmetic properties of y 2 = x3 − n2 x (a particular
example of an elliptic curve) to discover a previously unknown elementary necessary condition on congruent numbers and he was able to prove the condition is sufficient if a certain
other conjecture is true.
Theorem 4.11 (Tunnell). Let n be a squarefree positive integer. Set
f (n) = #{(x, y, z) ∈ Z3 : x2 + 2y 2 + 8z 2 = n},
g(n) = #{(x, y, z) ∈ Z3 : x2 + 2y 2 + 32z 2 = n},
h(n) = #{(x, y, z) ∈ Z3 : x2 + 4y 2 + 8z 2 = n/2},
k(n) = #{(x, y, z) ∈ Z3 : x2 + 4y 2 + 32z 2 = n/2}.
For odd n, if n is congruent then f (n) = 2g(n). For even n, if n is congruent then h(n) =
2k(n). Moreover, if the weak Birch and Swinnerton–Dyer conjecture is true for the curve
y 2 = x3 − n2 x then the converse of both implications is true: f (n) = 2g(n) implies n is
congruent when n is odd and h(n) = 2k(n) implies n is congruent when n is even.
The weak Birch and Swinnerton–Dyer conjecture, which we won’t describe here, is one
of the most important conjectures in mathematics. (It is on the list of Clay Millenium
Prize problems.) Several years before Tunnell proved his theorem, Stephens [7] showed the
weak Birch and Swinnerton–Dyer conjecture implies any positive integer n ≡ 5, 6, 7 mod 8
is a congruent number. Tunnell’s achievement was discovering the enumerative criterion
for congruent numbers and its relation to the weak Birch and Swinnerton–Dyer conjecture.
For background on the ideas in Tunnell’s theorem, see [4] and [5]. In [6, pp. 112–114] the
particular case of prime congruent numbers is considered.
Tunnell’s theorem provides an unconditional method of proving a squarefree positive
integer n is not congruent (show f (n) 6= 2g(n) or h(n) 6= 2k(n), depending on the parity of
n), and a conditional method of proving n is congruent (conditional, that is, on the weak
Birch and Swinnerton-Dyer conjecture for the curve y 2 = x3 − n2 x).
Example 4.12. Since f (1) = g(1) = 2 and f (3) = g(3) = 4, we have f (n) 6= 2g(n) for
n = 1 and 3, so Tunnell’s criterion shows 1 and 3 are not congruent.
2The two rational points on y 2 = x3 − 2102 x which correspond to the repetition of 210 in Table 1 are
independent in the group law: they do not have a common multiple.
THE CONGRUENT NUMBER PROBLEM
13
Example 4.13. Since h(2) = k(2) = 2, we have h(2) 6= 2k(2), so Tunnell’s criterion shows
2 is not congruent.
Example 4.14. Since f (5) = g(5) = 0 and f (7) = g(7) = 0, we have f (n) = 2g(n) for n = 5
and 7. Tunnell’s theorem says 5 and 7 are congruent if the weak Birch and Swinnerton-Dyer
conjecture is true for y 2 = x3 − 25x and y 2 = x3 − 49x. Unconditionally, we saw earlier
that 5 and 7 are congruent.
Example 4.15. Since h(10) = 4 and k(10) = 4, h(10) 6= 2k(10), so Tunnell’s theorem says
10 is not a congruent number.
Example 4.16. We will show (conditionally) that any squarefree positive integer n satisfying n ≡ 5, 6, 7 mod 8 is a congruent number. Tunnell’s theorem tells us to check
that f (n) = 2g(n) when n ≡ 5, 7 mod 8 and h(n) = 2k(n) when n ≡ 6 mod 8. Since
x2 + 2y 2 6≡ 5, 7 mod 8 for any integers x and y, f (n) = 0 and g(n) = 0 when n ≡ 5, 7 mod 8,
so f (n) = 2g(n). When n ≡ 6 mod 8 we have n/2 ≡ 3 mod 4, so x2 6≡ n/2 mod 4 for any integer x. Therefore h(n) = 0 and k(n) = 0 when n ≡ 6 mod 8, so h(n) = 2k(n). This shows
n is congruent if the weak Birch and Swinnerton-Dyer conjecture is true for y 2 = x3 − n2 x.
Appendix A. Discovering Theorem 4.1
Fix a real number n 6= 0. The real solutions (a, b, c) to each of the equations
1
ab = n,
2
describe a surface in R3 , so it is reasonable to expect these two surfaces intersect in a curve.
We want an equation for that curve, which will be y 2 = x3 − n2 x in the right choice of
coordinates. Two approaches will be described, one algebraic and the other geometric. The
sign on n will be irrelevant, so we allow any n 6= 0 rather than n > 0.
The algebra is simplified by introducing a cross-term in the equation a2 + b2 = c2 . Let
c = t + a, which turns this equation into b2 = t2 + 2at, or equivalently
a2 + b2 = c2 ,
(A.1)
2at = b2 − t2 .
(A.2)
Since ab = 2n is nonzero, neither a nor b is 0, so we can write a = 2n/b and substitute it
into (A.2):
4nt
= b2 − t2 .
b
Multiplying through by b makes this
4nt = b3 − t2 b.
Divide by t3 (t 6= 0, as otherwise a = c and then b = 0, but ab = 2n 6= 0):
3
4n
b
b
=
− .
t2
t
t
Multiply through by n3 :
2n2
t
2
=
nb
t
3
−n
2
nb
t
.
Set x = nb/t and y = 2n2 /t, so y 2 = x3 − n2 x. Then x = nb/(c − a) and y = 2n2 /(c − a),
as in Theorem 4.1.
14
KEITH CONRAD
We now turn to a geometric explanation of Theorem 4.1, taking greater advantage of the
interpretation of the two equations in (A.1) as surfaces which meet in a curve. Rather than
working with the equations as surfaces in R3 , we will work in the projective space P3 (R)
by homogenizing the two equations. This doesn’t change the first equation in (A.1), but
makes the second one (1/2)ab = nd2 .
Letting [a, b, c, d] be the homogeneous coordinates of a typical point in P3 (R), the two
equations
(A.3)
a2 + b2 = c2 ,
1
ab = nd2
2
each define surfaces in P3 (R). Let C be the intersection of these surfaces (a curve). There
are points on C with b = 0, namely [a, b, c, d] = [1, 0, ±1, 0]. These points are not in the usual
affine space inside P3 (R), and we will use one of these points in a geometric construction.
Let’s project through the point P := [1, 0, 1, 0] to map C to the plane
Π := {[0, b, c, d]}
and find the equation for the image of C in this plane. The point P lies on C and not in Π.
For each Q ∈ C other than P , the line P Q in P3 (R) meets Π in a unique point. Call this
point f (Q). When Q = P , intersect the tangent line to C at P with the plane Π to define
f (P ). We have defined a function f : C → Π.
Computing a formula for f necessitates a certain about of computation to see what
happens. Suppose first that Q = [a, b, c, d] is not P . The line through P and Q is the set of
points
[λ + µa, µb, λ + µc, µd],
which meets Π where λ = −µa, making
f (Q) = [0, µb, µ(c − a), µd] = [0, b, c − a, d].
As for f (P ), the tangent planes to each of the surfaces a2 + b2 = c2 and (1/2)ab = nd2 in
P3 (R) at the point P are the planes a = c and b = 0, so the tangent line at P is the set of
points
[a, 0, a, d],
which meets Π in [0, 0, 0, 1], so f (P ) = [0, 0, 0, 1]. Thus
(
[0, b, c − a, d], if [a, b, c, d] 6= [1, 0, 1, 0],
f ([a, b, c, d]) =
[0, 0, 0, 1],
if [a, b, c, d] = [1, 0, 1, 0].
As an exercise, check f is injective. (Hint: Since (1/2)ab = nd2 , b and d determine a if
b 6= 0.)
All points in the plane Π have first coordinate 0. Identify Π with P2 (R) by dropping
this coordinate, which turns f into the function g : C → P2 (R) where
(
[b, a − c, d], if [a, b, c, d] 6= [1, 0, 1, 0],
(A.4)
g([a, b, c, d]) =
[0, 0, 1],
if [a, b, c, d] = [1, 0, 1, 0].
We have mapped our curve C to the projective plane P2 (R). What is an equation for
the image g(C)? For Q = [a, b, c, d] on C, write g(Q) = [x, z, y]. (This ordering of the
coordinates will make formulas come out in close to the expected way more quickly.) When
THE CONGRUENT NUMBER PROBLEM
15
Q 6= [1, 0, 1, 0] (that is, a 6= c), (A.4) says we can use x = b, y = d, and z = c − a 6= 0.3 The
equations in (A.3) become a2 + x2 = (a + z)2 and (1/2)ax = ny 2 , so
x2 = 2az + z 2 , ax = 2ny 2 .
Since z 6= 0, we can solve for a in the first equation, so a is determined by x, y, and z.
Multiplying the first equation by x and the second by 2z, x3 = 2axz + xz 2 = 4ny 2 z + xz 2 .
Thus
4ny 2 z = x3 − xz 2 .
Set X = x, Y = 2ny, and Z = z/n to find Y 2 Z = X 3 − n2 XZ 2 , which is the homogeneous
form of Y 2 = X 3 − n2 X.
Tracing this correspondence out explicitly from the start, if we begin with [a, b, c, d] on
C where d 6= 0 (the standard affine part of C), its image [X, Z, Y ] in P2 (R) is
nb
2n2 d
c−a
2
.
, 2nd = [nb, c − a, 2n d] =
, 1,
b,
n
c−a
c−a
Since d 6= 0 implies a 6= c, using inhomogeneous coordinates with middle coordinate 1 in
P2 (R) the point (a, b, c) goes to (nb/(c − a), 2n2 /(c − a)), which is the transformation in
Theorem 4.1.
As an exercise in these techniques, consider the problem of classifying triangles with a
given area n > 0 and a given angle θ. (Taking θ = π/2 is the congruent number problem.)
Let a, b, c be the side lengths of the triangle, with c the length of the edge opposite the
angle θ. The equations in (A.1) are replaced by
1
(A.5)
a2 + b2 − 2ab cos θ = c2 ,
ab sin θ = n.
2
(If there is a solution with rational a, b, c, and n then cos θ and sin θ must be rational.)
Show the solutions (a, b, c) of (A.5) are in one-to-one correspondence with the solutions
(x, y) of the equation
2n cos θ 2
y 2 = x3 +
x − n2 x
sin
θ
cos θ + 1
cos θ − 1
= x x+n
x+n
,
sin θ
sin θ
with y 6= 0. The correspondence should specialize to that in Theorem 4.1 when θ = π/2.
Appendix B. Other Diophantine Equations
In Table 5, the first two columns show how to convert the sides (a, b, c) of a rational
right triangle with area 1 into a positive rational solution of the equation y 2 = x4 − 1 and
conversely. (These correspondences are not inverses, but they do show a positive rational
solution in the first column leads to a positive rational solution in the second column, and
conversely.) The last two columns give a (bijective) correspondence between rational right
triangles with area 2 and positive rational solutions of y 2 = x4 + 1. So showing 1 and 2
are not congruent numbers is the same as showing the equations y 2 = x4 ± 1 don’t have
solutions in positive rational numbers.
A positive rational solution (x, y) to y 2 = x4 ± 1 can be turned into a positive integral
solution (u, v, w) of w2 = u4 ± v 4 by clearing a common denominator, and we can go in
3The cross term t = c − a in the algebraic method is precisely z, so now we get a geometric interpretation
of this cross term as a coordinate in a projection map to a plane.
16
KEITH CONRAD
a2 + b2 = c2 ,
1
2 ab = 1
x = c/2
y = |a2 − b2 |/4
y 2 = x4 − 1
a2 + b2 = c2 , y 2 = x4 + 1
1
2 ab = 2
a = y/x
x = a/2
a = 2x
b = 2x/y
y = ac/4
b = 2/x
c = (x4 + 1)/xy
c = 2y/x
Table 5. Correspondences between rational right triangles with area 1 and
y 2 = x4 ± 1.
reverse by dividing by v 4 . That 1 and 2 are not congruent is therefore the same as the
equations w2 = u4 ± v 4 having no positive integer solutions. The reader is referred to [1,
pp. 252–256] for a proof by descent that w2 = u4 ± v 4 has no positive integer solutions.
That the congruent number property for 1 and 2 is equivalent to the solvability of a single
equation in positive rational numbers (y 2 = x4 − 1 for 1 and y 2 = x4 + 1 for 2) generalizes:
n is congruent if and only if y 2 = x4 − n2 has a positive rational solution and if and only
if y 2 = x4 + 4n2 has a positive rational solution. See Table 6, where the first two columns
turn rational right triangles with area n into positive rational solutions of y 2 = x4 − n2 and
conversely, and the last two columns do the same with y 2 = x4 + 4n2 . As in Table 5, the
correspondences in the first two columns of Table 6 are not inverses of each other, but the
correspondences in the last two columns are inverses. (When n = 2 the equation in Table
6 is y 2 = x4 + 16 rather than y 2 = x4 + 1 as in Table 5. We can easily pass from the former
to the latter by replacing y with 4y and x with 2x.) The equivalence of n being congruent
with y 2 = x4 − n2 having a positive rational solution is due to Lucas (1877).
a2 + b2 = c2 , y 2 = x4 + 4n2
1
2 ab = n
a = y/x
x=a
a=x
b = 2nx/y
y = ac
b = 2n/x
c = (x4 + n2 )/xy
c = y/x
Table 6. More correspondences between rational right triangles and Diophantine equations.
a2 + b2 = c2 ,
1
2 ab = n
x = c/2
y = |a2 − b2 |/4
y 2 = x4 − n2
We pulled the equations y 2 = x4 − n2 and y 2 = x4 + 4n2 out of nowhere. How could
they be discovered? The arithmetic progression viewpoint on congruent numbers (Theorem
3.1) leads to one of them. If n is congruent, there are rational squares r2 , s2 , and t2 with
s2 − r2 = n and t2 − s2 = n. Then r2 = s2 − n and t2 = s2 + n, so multiplication gives
(rt)2 = s4 − n2 and we’ve solved y 2 = x4 − n2 in positive rational numbers.
Remark B.1. For t 6= 0, solutions to y 2 = x4 + t and to Y 2 = X 3 − 4tX are in a one-toone correspondence, by (x, y) 7→ (2t/(y − x2 ), 4tx/(y − x2 )) and (X, Y ) 7→ (Y /2X, (Y 2 +
8tX)/4X 2 ). In particular, solutions to y 2 = x4 − n2 correspond to solutions to Y 2 =
X 3 + (2n)2 X, which is not the equation Y 2 = X 3 − (2n)2 X and thus isn’t related to
whether or not 2n is a congruent number. Explicit examples show the lack of a general
connection between n and 2n being congruent: 5 is congruent but 10 is not, while 3 is not
congruent but 6 is.
THE CONGRUENT NUMBER PROBLEM
17
References
[1] D. M. Burton, “Elementary Number Theory,” 6th ed., McGraw-Hill, New York, 2007.
[2] W. A. Coppel, “Number Theory: An Introduction to Mathematics. Part B,” Springer-Verlag, New York,
2006.
[3] L. E. Dickson, “History of the Theory of Numbers,” Vol. II, Chelsea, New York, 1952.
[4] G. Henniart, Congruent Numbers, Elliptic Curves, and Modular Forms, translation by F. Lemmermeyer
at http://www.fen.bilkent.edu.tr/∼franz/publ.html.
[5] N. Koblitz, “Introduction to Elliptic Curves and Modular Forms,” 2nd ed., Springer–Verlag, New York,
1993.
[6] A. Knapp, “Elliptic Curves,” Princeton Univ. Press, Princeton, 1992.
[7] N. M. Stephens, Congruence properties of congruent numbers, Bull. London Math. Soc. 7 (1975), 182–
184.
[8] J. Tunnell, A Classical Diophantine Problem and Modular Forms of Weight 3/2, Invent. Math. 72 (1983),
323–334.
Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertising