I discovered this variation of the classic congruent number problem in a streaming-video of a talk at MSRI by Noriko Yui.

Consider a rational triangle with sides a, b, c and opposing angles A, B, and C = 2 π / 3.

Then

c2  =  a2  +  b2 - 2 a b cos C   =   a2  +  b2  +  a b

and the area Δ satisfies

2 Δ   =  a  b  sin C  =  a  b  √ 3 / 2

Now, suppose the area = N √ 3 for some integer N, where N can be assumed to be squarefree.

Then a b = 4 N and so

c2  =   a2   +  16 N2  a2   +  4 N

which simplifies to

c2 a2   =   a4   +   4 N a2   +   16 N2    =   d2

if we define d = a c .

This quartic has an obvious rational solution a = 0 , d = 4 N, so is birationally equivalent to an elliptic curve. Standard methods show that this curve is

S2  =  T ( T + N ) ( T - 3N )

with the transformation  a = S / T.

For example, if N = 10, we find that T = 54,  S = 288 lies on the curve, giving  a = 16 / 3. From a b = 4 N , we get  b = 15 / 2 and thus  c = 67 / 6.

There are 4 points of order 2, T = 0, T = -N, T = 3 N, and the pt at infinity. It is reasonably easy to show that there cannot be points of order 4. Thus the torsion subgroup is isomorphic to Z2×Z2 or Z2 ×Z6. Numerical evidence suggests the former. Anybody able to prove this in a simple manner?

If there are only 4 torsion points, then a practical solution can only come from curves which give rank > 0.

If we change the angle to p /3, then the curve changes to

S2  =  T ( T - N ) ( T + 3N )

with c² = a² + b² - a b.

For both types of curve, we used the Birch and Swinnerton-Dyer conjecture to estimate the rank for 1 £ N £ 999, and hence to predict those values of N giving a soluion. The results found so far can be found in

(a) 2p /3 congruent number solutions

(b)   p/3 congruent number solutions

If we investigate the solutions found we can arrive at the following conjectures:

(1) N (squarefree) is a 2p/3 congruent number if N is congruent to one of 5, 9, 10, 15, 17, 19, 21, 22, 23 mod 24.

(2) N (squarefree) is a p/3 congruent number if N is congruent to one of 6, 10, 11, 13, 17, 18, 21, 22, 23 mod 24.

Yui's lecture contains some results from M. Yoshida at Chiba University on such conjectures, but most are unproven - can they be proven by variations of Tunnell's approach to the congruent number problem?

There are still some curves, which we think have rank 1, but no rational point is known. The nature of the curves means that we can apply Elkies' Heeger-point method to this problem. This is excellently described in the reference [elk2] . I should state that this is not the easiest paper for the mathematically challenged to read.

August 2005:

For some reason the back-up files on this problem disappeared from my laptop, so I had to spend some time restoring them.

I have managed to find ALL the solutions in the [1,999] range for both types of curve.