CONGRUENT
NUMBERS
The most famous right-angled triangle is the (3,4,5) triangle
which clearly has area 6.
Consider the triangle with sides a=20/3 and b=3/2,
then a2+b2=1681/36=(41/6)2,
so we have a right-angled triangle
with rational sides and area 5.
It is easy to work out that area 5 can not occur with integer sides.
Is it possible for other integers N to occur as the areas of rational-sided right-angled triangles?
Integers that do are called CONGRUENT NUMBERS.
Suppose a2+b2=h2 with a,b,h rational and area=N=ab/2. Then the triangle with sides (da, db, dh) is right-angled and area = Nd2. Thus we can generate triangles with area having a square factor from triangles with area without the square-factor.
We assume from now on that N is square-free.
Now b=2N/a, giving
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Define g = h + a giving h -
a = 4N2/(a2g), and so
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Multiply by a3g2/8 so that
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Now,define Y=a2g/2 and X=a g/2 giving
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The elliptic curve EN has 3 points of order 2 when X = 0, N, -N and Y=0, all giving a=0.
Points of order 3 only occur at points of inflexion. These
occur
where
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If (p,q) were a point of order 4, then 2(p,q) would have X-coordinate (n4+2p2n2+p4)/(4q2) which is a positive square, so we would require N=m2. If we try and solve for p we find terms involving √ 2 which give irrational points.
So there are no points of order 3 or 4, and thus the 3 points of order 2 are the only torsion points.
So to get a real triangle we need the curves to have rank 1.
For example, E6: Y2 = X3 - 36 X has rank 1 with points (12,36) and (18,72) on it, corresponding to a=3 or a=4. If we double (12,36) we get (25/4,-35/8) giving a=7/10, b=120/7, but area still 6.
Thus to find out if N is a congruent number we need to show EN has rank greater than 0, and then find a point on the curve.
In a classic paper, Tunnell proves that if N is congruent to 5 or 6 or 7 modulo 8 then the rank is odd and N is a congruent number. If N is congruent to 1 or 2 or 3 then the rank is even so might be 2,4,.. but usually is 0.
We can use the Birch and Swinnerton-Dyer conjecture to provide an estimate of the rank and then try to find a point and hence a triangle.
We have done this for 1 < = N < = 9999. For ranks equal to 1, the BSD conjecture also provides an estimate of the height of a rational point, which gives a rough idea of the size in terms of number of digits.
Some of the heights were huge, being greater than 500, indicating rational points with several hundreds of digits in the numerator and denominator.
Such heights are currently way beyond practical computation, but the congruent number curve has a wonderful property that allows us to find such points, sometimes in minutes.
Define X = -N U and Y = N √(-N) V, then V2 = U3 - U. In a superb paper in 1994, Noam Elkies of Harvard showed how to use this relationship together with the idea of Heegner points to derive an extremely efficient method for finding X and Y. This is especially true for N ≡ 7 modulo 8.
I have used this method along with probably every other method ever described to find solutions for all the curves up to N=10000. I am fairly certain that this is NOT the first time this has been done so no claims for originality are made.
The results can be found in the following files where we give N and a - easy to get b = 2N/a.