| With hypotenuse z, inradius r, some elementary geometry gives:
x^2 + y^2 = z^2
z = (x - r) + (y - r) = x + y - 2r
Eliminating z gives:
(x - 2r)(y - 2r) = 2r^2
If r's prime decomposition is:
2^A 3^B 5^C ...
then x - 2r must have the decomposition:
2^a 3^b 5^c ... (ie powers of the SAME primes)
and to ensure that x and y are coprime we must have:
a = 0 or 2A + 1,
b = 0 or 2B, c = 0 or 2C ...
This gives 2^n possibilities for x - 2r, and hence x. I have assumed that we
can't turn triangles over: if we can then we just have to observe that this
enumeration of solutions will cover them all exactly twice.
Dick
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