Originally posted by talzamir
That is helpful indeed. So, using the buffed-up version of Euclid's method of finding Pythagorean triplets,
a = k (m^2 - n^2)
b = k (2mn)
c = k (m^2 + n^2)
where k, m , n are all integers and m > n, what are the values that a + b + c can get?
a + b + c = k(m^2 - n^2 + 2mn + m^2 + n^2) = 2 k m (m n)
So the possible values are even, and are divisible by a square.
n = 2, m = 5, k = 10 would give 2 x 10 x 5 x 5 x 2 = 1,000.
Slight mistake, your formule should read:
a+b+c = k(m^2 - n^2 + 2mn + m^2 + n^2) = k(2m^2 + 2mn) = 2km(m+n)
Then choose: k = 25, m = 4, n = 1
--> a = 375, b = 200, c = 425