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1. a b c ->
c = i * i + j * j;
b = 2 * j;
a = i * i - j * j;
-> a * a + b * b = c * c;
if i and j only have a common divisor of 1 and meanwhile (a + b) % 2 ==
1 then a, b and c are relatively prime.
ATTENTION: the method above can not find all the a, b and c.
It just make sure that if a, b and c are from i and j, then they must
meet
a * a + b * b = c * c; So point 2 must be added.

2. (a * k) * (a * k) + (b * k) * (b * k) == (c * k) * (c * k); The
multiple of k.