Quote:
Originally Posted by CRGreathouse
Ah, you think he meant c^2 = u^2 + v^2 rather than c = u^2 + v^2? That doesn't match either, though: I get

Quote:
Originally Posted by A008846
Not only the square of these numbers is equal to the sum of two nonzero squares, but the numbers themselves also are; this sequence is then a subsequence of A004431.  JeanChristophe Hervé, Nov 10 2013

so c=v^2+u^2 works. for c<2^n. you still get the same numbers I think but the statement is about that hitting 2 pi in ratio with 2^n ( aka that ratio ((2^(n1))/(number of prim. pythag triples) tends to pi as n tends to infinity)