First, suppose n is the shortest side of the triangle, m, k are other two sides. According to Pythagorean Theorem, we know n² + m² = k²

just do a change k² - m² = n²

futherly (k + m)(k - m) = n²

as we know, n² × 1 = n² so we can suppose that k + m = n², k - m = 1 [because we have SPJ]

easily, we get

because the side is a interger, so this solution can only be used when n is a odd.

So how to deal with even? we find that if (k-m) is odd, the solution is suitable for odd. On the other hand, we guess that if(k-m) is even, the solution is suitable for even.

just as this,

so, we get

this is an O(1) algorithm.

So if you want to see more clearly, you can see the formual on this page. if you have some questions, i am willing to help you.:) [my english is not very good, so please pass some grammer fault.:P]