Finding another set of pythagorian triples

We look at the difference between squares of numbers differing by 2, and set it to be a square itself.

So, if n and n+2 are the two numbers separated by 2, the difference of their squares, 4n+4, must be a square number. Since it is clearly even, let us set it to the square of an even number, 2m. So

4n+4 = (2m)²

and hence n=((2m)²-4)/4 = m²-1

So the triplet is 2m, m²-1, m²+1

However, if m is odd then all three terms are even, and this cannot be a primitive triple. In fact, if we substitute m=2k+1 we get

2(2k+1), 4k²+4k, 4k²+4k+2

which is precisely twice the last sequence. We can discard these and keep only those with even m, which can set to m=2k. This gives

2(2k+1), 4k²+4k, 4k²+4k+2

4k, 4k²-1, 4k²+1