if a and b are legs in a pythagorean triple and a=2mn and b=m^2-n^2 proof a = b ± 1 implies 2n^2 ± 1 is a perfect square.

Question

If $m > n$ and $a = 2mn$, $b = m^2 − n^2$ and $c = m^2 + n^2$ then $(a, b, c)$ is a Pythagorean triple.

Show that triples where $a = b\pm 1$ will only occur if $2n^2\pm 1$ is a perfect square.

For the life of me, I can't figure it out. what am I supposed to plug in to get this?

Answer 1

First assume $a=b+1$. Plug in the expressions you have for $a,b$ and get $$a=b+1\\2mn=m^2-n^2+1\\n^2=m^2-2mn+1\\n^2=(m-n)^2-n^2+1\\2n^2-1=(m-n)^2$$ The other case is similar. Note that you would more correctly write that $a=b\pm 1 \implies 2n^2 \mp 1$ is a perfect square because the signs are opposite.