I came across the following problem and wrote up the following proof a few years ago - what do you think?
It is required to demonstrate the truth of Statement S1: "If $a^2 + b^2$ is a perfect square, then $a^2 - b^2$ can't be one.
Assume that S1 is false and that $a^2 - b^2$ is a perfect square. Then there exists $m < a$ such that $a^2 - b^2 = m^2$. This implies that $b^2 + m^2$ is a perfect square. Again, with S1 assumed false, the difference between $b^2$ and $m^2$ would be a perfect square $n^2$ smaller than at least one of them. We should be able to continue this process, getting smaller and smaller perfect squares $m^2$, $n^2$...ending at $3^2$, corresponding to $5^2 - 4^2 = 3^2$. However, $5^2 + 4^2 = 41$ is not a perfect square. Hence S1 must be true. QED?
PS: The above argument seems to be Fermat's method of infinite descent. I wonder if Fermat's method of infinite descent can be used to prove his last theorem. Incidentally, I think the following "proof" of Fermat's Last Theorem holds only for Pythagorean triples: http://script864.rssing.com/chan-14851556/latest.php#item23