TITLE: Is it possible to have $a^2 + b^2 = c^2 + 1$ for $a$, $b$, $c$ being coprime integers?
QUESTION [5 upvotes]: As stated above. I'm working on a possible proof. It appears that $$(b+1)(b-1)=(c+a)(c-a)$$ That's where I'm stuck. Any help please? A clear, simple proof desired, thanks!

REPLY [2 votes]: Every number whose prime divisors are of the form $4k+1$ can be written as the sum of two coprime squares. For sure, if we take $c=2^m$, every prime divisor $p$ of $c^2+1=4^m+1$ is of the form $4k+1$, because
$$ 4^m+1\equiv 0\pmod{p}$$
implies:
$$ -1\equiv (2^m)^2,\pmod{p}$$
hence $-1$ is a square $\pmod{p}$. Moreover, if $4^m+1$ is not a prime, it can be written as a sum of two coprime squares in at least three different squares. So we have a solution for every $c=8^n$, for instance:
$$ 4^2+7^2 = 8^2+1, $$
$$ 31^2+56^2 = 64^2+1, $$
$$ \ldots $$