Is it possible to have $a^2 + b^2 = c^2 + 1$ for $a$, $b$, $c$ being coprime integers? 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!
 A: $7^2+11^2=13^2+1$
They all have to be odd, otherwise two are even and not coprime.
Look for values $c^2+1$ for which $(c^2+1)/2$ is composite.  There will always be another $a^2+b^2=c^2+1$
A: 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:
$$ 4^2+7^2 = 8^2+1, $$
$$ 31^2+56^2 = 64^2+1, $$
$$ \ldots $$
A: HINT:
Using Brahmagupta-Fibonacci Identity,
$$(ab+cd)^2+(ad-bc)^2=(ab-cd)^2+(ad+bc)^2$$
Set $ad-bc=\pm1\implies (ad,bc)=1\implies(a,b)=(a,c)=(c,d)=(b,d)=1$
Choose arbitrary $a,b$ such that $(a,b)=1;$  then use the method described in my answer here: Solving a Linear Congruence  to determine $c,d$
