<s>[INST] Integer solutions of $(b^2+1)(c^2+1)=a^2+1$ As the title says, I'm interested in integer solutions of the equation $(b^2+1)(c^2+1)=a^2+1$. Is it possible to parametrize the solutions as in the case of Pythagorean triples? If yes, then how would one proceed to find a parametrization in this case?

As suggested in the comments, the Pythagorean quadruples could be a good idea: we find that 
$c = \frac{p^2-n^2-m^2}{2mp},\ b = \frac{p^2-n^2-m^2}{2np},\ a = \frac{(p^2+n^2+m^2)(p^2-n^2-m^2)}{4mnp^2}$ where $m,n,p$ are coprime and all the fraction are integers... I don't know if we can eliminate one of the $m,n,p$ using this...
Another take, which shows that there are infinitely many solutions: for fixed $b$ we can write $a^2-(b^2+1)c^2=b^2$ which is a Pell equation. This has a solution $a=b,c=0$ so it has infinitely many... Parametrizations, algorithms are available...
 [/INST] Partial answer :
Let $n\in\mathbb{N}$ and :
$$(a,b,c)=(n,n+1,n^2+n+1)$$
We have :
$$(a^2+1)(b^2+1)=(n^2+1)(n^2+2n+2)=n^4+2n^3+3n^2+2n+2=(n^2+n+1)^2+1$$
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