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...
 A: $(n^2+1)((n+1)^2 +1) = (n(n+1)+1)^2 + 1$. You can check this by simple expansion.
Hence, the cases such as $(3,4,13)$, $(4,5,21)$ etc. will come under this.
This is only an example. As I will get more I will keep updating this answer.
A: Partial answer :
Let $n\in\mathbb{N}$ and :
$$(a,b,c)=(n,n+1,n^2+n+1)$$
We have :
$$(a^2+1)(b^2+1)=(n^2+1)(n^2+2n+2)=n^4+2n^3+3n^2+2n+2=(n^2+n+1)^2+1$$
A: Show how the General formula allows us to solve this equation.
$$(a^2+1)(b^2+1)=c^2+1$$
We write this equation differently.
$$c^2-(a^2+1)b^2=a^2$$
Will do the replacement.  $c=q+(k-1)b$
Then the equation takes the form.
$$q^2+2(k-1)qb+(k^2-2k-a^2)b^2=a^2$$
Now you can use the General formula.  http://www.artofproblemsolving.com/community/c3046h1048219
Let the root equal to 1.  This means that using the solutions of the equation Pell.
$$k^2-2a^2=1$$
Knowing the first solution.
$$(k_0 ; a_0) - (3 ; 2)$$
You can find the following solution to the formula.
$$k=3k_0+4a_0$$
$$a=2k_0+3a_0$$
Using the General formula and the solution of the equation Pell.
$$p^2-(a^2+1)s^2=\pm1$$
You can write the solution of the equation in this form.
$$c=kp^2\pm2(a^2+1)ps+k(a^2+1)s^2$$
$$b=p^2\pm2kps+(a^2+1)s^2$$
