Solutions to a particular equation of perfect squares Can we prove (or is it otherwise known) that for positive integers $a>b>c>1$, the only solutions to the equation:
$$a^2+1=b^2(c^2+1)$$
is given by the following family for any fixed $c$:
$$   a_1 = 4 c^3 + 3 c, \; \; \;  b_1 = 4 c^2 + 1,$$
$$   a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \;  b_2 = 16 c^4 + 12 c^2 + 1,$$
$$ a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \;  b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n  $$

Yes.  Given $c$, this is a "Pell equation" $a^2 - (c^2+1) b^2 = -1$ 
whose general solution is
$$
a_n + b_n \sqrt{c^2+1} = (c + \sqrt{c^2+1})^{2n-1}
$$
because $c + \sqrt{c^2+1}$ is clearly the fundamental unit in
${\bf Z}[\sqrt{c^2+1}]$.
I just answered this on stackexchange.
