<s>[INST] Perfect Square relationship with no solutions I would like to show that for positive integers $a>b,c$ all greater than 1 such that $c\nmid a$, there are no solutions to the following equation:
$$a^2+1=b^2(c^2+1)$$
As was pointed out in the comments, it might help to rearrange and factor:
$$(a-b)(a+b)=(bc+1)(bc-1)$$
Another interesting approach to the question, if we rewrite:
$$a^2-b^2c^2=b^2-1$$
Then we are trying to argue that $b^2-1$ cannot be squarefree (in particular, $c^2|b^2-1$)
 [/INST] Unclear. 
This equation is equivalent to the following : $a^2-qb^2=a^2-(c^2+1)b^2=-1$
You can use formulas solutions binary quadratic forms. It can be seen that the solutions of this equation is always there.
Though it is necessary to bring the decisions some pretty simple solutions:
the equation:  $aX^2+bXY+cY^2=f$
If the root of the whole:  $\sqrt{\frac{f}{a+b+c}}$
Then use the solution of Pell's equation:  $p^2-(b^2-4ac)s^2=1$
Solutions can be written:
$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
If a root: $\sqrt{fa}$
then the solutions are of the form:
$Y=4ps\sqrt{fa}$
$X=(-2bps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a}}$
Although it should be mentioned, and the equation: $aX^2-qY^2=f$
If the root of the whole: $\sqrt{\frac{f}{a-q}}$
Using equation Pell:  $p^2-aqs^2=1$  solutions can be written:
$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$
$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$
And for that decision have to find double formula.
$Y_2=Y+2as(qsY-pX)$
$X_2=X+2p(qsY-pX)$
So, take that formula and believe. Hopefully this will be enough.
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