If $ab^2+1 = c^2+d^2$ with $a$ squarefree, what [else] can be said about $a$? What is known about squarefree integers $a$ where there exist non-zero integers $b$, $c$, and $d$, with $\gcd(b,c)=\gcd(b,d)=1$, such that 
$$ab^2+1=c^2+d^2$$
?
EDIT: As pointed out by individ, if an integer $g>1$ exists such that $g \mid \gcd(b,c)$, then $b=gb_1$ and $c=gc_1$ yields
\begin{align}
a(gb_1)^2 + 1 &= (gc_1)^2+d^2  \\
d^2 - (ab_1^2-c_1^2)g^2 = 1.
\end{align}
Now if $ab_1^2-c_1^2$ is not a square, there are infinite solutions. So to make the question less trivial, we may assume $\gcd(b,c)=\gcd(b,d)=1$.
 A: Generally speaking for the equation. $$x^2+y^2=qz^2+1$$ For all values of the coefficient  $q$  There are solutions. It is easy to show.  
Will make a replacement.  $y=as$ ; $z=bs$  Then.
$$x^2-(qb^2-a^2)s^2=1$$
This equation Pell. And we can always choose such numbers $a,b$ . To the expression  $(qb^2-a^2)$ It was not square.
A: Under the condition $\gcd(b,cd)=1$, one thing we can say for sure is that, if there is a solution to $ab^2+1=c^2+d^2$, then $a\equiv0$, $1$, $3$, $4$, or $7$ mod $8$.  This is because $b$ can never be even, since $2\mid b$ implies $c$ and $d$ are both odd, which makes $c^2+d^2\equiv2$ mod $4$ whereas $ab^2+1\equiv1$ mod $4$.  But if $b$ is odd, then $ab^2+1\equiv a+1$ mod $8$, whereas $c^2+d^2\equiv1$, $2$, $4$, $5$, or $0$ mod $8$.
Another way to say this is that there is definitely no solution if $a\equiv2$ mod $4$ or $5$ mod $8$.  If I had to guess, I'd say there probably are solutions for all $a$ in the other congruence classes.
Remark:  The condition $cd\not=0$ as well as $\gcd(b,cd)=1$ makes life a little more interesting than it would be if we allowed solutions of the form $(b,c,d)=(1,c,0)$.  For example, for $a=8$, $(b,c,d)=(1,3,0)$ would be an easy solution, but if it's disallowed, the next solution (i.e., the smallest value of $b$ that works) is $8\cdot17^2+1=48^2+3^2$.
