Characterizing integer solutions to $a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr)$ I’m looking to characterize all integers $a$ and $b$ satisfying $$a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr), \qquad a > b \ge 1.$$
Brute force searches have so far turned up the two solutions $(a,b) = (13,7)$ and $(a,b)=(9601,5441)$. I’m guessing there are an infinite number, but I’d like to reduce the numbers I have to check. Obviously, $a$ must be odd and the sum of two squares, so I’m already taking that into account.
Any thoughts about what the next solution might be, or how to attack the characterization of any/all solutions?
If $b$ is odd, then there exists an integer $c$ such that $a=b+2c$. Substituting, we find $a \parallel (32c^4+8c^2+1) = \bigl((4c^2)^2+(4c^2+1)^2\bigr)$, and then $a^2$ divides a related quotient. Based on that, I feel like there might be a Vieta-jumping solution… but I still can’t quite find it.

This problem in a way is related to equation $x^2+(x+1)^2=y^2$ which has infinite solution.There is a proof in this site.I will try to show this relation by an examples:
$119^2+120^2=13^4$
$119=7\times 17$
$(7\times17)^2+(7\times17+1)=13^4$$\space\space\space\space\space\space(1)$
and we see that:
$7^4+(7^2+1)^2=29\times 13$
Relation (1) is symmetric for 7 and 17 so relation $b^4+(b^2+1)^2= km^2$ must also be true for 17, we check this:
$17^4+(17^2+1)^2=86021=509\times 13^2$
that is if $b$ is a primes multiple of a primitive solution, that prime is also a solution. $x$ and $x+1$ make following sets:
$x\in\{3, 119, 4059, 137903\cdot\cdot\cdot\}$
$(x+1)\in\{4, 120, 4060, 137904\cdot\cdot\cdot\}$
