Post Number:#2 by ahmedittihad » Tue Jul 04, 2017 11:53 pm
$-a \equiv b^2 (moda+b^2) $.
So, $-a^3 \equiv b^6 (moda+b^2) $.
Which implies that $a+b^2 | b^6-b^3$.
Now fix $b$ and notice that for every divisor $x $ of $b^6-b^3$ there exists a $a $ such that $a+b^2=x $. So we get infinite solutions.
Frankly, my dear, I don't give a damn.