Baby step and giant step?
$(a)-$ Let $p$ be prime. Then $\sqrt{p}&lt;A&lt;\sqrt{p}+1 $, so $0&lt;A^2-p&lt;2\sqrt{p}+1$. This means that, prime divisor of $A^2-p$ is less than $3\sqrt{p}$.

Now try to solve $(b)$ for next $10$ points.