Baby step and giant step? When i see this problem,
 i think that it is similar to 'baby step, giant step' because of gauss symbol.

But in here, i don't know how to prove both (a) and (b).

What i have done is that $A= [p]$ or $A=[p]+1$. here $[p]$ is the lower integer version of gauss symbol.

Also, $$A=p^{1/2}$$ (when this is integer) or $$p^{1/2}&lt;A&lt;p^{1/2}+1$$ (when the square root is not an integer)
$(a)-$ 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.