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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 $$p^{1/2}<A<p^{1/2}+1$$ (when the square root is not an integer)

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$(a)-$ Let $p$ be prime. Then $\sqrt{p}<A<\sqrt{p}+1 $, so $0<A^2-p<2\sqrt{p}+1$. This means that, prime divisor of $A^2-p$ is less than $3\sqrt{p}$.

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

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  • $\begingroup$ Can you tell me about (b)? I tried but it is quite complicated to me. $\endgroup$ – nien Dec 12 '16 at 5:31

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