Reminds me of Fermat's Little Theorem

It is clear that $p$ is an odd prime. let $\frac{2^{p-1}-1}{p}=n^2$ $2^{p-1}-1=pn^2$ $\left( 2^{\frac{p-1}{2}}-1\right)\left( 2^{\frac{p-1}{2}}+1\right)=pn^2$ We have two consecutive odd numbers in the LHS of last equation which are coprime and also LHS is divisible by $p$ , it follows that $p| 2^{\frac{p-1}{2}}-1$ or $p| 2^{\frac{p-1}{2}}+1$ .

Case 1) $p| 2^{\frac{p-1}{2}}-1$ and equation becomes $\left( \frac{2^{\frac{p-1}{2}}-1}{p}\right)\left( 2^{\frac{p-1}{2}}+1\right)=n^2$ Its easy to see that $\gcd \left( \frac{2^{\frac{p-1}{2}}-1}{p} , 2^{\frac{p-1}{2}}+1 \right)=1$ And now we make use of the fact that for two coprime integers $a$ and $b$ , if $ab=m^2$ then $a=k^2$ and $b=l^2$ for some integers $l$ and $k$ . So we can write $2^{\frac{p-1}{2}}+1=m^2$ $2^{\frac{p-1}{2}}=(m-1)(m+1)$ It is clear that $m$ is odd. After substituting $m=2k+1$ the only possibility for $(m-1)(m+1)=4k(k+1)$ to be a power of 2 is $k=1$ . It follows that $2^{\frac{p-1}{2}}=8$ and $p=7$ .

Case 2) $p| 2^{\frac{p-1}{2}}+1$ , with a similar procedure you can conclude that $2^{\frac{p-1}{2}}-1$ must be a perfect square. That means $2^{\frac{p-1}{2}}-1=m^2$ $2^{\frac{p-1}{2}}=m^2+1$ for $\frac{p-1}{2} \ge 2$ LHS is divisible by 4, but for RHS we have $m^2 \stackrel{4}{\equiv} (0,1)$ $m^2+1 \stackrel{4}{\equiv} (1,2)$ which is contradiction and $\frac{p-1}{2} < 2$ , so $\frac{p-1}{2} =1$ and $p=3$ which satisfies the condition of problem. Therefore the answers are $p=3,7$ and you must enter 10 to see Correct!