TamreF

$\frac{2^{p-1}-1}{p}=n^2 \ \ \Rightarrow \ \ \left( 2^{\frac{p-1}{2}}-1\right)\left( 2^{\frac{p-1}{2}}+1\right)=pn^2$ We have two consecutive odd numbers in the LHS, 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 two factors in the RHS are coprime 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 $2^{\frac{p-1}{2}}+1=m^2 \ \ \Rightarrow \ \ 2^{\frac{p-1}{2}}=(m-1)(m+1)$ 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 $2^{\frac{p-1}{2}}=m^2+1$ Note that $m$ is odd and for $\frac{p-1}{2} \ge 2$ LHS is divisible by $4$ , but for RHS we have $m^2 \stackrel{4}{\equiv} 1$ and $m^2+1 \stackrel{4}{\equiv}2$ , which is contradiction and $\frac{p-1}{2} < 2$ , so $\frac{p-1}{2} =1$ and $p=3$ . See that both $p=3, 7$ indeed satisfy the condition of problem. Therefore the answers are $p=3, 7$ and you must enter 10 to see Correct!