16th Turkish Mathematical Olympiad

Since $(q^{(p-1)/2}-1,q^{(p-1)/2}+1)=2$ for any odd integer $q$, if $p{x}^2=q^{p-1}-1$ for an integer $x$ and an odd prime $p$, then either

Case 1: $q^{(p-1)/2}-1=2p{y}^2$ and $q^{(p-1)/2}+1=2z^2$

or

Case 2: $q^{(p-1)/2}-1=2y^2$ and $q^{(p-1)/2}+1=2p{z}^2$

for some integers $y,z$.

a. Let $q=7$. Since $2$ is a quadratic residue mod $7$, but $-1$ is not; Case 2 is not possible. In Case 1, $6\mid 7^{(p-1)/2}-1=2p{y}^2\Rightarrow 3\mid p{y}^2$. When $p=3$ we have $(7^2-1)/3=4^2$. If $p\ne 3$, then $3\mid y^2\Rightarrow 9\mid 7^{(p-1)/2}-1\Rightarrow 3\mid (p-1)/2$. Let $k=(p-1)/6$. Then $2z^2=7^{3k}+1=(7^k+1)(7^{2k}-7^k+1)$ implies that $7^{2k}-7^k+1$ is a perfect square. But this is not possible as $(7^k-1{)}^2<7^{2k}-7^k+1<(7^k{)}^2$. We conclude that $p=3$ is the only prime for which $(7^{p-1}-1)/p$ is a perfect square.

b. Let $q=11$. This time, $2$ is not a quadratic residue mod $11$, and we have only Case 2 to consider. $11^{(p-1)/2}+1=2p{z}^2\Rightarrow 11^{(p-1)/2}\equiv -1(\mathrm{mod}p)\Rightarrow$

$2y^2\equiv 11^{(p-1)/2}-1\equiv -2(\mathrm{mod}p)⟹y^2\equiv -1(\mathrm{mod}p)⟹p\equiv 1(\mathrm{mod}4)$.

Therefore, $11^{(p-1)/2}-1=2y^2$ implies $11^{(p-1)/4}+1=u^2$, which is impossible as $u-1$ and $u+1$ cannot be both powers of $11$; or $11^{(p-1)/4}+1=2u^2$, which is impossible as $2$ is not a quadratic residue mod $11$. So there is no prime $p$ for which $(11^{p-1}-1)/p$ is a perfect square.