<s>[INST] Finding all primes $p$ such that $\frac{(11^{p-1}-1)}{p}$ is a perfect square How to Find all primes $p$ such that $\dfrac{(11^{p-1}-1)}{p}$ is a perfect square
 [/INST] If $p=2$, $\frac{11^{p-1}-1}{p}$ is not a perfect square. In the following, we assume that $p\neq 2$.
Suppose that $\frac{11^{p-1}-1}{p}$ is a perfect square and write $p=2k+1$. 
$$11^{2k}-1 = (11^k-1)(11^k+1)= p n^2 $$
Now, as $gcd(11^k-1, 11^k+1)=2$, we have one of the following cases
$$ 11^k-1 = p a^2\text{ and }11^k +1 = b^2 $$
$$ 11^k-1 = a^2\text{ and }11^k +1 =p b^2 $$
$$ 11^k-1 = 2 p a^2\text{ and }11^k +1 = 2 b^2 $$
$$ 11^k-1 = 2 a^2\text{ and }11^k +1 =2 p b^2 $$
In the first case, $11^k=(b-1)(b+1)$ which is impossible.
In the second case, $11^k-1 = a^2$ so $a$ is even and $4$ divides $a^2$. But $a^2=pb^2-2$ so, $pb^2=2$ modulo $4$, and $b^2=2$ modulo $4$, which is impossible. 
In the third case, $2 b^2 = 1$ modulo $11$ which is impossible, as it is easy to verify that the squares modulo $11$ are 0, 1, 4, 9, 5, 3.
The fourth case remains to be examined.
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