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
 A: Let $\frac{a^{p-1}-1}{p}=b^2$ for some integer b.
If p=2, $a=2b^2+1$(11 is not of the form).
If p>2 & prime ,so must be odd=(2k+1) (say)
$(a^k+1)(a^k-1)=b^2(2k+1)$
Now, if a is odd (like 11), $a^k±1$ is even => b is even =2d(say).
$\frac{a^k+1}{2}\frac{a^k-1}{2}=d^2(2k+1)$
Now, $(\frac{a^k+1}{2},\frac{a^k-1}{2})$=1
So, either $\frac{a^k+1}{2}=m^2p$   and  $\frac{a^k-1}{2}=n^2$ or vice versa.
Now, $n^2$≡0,1,4,9,5,3(mod 11) =>$11∤(2n^2-1)$
In other case,
If $11^k=2n^2+1$ 
n must be even=2m (say)=>  $2n^2+1≡1(mod\ 8)$.
Now, 11≡3(mod 8) =>$11^2≡1(mod\ 8)$=>$11^{2r}≡1(mod\ 8)$  and $11^{2r+1}≡3(mod\ 8)$ 
So, k must even=2r(say).
The part problem becomes $(11^r)^2-2n^2=1$ which is well known Pell equation. 
The minimum solution in natural number for $A^2-2B^2=1$ is (3,2).
So, A is $\sum_{s≥2t≥0}sC_{2t}3^{s-2t}(2\sqrt2)^{2t}$ where s is a natural number, can this be a power of 11? 
Find here
An observation:
$a^k≡-1(mod\ p)=>a^{2k}≡1(mod\ p) $
Let $d=ord_pa$ => d|2k,but d∤k and d∤2 as k is even
=> a(=11) must be a primitive root of p to admit any solution.
A: 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.
A: In the solution proposed by saposcat, the fourth case remained open. Here is the continuation of the solution in that case.
Let we have $11^k−1=2a^2$ and $ 11^k+1=2pb^2.$ 
If $3\not|a$, then $11^k≡0(mod \ 3)$, which is not possible, so $3|a$. Therefore $11^k≡1(mod\ 3)$, that's why k should be even. Let's suppose $k=2l$. Then we have $11^{2l}-1 = 2a^2$, so $(11^l-1)(11^l+1) = 2a^2$. We get that $11^l-1 = 2m^2$ and $11^l+1 = n^2$ or $11^l-1 = m^2$ and $11^l+1 = 2n^2$, for both cases saposcat got contradictions (like 1st and 3rd cases).
