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

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.
