What primes $p$ satisfying $11 p+1$ to be a perfect square?

Question

I have tried to get prime values for which 11 p+1 is a perfect square but i didn't succeed to get a solution, I have started from :$(11p+1) \mod 2=0=m^2$ this means $11p+1 \bmod 2=0 $ implies $11 p\bmod2=1$ and this implies $p=3$ but $3$ is not a solution , any way ?

Answer 1

Following from what Barry said we have:

11=(n-1)(n+1) which means that for to be prime one of the two factors must be p and the other must be 11.

When n-1=11, then n=12 and indeed we have n+1=13=p so p is prime. If n+1=11, then n=10, but n-1=9 and 9 is not prime, so =13 is the only solution.

Answer 2

If $11p+1=n^2$, then $11p=(n+1)(n-1)$. Since the LHS has only two prime factors, $11 \text{ and } p$, the RHS must have two factors as well. Assuming $n-1=11\implies p=13$. Assuming $n+1=11 \implies p=9$ which is a contradiction. Hence only one prime, $p=13$ satisfies the condition.