jmc

A solution exists if and only if $8$ is invertible modulo $p$. In other words, $\gcd (8,p)=1$. Since $8=2^3$ is a power of $2$, $8$ is invertible modulo $q$ if and only if $q$ is an odd integer. All primes except for $2$ are odd, so the number we are looking for is $\boxed{2}$.