Never prime!

If n is odd and p^2 odd prime then,p^2+n will never be a prime.Hence,we just need 4+n not to be a prime with odd n which gives n=5. Now,for smaller even numbers 2 and 4 we have $7^2+2=51=3*17$ and $11^2+4=125=5^3$ .

Hence,the answer is $\boxed{5}$

Let $q(p,n) = p^2 +n$ . Let us start with the smallest $n=1$ , then $q(2,1) = 2^2+1 = 5$ , which is a prime. Therefore, $n \ne 1$ . We note that $q(3,2) = 3^2 + 2 = 11$ , a prime, then $n \ne 2$ . $q(2,3) = 7$ , a prime, $n \ne 3$ . $q(3,4) = 13$ , a prime, $n \ne 4$ , When $n=5$ , $q(2,5) = 9$ , not a prime, and for $p>2$ , $p$ is odd, so is $p^2$ and $q(p,5) = p^2+ 5$ is even and larger than 2 and hence not a prime. Therefore, the smallest $n$ is $\boxed{5}$ for all $p^2 + n$ to be never a prime.