Note that the remainder when $n^{2}$ is divided by 5 must be 0,1 , or 4 . Then we have that $25 p=n^{2}$ or $25 p=n^{2}-1$ or $25 p=n^{2}-4$. In the first case there are no solutions. In the second case, if $25 p=(n-1)(n+1)$, then we must have $n-1=25$ or $n+1=25$ as $n-1$ and $n+1$ cannot both be divisible by 5 , and also cannot both have a factor besides 25 . Similarly, in the third case, $25 p=(n-2)(n+2)$, so we must have $n-2=25$ or $n+2=25$. Therefore the $n$ we have to check are $23,24,26,27$. These give values of $p=21, p=23, p=27$, and $p=29$, of which only 23 and 29 are prime, so the answer is $23+29=52$.