About prime

First consider the specific case of $n = 1$ . Since the sum of the first k integers is $\frac{k(k-1)}{2}$ , we are trying to find $p$ values for when $\frac{(p+1)(p+2)}{2}$ is divisible by $p-1$ , or integer solutions to $\frac{(p+1)(p+2)}{2(p-1)}$ . Now $\frac{(p+1)(p+2)}{2(p-1)}$ $=$ $\frac{1}{2}(\frac{(p^2+3x+2)}{p-1})$ $=$ $\frac{1}{2}(\frac{p^2+3x-4+6}{p-1})$ $=$ $\frac{1}{2}(\frac{(p+4)(p-1)+6}{p-1})$ $=$ $\frac{1}{2}(p+4+\frac{6}{p-1})$ , so possible $p$ values are limited to when $p-1$ are factors of $6$ , which is when $p-1$ is $1$ , $2$ , $3$ , and $6$ , or when $p = 2$ , $p = 3$ , $p = 4$ , and $p = 7$ . Since $p$ is prime, $p$ values are further limited to $p = 2$ , $p = 3$ , and $p = 7$ .

Now consider all positive integers of $n$ . One of the prime numbers that satisfies the given conditions is $p = 2$ because $p-1=1$ and $1$ divides any integer.

Another prime number that satisfies the given conditions is $p = 3$ , since an odd number to any integer exponent is odd and an even number to any integer exponent is even, the sum $1^n + 2^n + 3^n + 4^n$ is the sum of $2$ odds and $2$ evens which is even and therefore always divisible by $p-1=2$ .

Finally, $p = 7$ satisfies the given conditions, since the sum $1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n$ is the sum of $4$ odds and $4$ evens which is even, and since there are $3$ numbers that are $1$ more than a multiple of $3$ ( $1$ , $4$ , and $7$ ), and there are $3$ numbers that are $2$ more than a multiple of $3$ ( $2$ , $5$ , and $8$ ), the sum $1^n + 2^n + 3^n + 4^n + 5^n + 6^n + 7^n + 8^n$ is also divisible by $3$ , and therefore always divisible by $p-1=6$ .

Therefore, the prime numbers that satisfy the given conditions are $p = 2$ , $p = 3$ , and $p = 7$ , and $2+3+7=\boxed{12}$ .

A quick search using $Mathematica$

Intersection@@Table[Select[Prime@Range@100,Mod[Sum[n^k,{n,#+1}],#-1]==0&],{k,100}]

returns {2,3,7}

(tested first 100 primes p and first 100 integers n) and took the intersection of the results