prove/disprove: for all odd $n \geq 3$ we have $\sum_{k=1}^{n-1} p_k(n-k) \not \equiv 0\mod n$

Question

Assume $p_k$ is the $k$th prime. I really don't know where to start except the fact I know that the numbers modulo $n$ form a group with the addition operator. All I know about the primes is multiplication related. Is there any theorem that I am missing out here?

By the way, I know this is true for all even $n \geq 2$ but I can't prove/disprove for odd $n$'s.

Thanks for any help.

Answer 1

For $n=21$, we have $$\sum_{k=1}^{n-1}p_k(n-k)=4200 = 21\cdot 200 $$ and for $n=98$, we have $$\sum_{k=1}^{n-1}p_k(n-k)=687078 = 98\cdot 7011. $$ The sequence of $n$ for which $n\mid \sum_{k=1}^{n-1}p_k(n-k)$, begins $$1,2, 21, 31, 39, 98, 343, 889, 891, 1957, 3250, 4493, 9182, 19587, 97017, 112339, 249617, 259895, 497901, 960730, \ldots $$ (and is not yet known to the OEIS)