You are given positive integers N, m, and k. Is there a way to check if $$\sum_{\stackrel{p\le N}{p\text{ prime}}}p^k\equiv0\pmod m$$ faster than computing the (modular) sum?

For concreteness, you can assume $e^k<m<N.$

I don't know of a way, but it's not obvious to me that no method exists. Fast ways to prove or refute the equivalence would be of interest. You can assume that the particular instance of the problem is 'hard', that is, the modulus is not close enough to N so that Rosser-type bounds on the sum would rule it out.

With $k=0$ this is just asking if $m|\pi(N)$, so it is possible to compute the sum in time $O(N^{1/2+\varepsilon})$ using the Lagarias-Odlyzko method. (Or more practically, one of the combinatorial $\pi(x)$ methods.) For $k>0$ the sum is superlinear and so cannot be stored directly (without, e.g., modular reduction) but it's not clear whether a fast algorithm exists.

You can think of the problem as "Your friend, who has access to great computational resources, makes the claim (N, m, k). If her claim is true, can you prove it? If her claim is false, can you refute it?".

Edit: I posted a related problem on cstheory, asking if there is a short proof or interactive proof that the sum is correct.