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Can you give a counterexample for the following conjecture :

For $m \ge 1$ and $n>2$ , $n$ is a prime iff

$$\displaystyle\sum_{k=1}^{n^m-1} k^{n^m-1} \equiv (n-1) \cdot n^{m-1} \pmod{n^m} $$

You can run this test here .

P.S.

For $m=1$ we have Giuga's conjecture .

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    Even if this test is correct, there's no significant value over simply testing the divisibility of $n$ with all numbers smaller than $n$. There are plenty of prime formulas, but their computation complexity is not significantly lower than brute-force. For example, consider: $F_n=\left\lfloor\frac{\left(\sum\limits_{k=2}^{n-1}\left\lceil\frac{{n}\bmod{k}}{n}\right\rceil\right)+2\cdot\left\lceil\frac{n-1}{n}\right\rceil}{n}\right\rfloor$. If $F_n=0$ then $n$ is not prime, if $F_n=1$ then $n$ is prime. – 2017-01-24
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    @barakmanos You are correct...this test has no practical value . – 2017-01-24
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    Isn't this just trying to use Fermat's little theorem while guarding against Fermat pseudoprimes? – 2017-01-25
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    @Mr.Brooks Essentially yes... – 2017-01-26

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