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Fermat's little theorem says:

$a^{p-1} \equiv 1 \pmod p$

I have a equation which is similar to this but it's not covered by Fermat's theorem nor by Euler's theorem.

My equation is:

$2^{(p-1)/3} \equiv 1 \pmod p$

Of course this has to be true:

$1\equiv p\pmod 3 $

So is it possible to say when my equation is true? Which primes apply to my equation?

For the prime numbers 31, 43, ... the equation is correct.

For the prime numbers 7, 13, 19, 37, ... the equation is incorrect.

It looks randomly for me, which prime numbers apply.

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    $\begingroup$ $2^{(p-1)/3}\equiv 1\pmod p$ iff $2$ is a cubic residue, which happens iff $p$ is of the form $x^2+27y^2$. Look up cubic reciprocity. $\endgroup$
    – Wojowu
    Commented Jun 4, 2020 at 13:15
  • $\begingroup$ That's great! Maybe you can also help me with n > 4 here $\endgroup$
    – zomega
    Commented Jun 4, 2020 at 17:01

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