prime numbers - If $qmid 2^p + 3^p$ then $q gt p$

Let $p, q$ positive prime numbers, $q > 5$. Prove that if
$q \mid \left(2^{p} + 3^{p}\right)$ then $q > p$.

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First, it's clear that $p \ne q$ because, using Fermat's little theorem, $2^p = 2 \mod p$ and $3^p = 3 \mod p$, therefore $5=0 \mod p$ false because $q \gt 5$.

Now suppose $q \lt p$. I tried to analyse $2^p +3^p \mod q$ but I didn't get a contradiction.

Answer

HINT:

For $p>2$

$$\left(-\dfrac32\right)^p\equiv1\pmod q$$

$\implies$ord$_q(-3/2)|(p,q-1)$

But As $p$ is prime, ord$_q(-3/2)=1$ or $p$

ord$_q(-3/2)=1\implies-3\equiv2\pmod q\iff5\equiv0\iff q|5\implies q=5$

Else $p|(q-1)$