If $q\mid 2^p + 3^p$ then $q \gt p$

Question

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

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.

Are you familiar with the notion of "order"? (as in Lab's proof) If not, I can give a more elementary proof. — Bill Dubuque, Jul 16 '16 at 16:13
Great, then using the order as in Lab's answer is probably the simplest way to proceed. See here for a generalization. — Bill Dubuque, Jul 16 '16 at 16:47

score 5 · Accepted Answer · answered Jul 16 '16 at 15:29

5

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 $p|(q-1)$

answered Jul 16 '16 at 15:29

lab bhattacharjee

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In fact, one can from here prove the stronger statement that $q > 2p$. – Plato Aug 25 '18 at 18:41

If $q\mid 2^p + 3^p$ then $q \gt p$

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