If $q|\frac{x^5-2^5}{x-2}$ then $q=1 \mod5$

Question

Prove if $q$ is prime and in the form of $4k-1$ and $q\mid\frac{x^5-2^5}{x-2} $ then $ q\equiv1 \mod 5$

Answer 1

Clearly, $x\not\equiv2\pmod q\ \ \ \ (1)$

Let $x\equiv2y\pmod q\implies y^5\equiv1\pmod q$ for odd prime $q$

$\implies(5,q-1)\mid$ord$_qy$

If $5\nmid(q-1),(5,q-1)=1\implies1\mid$ord$_qy\implies$ord$_qy=1\implies y^1\equiv1\pmod q$

$\implies x\equiv2q\equiv2\pmod q$ which contradicts $(1)$

Answer 2

Clearly, $q\mid x^5-2^5$, that is, $x^5\equiv 2^5\pmod{q}$. Now, observe that, if $x\equiv 2\pmod{5}$, then $q\mid 80$, which is not possible as $q\equiv -1\pmod{4}$. Hence, $x\not\equiv 2\pmod{5}$. This means, if we let $a\equiv x2^{-1}\pmod{q}$, then $a^5\equiv 1\pmod{q}$, and $a\not\equiv 1\pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^d\equiv 1\pmod{q}$.

It is well known that $d\mid 5$ and $d\nmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}\equiv 1\pmod{q}$, hence, $5=d\mid q-1$, which is the claim.