Let $p$,$\ q$ be prime numbers and $n\in \mathbb{N}$ such that $p \nmid n-1$. If $p \mid n^q-1$, then $q \mid p-1$.

Question

Attempted Proof:

Let $S = \{k: n^k-1 \equiv 0 \pmod{p}, k\in \mathbb{N}\}$. Minimal element $r$ exists by Well-Ordering. we show $r=q$. We note, $\gcd(p,n)=1$, otherwise $p \mid n^q-1 \implies p \mid 1$.

We have:

$n \not\equiv 1 \pmod{p}\implies n^2 \not\equiv n \pmod{p}$ ; [since $p , n$ are coprime and $p$ prime ]. Now, for any $0<j< i < r,$ $n^i \equiv n^j \pmod{p} \implies n^j(n^{i-j}-1) \equiv 0 \pmod{p} \implies (n^{i-j}-1) \equiv 0 \pmod{p} $, contradicting the minimality of $r$.

Now, we proceed to show that, all elements of $S$ are integral multiples of $r$. (Claim 1)

We have, $n^{tr} \equiv 1 \pmod{p} \implies n^{tr+l} \equiv n^l \pmod{p} $, for $0\leq l <r$. Now, for any two such $l$, say $c$ and $d$, with $d<c$ , $n^{tr+c} \equiv n^{tr+d} \pmod{p} \implies n^{tr+d}(n^{c-d}-1) \equiv 0 \pmod{p}$, which again contradicts our assumption that $r$ is the least element of $S$. Hence, Claim 1 is established. But, it implies that $q$ is composite (an impossibility); hence $r$ must be equal to $q$ .

Finally, by Fermat's Little Theorem, we have $n^{p-1} \equiv 1 \pmod{p}$ [ it holds, since $\gcd (p, n) =1$].

Therefore, $p-1$ must be an integral multiple of $q$, i.e. $q \mid p-1$.

PS: I am aware of the duplicates, but I want my solution checked. Thank you.

Answer 1

You appear to be reinventing the wheel (here $\rm ord$ = order of an element). Using this it's a one-liner

$\!\bmod p\!:\ n^{\large q}\equiv 1\,\Rightarrow\, {\rm ord}\, n\mid q\,$ prime, so $\, {\rm ord}\,n = q\,$ by $\,n\not\equiv 1$, so $\ n^{\large p-1}\equiv 1\,$ implies $\, q\mid p\!-\!1$

Said more structurally: $S$ is a nontrivial set of integers closed under subtraction, so a one-line Euclidean descent proof shows its least positive element $\,r := {\rm ord}\, n\,$ divides every element of $S$.