# Which one of the option/s is/are correct? [duplicate]

Let $$p,q$$ be primes and $$n \in \Bbb N,$$ such that $$p \nmid\ n-1.$$ If $$p\ |\ n^q-1$$ then which one of the following option/s is/are correct?

$$(1)$$ $$p\ |\ q-1.$$

$$(2)$$ $$q\ |\ p-1.$$

$$(3)$$ $$p\ |\ (p-1)(q-1).$$

$$(4)$$ $$q\ |\ (q-1)(p-1).$$

My attempt $$:$$ If I take $$p=3, q=2$$ and $$n=2$$ then I find that the given condition holds but option $$(1)$$ and hence option $$(3)$$ fails to hold. What about options $$(2)$$ and $$(4)$$? It is quite clear that either both the options $$(2)$$ and $$(4)$$ are correct or both the options are false. How to prove or disprove $$(2)$$? Any help in this regard will be highly appreciated.

Thank you very much for your valuable time.

• What have you edited @janmarqz? – math maniac. Mar 30 at 15:42
• Changed the | to \mid in the first line. It makes the slash look nicer – Ross Millikan Mar 30 at 15:44
• better latex for $p$ divides not $n-1$ – janmarqz Mar 30 at 15:44
• Dub Dub dubuque @Bill Dubuque. – math maniac. Mar 30 at 17:36

If $$p$$ divides $$n^{q} - 1$$ then $$n^{q}\equiv 1$$ ($$\mathrm{mod}$$ $$p$$). So the order of $$n$$ modulo $$p$$ divides $$q$$. Hence the order is $$1$$ or $$q$$, and yet $$p$$ does not divide $$n -1$$ so the order is $$q$$.
Now apply Euler's theorem and once more use the fact that if $$n^{k} \equiv 1$$ ($$\mathrm{mod}$$ $$p$$) then the order of $$n$$ modulo $$p$$ divides $$k$$.