# How to prove Lucas's Converse of Fermat's Little Theorem without using primitive root?

Problem:
If $x^{n-1} \equiv 1 \pmod{n}$, and for all divisors $q$ of $n - 1$, $a^{q} \not\equiv 1 \pmod{n}$, then $n$ is prime. $(n > 1)$

I read the proof in the book and it was very straightforward; however, I wonder is there a way to prove it by just using congruence property?

And another related question about power residue:
If we have $a^{n - 1} \equiv 1 \pmod{n}$. Is there any relation between $n - 1$ and $\phi(n)$? Because I thought of $a^{\phi(n)} \equiv 1 \pmod{n}$, when $(a, n) = 1$. I try to think of away to connect these two ideas, but I still cannot see it.
Any idea?

Thanks,

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HINT $\ \$ The conditions imply that the order $\rm\:k\:$ of $\rm\:x\:$ is a divisor of $\rm\:n-1\:$ but not a proper divisor, therefore $\rm\: k = n-1\:.\$ By Euler, $\rm\ k\ |\ \phi(n)\$ so $\rm\ n-1\: \le\: \phi(n)\:.\$ This implies that $\rm\:n\:$ is prime, since $\rm\ \phi(n) \:\le\: n-\color{red}{2}\$ for composite $\rm\:n\:,\:$ since they have at least $\:\color{red}2\$ smaller naturals non-coprime to $\rm\:n\:.$

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When you say order $k$ of $x$, do you mean $ord_{k}x$? – Chan Mar 10 '11 at 6:44
@Chan: $\rm\ mod\ n\::\ k\$ is the order of $\rm\ x\:,\$ so $\rm\ x^j\equiv 1\ \iff\ k\ |\ j\:$ – Bill Dubuque Mar 10 '11 at 6:47
Thanks, sorry my mathematics vocabulary are very limited. I tried to understand your hint, but I was confused because the argument that you used is very similar to the way the book proved using primitive root. – Chan Mar 10 '11 at 6:51
@Chan: What book? – Bill Dubuque Mar 10 '11 at 6:54
I'm currently reading "Elementary Number Theory and Its application" by Kenneth H. Rosen 6th edition. Thanks. – Chan Mar 10 '11 at 7:07

Do you understand the definition of $\phi(n)$? It's the number of natural numbers less than $n$ which are relatively prime to $n$. If $n$ is prime, then $\phi(n)$ is necessarily $n-1$ (since only 1 is relatively prime with $n$). Likewise, if $\phi(n) = n-1$ then only one number less than $n$ is relatively prime to it, which must be the number 1, so $n$ is prime. So that would be the relationship between them.

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The above definition of $\rm\:\phi(n)\:$ is incorrect. $\rm\:\phi(n)\:$ is the number of units (invertibles) modulo $\rm\:n\:$ or, equivalently, the number of naturals smaller than $\rm\:n\:$ that are coprime to $\rm\:n\:.\:$ Also it is not true that $\rm\ a^{\phi(n)}\equiv 1\ (mod\ n)\$ for all $\rm\ a\neq 0$. – Bill Dubuque Mar 10 '11 at 18:43
Yeah, I meant to say "less than n which are relatively prime to n", I just had a little brain-fart and typed "divides" instead. And you're also correct about the second mistake. Clearly I wasn't at my best. I'll edit the answer substantially. – Keith Irwin Mar 11 '11 at 6:17