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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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2 Answers 2

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Hint $ $ The hypotheses imply the order $\,k\,$ of $\,x\,$ is a divisor of $\,n-1\,$ but not a proper divisor, so $\, k = n-1.\, $ By here and Euler, $\,n-1\mid \phi(n)\ $ so $\ n-1\, \le\, \phi(n).\, $ This implies $\,n\,$ is prime, by$\ \phi(n) \,\le\, n-\color{red}{2}\ $ for composite $\,n,\,$ by at least $\color{red}2\,$ smaller naturals $1<n\!-\!1$ are noncoprime to $\,n$.

Optimization $ $ We can we can restrict to maximal proper divisors $\,q\,$ using the following.

Order Test $\ \,a\,$ has order $\,n\iff a^n \equiv 1\,$ but $\,a^{n/p} \not\equiv 1\,$ for every prime $\,p\mid n.\,$

Proof $\ (\Leftarrow)\ $ If $\,a\,$ has $\,\color{#c00}{{\rm order}\ k}\,$ then $\,k\mid n\,$ (proof). If $\:k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ therefore $\,k\,$ arises by deleting at least one prime $\,p\,$ from the prime factorization of $\,n,\,$ hence $\,k\mid n/p,\,$ say $\, kj = n/p,\ $ so $\ a^{n/p} \equiv (\color{#c00}{a^k})^j\equiv \color{#c00}1^j\equiv 1,\,$ contra hypothesis. $\ (\Rightarrow)\ $ Clear.

Remark $ $ This is frequently called the Lucas Primality Test.

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  • $\begingroup$ When you say order $k$ of $x$, do you mean $ord_{k}x$? $\endgroup$
    – roxrook
    Commented Mar 10, 2011 at 6:44
  • $\begingroup$ @Chan: $\rm\ mod\ n\::\ k\ $ is the order of $\rm\ x\:,\ $ so $\rm\ x^j\equiv 1\ \iff\ k\ |\ j\:$ $\endgroup$ Commented Mar 10, 2011 at 6:47
  • $\begingroup$ 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. $\endgroup$
    – roxrook
    Commented Mar 10, 2011 at 6:51
  • $\begingroup$ @Chan: What book? $\endgroup$ Commented Mar 10, 2011 at 6:54
  • $\begingroup$ I'm currently reading "Elementary Number Theory and Its application" by Kenneth H. Rosen 6th edition. Thanks. $\endgroup$
    – roxrook
    Commented Mar 10, 2011 at 7:07
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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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  • $\begingroup$ 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 $\rm\ a\neq 0$. $\endgroup$ Commented Mar 10, 2011 at 18:43
  • $\begingroup$ 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. $\endgroup$ Commented Mar 11, 2011 at 6:17

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