# If $n \geq 1$ is not prime and $x \in\mathbb{Z}_n$ such that $\gcd(x, n) \neq 1$, prove that $x^{n-1} \bmod n \not\equiv 1$.

If $n \geq 1$ is not prime and $x \in\mathbb{Z}_n$ such that $\gcd(x, n) \neq 1$, prove that $x^{n-1} \bmod n \not\equiv 1$.

I am not sure why this would be true. So, letting $n$ be a nonprime and $x$ being in $\mathbb{Z}_n$, and with $\gcd(x, n)\neq 1$, both $x$ and $n$ share some common factor. I need to relate this to $x^{n-1}$ having a non-$1$ remainder when divided by $n$, but I don't see any apparent connection that makes this so.

I would really appreciate guidance or clarity, but would not like the problem fully solved for me.

• To get subscripts, use the underline, so a_n gives $a_n$. To get multicharacter exponents, enclose them in braces, so x^{(n-1)} gives $x^{(n-1)}$. This works everywhere you need to treat a number of characters as one. Finally, for mod n! use \pmod {n!} and for gcd use \gcd to get $\pmod {n!}$ and $\gcd$ – Ross Millikan Mar 5 '13 at 1:53
• Welcome to math.SE. You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Mar 5 '13 at 1:53
• Thank you both for the help and info on formatting! – samxx Mar 5 '13 at 2:02

Hint $\rm\,\ d\mid x,\,\ d\mid n\mid x^{n-1}\!-\!1\:\Rightarrow\:d\mid 1$
Let $d$ be the common factor. Show $d$ is also a factor of $x^{n-1}$. Deduce that $x^{n-1}\not\equiv1\pmod n$.