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Let p be prime. For $n|p-1$, let $f_n(x)=(x-a_1)(x-a_2)\cdots(x-a_s)$ where $a_1$ through $a_s$ belong to the exponent $n \pmod{p}$ and $s=\phi(n)$. For sufficiently large x, $f_n(x)$ is positive. Prove that $f_n(x)=\prod_{m|n} (x^{n/m} - 1)^{\mu(m)} \pmod{p}$

I did prove (using logs) that if we let $f(n)$ be a positive function over the natural numbers and let $g(n) = \prod_{m|n} f(m)$,

then $f(n)=\prod_{m|n} g(n/m)^{\mu(m)}.$

So, if I can prove $x^n-1=\prod_{m|n} f_m(x) \pmod{p}$, then I can use what I have already proven to finish the proof. I'm not sure where to go from here, though.

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The set of solutions of $x^n-1$ are all elements of order $d$ with $d|n$.

If $f_d(x)$ is just the elements of order $d$ then $f_n(x) = \frac{x^n-1}{\prod_{d|n, d \not = n}f_d(x)}$.

Rearranging to get the formula which you can apply Möbius inversion to. By the way, you don't have to use logarithms because this is just the normal Möbius inversion written in multiplicative notation instead of additive.

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Check Serre's book "A course in arithemetic". This is like in the second page. You will learn a lot by reading that book.