# Let $p$ be a prime. Prove that $x^m \bmod p = 1$ has at most $m$ solutions for $x \in U(p)$ using only group theory (no polynomials allowed)

This is part of a homework problem I am working on. In light of this, please do not post an entire solution. I've been told that we do not have to use anything but Fermat's little theorem and a bit of group theory to prove this result. And I'm honestly out of ideas on how to do it. Every resource I've found seems to suggest that polynomial roots and field theory is necessary.

• What is $U(p)$? $\mathbb Z/p\mathbb Z$ or the group of $p^{\mathrm{th}}$ root of unity in $\mathbb C$? – Can I play with Mathness Nov 9 '18 at 15:22
• It is $(Z/pZ)^*$ – farleyknight Nov 9 '18 at 15:23

We can try to "hide" the polynomial part, for example: look at the Vandermonde determinant (the polynomial is hidden here) $$\det\begin{bmatrix}1&x_1&x_1^2&\dots&x_1^m\\1&x_2&x_2^2&\dots&x_2^m\\1&x_3&x_3^2&\dots&x_3^m\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&x_{m+1}&x_{m+1}^2&\dots&x_{m+1}^m\end{bmatrix}=\prod_{1\leq i If $$x_1,\dots,x_{m+1}$$ are solutions to $$x^m\equiv 1\pmod{p}$$, then the matrix has two identical columns (mod $$p$$) so LHS is $$\equiv 0\pmod{p}$$. So $$p$$ must divide one of the factor on the RHS.
Rule Number 1. $$(\mathbb Z/p\mathbb Z)^\times$$ is cyclic, generated by some element $$a$$ of order $$p-1$$. So whatever solution you are after, it must be a power of $$a$$.
• Can you prove that $(Z/pZ)^*$ is cyclic without using polynomials? – farleyknight Nov 9 '18 at 15:22