Uniqueness of congruences proof I want to prove a statement, let $p_{1}$ and $p_{2}$ be distinct odd primes such that $(p_{1}-1)|(p_{2}-1)$. Let $m=p_{1}p_{2}$.
Prove the equation $[x]^{p_{2}}+[x]-[1]=[0]$ has a unique solution in $\mathbb{Z}_{m}$. [x] is a congruence class.
I think I can use the Chinese Remainder Theorem, but how do I prove uniqueness?
 A: Hint: the polynomial $\,f(x)\,$ is congruent to $\,2x-1\,$ both mod $p_1$ and $q_1$ by little Fermat, so by CRT it boils down to computing its root $\,x\equiv 1/2\equiv (1\!+\!n)/2\pmod{\!n}\,$ for odd $\,n = p_1 p_2$.
i.e. let  $\,p_1,p_2 = p,q.\,$ $\,f(x) = x(x^{q-1}\!-\!1)+\!2x\!-\!1\,$ so by Fermat & $\,p\!-\!1\mid q\!-\!1^{\phantom{|^|}}\!\!$ & $ $ CRT: $\bmod p\ \&\ q\!:\ 0\equiv f(x)\equiv 2x\!-\!1^{\phantom{|^|}}\!\!\!$ $\iff \bmod{ pq}\!:\ 2x\equiv 1^{\phantom{|^|}}\!\!\!\equiv 1\!+\!pq\iff x\equiv (1\!+\!pq)/2$
A: I assume $\mathbb{Z}_m=\mathbb{Z}/m\mathbb{Z}$.
By the chinese remainder theorem, $\mathbb{Z}_m=\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$.
So the problem boils down to proving that there is a unique solution to $x^{p_2}+x-1=0$ on each $\mathbb{Z}_{p_i}$.

*

*On $\mathbb{Z}_{p_2}$, note that $x^{p_2}=x$ for each $x$ (Fermat's little theorem) and solve the equation.


*On $\mathbb{Z}_{p_1}$, note that $x=0$ is not a solution. For $x\neq 0$, use the hypothesis: $p_2-1=k(p_1-1)$ for some $k$. Then
$$x^{p_2}=x^{k(p_1-1)+1}=(x^{p_1})^kx^{-k}x$$
Similarly to the previous case, $x^{p_1}=x$. So
$$x^{p_2}=x$$
(in fact, this shows that $x^{p_2}=x$ for each $x\in\mathbb{Z}_{p_1}\setminus\left\{0\right\}$, and the same equation holds trivially for $x=0$). Solve the equation again.
