Is there a notion of double roots of polynomials in the ring $\mathbb{Z}/p^n\mathbb{Z}$? By that I mean if for a polynomial $P(x)$ we have that $p^n|P(a) $ and $p^n|P'(a)$, then $(x-a)^2$ divides $P(x)$ in $\mathbb{Z}/p^n\mathbb{Z}[X]$.
To be specific, I wanted to show that if a prime power $>3$ divides $(n^2+n+1)^2$ for $6|n-1$, then it divides $(n+1)^{n+1}-n^n$.
I wanted to prove it by showing that if a prime power $p^n$ divides $n^2+n+1$, then $n$ is a double root of $(x+1)^{n+1}-x^n$ in $\mathbb{Z}/p^n\mathbb{Z}$.
I tried this because if we have what I claim, then since $6|n-1$, we have that $(x+1)^{n+1}-x^n=(x^2+x+1)Q(x)$ for some $Q(x)$ belonging to $\mathbb{Z}[X]$, and the fact that $(x-n)^2 $ cannot divide $x^2+x+1$ in $\mathbb{Z}/p^n\mathbb{Z}$, we would have that due to the double root, $p^n|Q(n)$, which would give the desired.
If $f(x)\in \Bbb{Z}[x]$ and $f(a)\equiv f'(a)\equiv 0\bmod p^k$ then let $g(x) = f(x+a)=\sum_{n=0}^N c_n x^n$ so that $c_0= f(a)\equiv 0 \bmod p^k,c_1= f'(a)\equiv 0\bmod p^k$ ie. $g(x)=x^2 h(x)+p^k j(x)$ with $j\in \Bbb{Z}[x]$ and $f(x)=(x-a)^2 h(x-a)+p^k j(x-a)$.