Polynomial Version for Fermat Last Theorem over $\mathbb{Z}_p$ 
How to prove the polynomial version for Fermat Last Theorem over $\mathbb{Z}_p$, that is, prove there exists no non-constant coprime solution $f(x), g(x), h(x) \in \mathbb{Z}_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n.$$ 

I have already proven the polynomial version for Fermat Last Theorem in $\mathbb{C}[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance. 
 A: Let's derive it as a corollary to the Mason-Stothers theorem, which states

Let $F$ a field and $f, g, h \in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then
  $$\max\{\deg(f),\deg(g),\deg(h) \}\leq \deg\big(\text{rad} (fgh)\big)-1 $$
  where $\text{rad}(f) \equiv \prod\limits_{\text{irred } p \mid f } p$

In fields of characteristic $0$, like $\mathbb{C}$, having nonvanishing derivative is equivalent to being nonconstant.  But in fields of characteristic $p > 0$, this is not the case.  However, things aren't very much more complicated, for our purposes:  if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] \rightarrow F[x]$ is precisely $F[x^p]$.  We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.  
Now suppose we are in a field $F$ of characteristic $p > 0$.  If we have $f^n + g^n = h^n$, $\gcd(n,p) = 1$, then $(f^n)' = nf'f^{n-1} =0$ iff $f' = 0$ iff $f \in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.
We thus observe that 

There exists $m \in N$ such that, setting $\bar{f} = f(x^{1/p^m}), \bar{g} = g(x^{1/p^m}), \bar{h} = h(x^{1/p^m})$, we have that $\bar{f}, \bar{g}, \bar{h} \in F[x]$ are coprime, $\bar{f}^n + \bar{g}^n = \bar{h}^n$, and one of $\bar{f}, \bar{g}, \bar{h}$ has nonvanishing derivative.  

Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h \in F[x^p]$, and setting $\bar{f} = f(x^{1/p}), \bar{g} = g(x^{1/p}), \bar{h} = h(x^{1/p})$ (which are certainly in $F[x]$) we get a relation $\bar{f}^n + \bar{g}^n = \bar{h}^n$.  If all of $\bar{f}, \bar{g}, \bar{h}$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.  
The above reduction gives us a relation
$$\bar{f}^n + \bar{g}^n = \bar{h}^n$$
which satisfies all the conditions of Mason-Stothers as stated above.  The argument is now the same as in the case of $F$ having characteristic $0$.  The bounds $$\deg\big(\text{rad}(\bar{f}^n \bar{g}^n \bar{h}^n)\big) - 1 < \deg(\bar{f}) + \deg(\bar{g}) + \deg(\bar{h})$$ and 
$$\max\{ \deg( \bar{f}^n),  \deg( \bar{g}^n),  \deg( \bar{h}^n ) \} \geq \frac{n}{3}\big(\deg( \bar{f}) +  \deg( \bar{g}) +  \deg( \bar{h} ) \big) $$ 
are immediate.  When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.  
