$f^n+g^n=h^2\implies f,g,h$ all constant Let $k$ be a field with char$(k)=0$ and suppose for $f,g,h\in k[x]$ having gcd$(f,g,h)=1$ and $n\in\mathbb{N}_{\geqslant4}$ it holds that $f^n+g^n=h^2$. I want to show that $f,g,h$ are all constant. Of course, we go the Mason Stothers-way. Quick note: by gcd $1$ and the equation, we have that $f,g,h$ are pairwise co-prime.
For the sake of contradiction, suppose that $f,g,h$ are not all constant. By the other conditions listed above, we can apply Mason-Stothers like so:
$$\max(\deg f^n,\deg g^n,\deg h^2)\leqslant\deg(\text{rad}(fgh))-1.$$
Now let's look at two cases:
1. $f^n$ has maximal degree (and is non-constant).
2. $h^2$ has maximal degree.
A contradiction should appear in both cases. Let's begin with 1.:
Since $f^n$ has max degree and the polynomials are pairwise co-prime, we can say $$\begin{align*}4\leqslant\deg(f^n)=n\deg(f)&<\deg({\text{rad}(fgh)})\\
&= \deg(\text{rad}(f)\text{rad}(g)\text{rad}(h))\\
&=\deg(\text{rad}(f))+\deg(\text{rad}(g))+\deg(\text{rad}(h))\\
&\leqslant3\deg(f).
\end{align*}$$
 Now since $n>3$, this is an immediate contradiction.
Thanks for all the fast help!
 A: We don't know anything useful about $\deg(\text{rad}(f))$, but we do know that it's at most $\deg(f)$. 
Similarly, $\deg(\text{rad}(g)) \le \deg(g)$, and since $f^n$ has maximal degree, we know $\deg(g) \le \deg(f)$.
Finally, $\deg(\text{rad}(h)) \le \deg(h)$. We know that $\deg(h^2) \le \deg (f^n)$, so $\deg(h) \le \frac n2 \deg(h)$.
(Be careful: we don't know $\deg(h) \le \deg(f)$ just from $\deg(h^2) \le \deg(f^n)$.)
In the end, we get $$n \deg(f) < \deg(f) + \deg(f) + \frac n2 \deg(f)$$ from which we have $n < 4$.
The second case is similar, except instead we have $\deg(f^n) \le \deg(h^2)$, so $\deg (f) \le \frac 2n \deg (h)$. The inequality we get is $2\deg(h) < \frac2n \deg(h) + \frac2n \deg(h) + \deg(h)$ which also implies $n<4$.
A: Since $f^n$ has maximal degree, it follows that $\deg(g)\leq\deg(f)$.  Moreover, 
$$
2\deg(h)\leq n\deg(f).
$$
Since $\operatorname{rad}$ only lowers the degree, the RHS is at most
$$
\left(2+\frac n2\right)\deg(f),
$$
Which is at most $n\deg(f)$ since $n\geq4$.
