# $f^n+g^n=h^2\implies f,g,h$ 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!

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 $$n<4$$.

• Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks. Dec 15, 2018 at 22:47

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 $$n\geq4$$.