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!


2 Answers 2


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 $n<4$.

  • $\begingroup$ 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. $\endgroup$
    – Algebear
    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 $n\geq4$.


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