Let $n \geqslant 3$ be an odd integer. Determine all triplets $(A, B, C)$ of real polynomials such that

$$
A^{n}+B^{n}+C^{n}=0
$$
We notice that $C^{n}$ divides $A^{n}+B^{n}$, so by applying the lemma we find that $C^{n-1}$ divides $A^{\prime} A^{n-1}+B^{\prime} B^{n-1}$. We deduce by subtraction that $C^{n-1}$ divides $A^{n-1}\left(A^{\prime} B-B^{\prime} A\right)$.

Now that we have this key relation, we choose $C$ properly to have an argument on the degree:

We start by noting that if one of the polynomials is zero, say $A$, then we necessarily have $C=-B$. We thus obtain a family of solutions of the form $(0, B,-B)$ and its permutations (we check that they work). In the following, we assume that the polynomials are non-zero.

First, if $D$ divides $A$ and $B$, then $D$ divides $C$. Thus, up to dividing, we can assume that $A, B$ and $C$ are pairwise coprime. We then assume by symmetry that $\operatorname{deg} C \geqslant \operatorname{deg} A, \operatorname{deg} B$ and that $C$ is non-constant. Since $C^{n-1}$ divides $A^{n-1}\left(A^{\prime} B-B^{\prime} A\right)$ and $C$ is coprime with $A$, we have that $C^{n-1}$ divides $A^{\prime} B-B^{\prime} A$. If this term is not zero, we then have

$$
(n-1) \operatorname{deg}(C) \leqslant \operatorname{deg}\left(A^{\prime} B-B^{\prime} A\right) \leqslant \operatorname{deg}(A)+\operatorname{deg}(B)-1<2 \operatorname{deg}(C)
$$

which is absurd if $C$ is non-constant.

In the case where $C$ is non-constant, we deduce that $A^{\prime} B-A B^{\prime}=0$, which can be rewritten as $\left(\frac{A}{B}\right)^{\prime}=0$ or $A=c B$ for some constant $c$. Since the polynomials are assumed to be pairwise coprime, we deduce that $A$ and $B$ are constants. But then, by substitution, we have $C^{n}$ which is constant, so $C$ is constant (which contradicts what we assumed).

Finally, $C$ is constant and since it was of maximal degree, $A$ and $B$ are also constant. The triplet $(A, B, C)$ is of the form $(a, b, c)$ with $a^{n}+b^{n}+c^{n}=0$. By retracing, we find that the triplets are of the form $(a D, b D, c D)$ with $a^{n}+b^{n}+c^{n}=0$ and $D$ any real polynomial.

See how we combined differentiation with an argument on the degree.

The lemma generalizes the following notion: $a$ is a root of multiplicity $k$ of $P$ if and only if $P(a)=$ $P^{\prime}(a)=\ldots P^{(k-1)}(a)=0$. Therefore, $P^{\prime}$ contains information about the roots of $P!$ We can understand this through the following example: