. Find all polynomials $P(x)$ with real coefficients that satisfy the equality $P(a-$ $b)+P(b-c)+P(c-a)=2 P(a+b+c)$ for all real numbers $a, b, c$ such that $a b+b c+c a=0$.
. The solutions are polynomials of the form $P(x)=\alpha x^{2}+\beta x^{4}$.

First, we verify that $P(x)=x^{2}$ and $P(x)=x^{4}$ are valid. Suppose $a b+b c+c a=0$. Consider $P(x)=x^{2}$. By direct expansion, using $a b+b c+c a=0$, we obtain

$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(a+b+c)^{2}
$$

For $P(x)=x^{4}$, we justify the two equalities

$$
4(a+b+c)^{4}=\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)^{2}=2\left((a-b)^{4}+(b-c)^{4}+(c-a)^{4}\right)
$$

The first is obtained by squaring the case of $x^{2}$; the second comes from the relation $\left(u^{2}+v^{2}+w^{2}\right)^{2}=2\left(u^{4}+v^{4}+w^{4}\right)$ for $u+v+w=0$ (applied to $u=a-b, v=b-c, w=c-a$), which is proven by applying a classical identity to the difference of the two sides:

$-u^{4}-v^{4}-w^{4}+2 u^{2} v^{2}+2 v^{2} w^{2}+2 w^{2} u^{2}=(u+v+w)(-u+v+w)(u-v+w)(u+v-w)=0$.

We deduce that all polynomials of the form $P(x)=\alpha x^{2}+\beta x^{4}$ are solutions to the problem. We will show that these are the only ones.

First, consider a monomial $P(x)=x^{k}$ satisfying the condition in the statement. By setting $(a, b, c)=(0,0,0)$, we see that $k \neq 0$; with $(a, b, c)=(1,0,0)$, we see that $k$ is even. Finally, with $(a, b, c)=(6,3,-2)$, setting $u_{k}=3^{k}+5^{k}+(-8)^{k}-2 \cdot 7^{k}$, we get $u_{k}=0$. If $k$ is an even integer greater than or equal to 6, we observe that $u_{k}>8^{k}-2 \cdot 7^{k}>0$ because

$$
\left(\frac{8}{7}\right)^{k} \geqslant\left(\frac{8}{7}\right)^{6}=\frac{64^{3}}{49^{3}}>\frac{262144}{125000}>2
$$

Thus, the monomial solutions are $x^{2}$ and $x^{4}$.

Now, let $P(x)=\sum p_{k} x^{k}$ be a solution to the functional equation. Let $a, b, c$ be real numbers such that $a b+b c+c a=0$. For any real number $t$, the triplet $(t a, t b, t c)$ satisfies the same relation, and we have $P(t(a-b))+P(t(b-c))+P(t(c-a))=2 P(t(a+b+c))$ and thus

$$
\sum p_{k}\left((a-b)^{k}+(b-c)^{k}+(c-a)^{k}\right) t^{k}=2 \sum p_{k}(a+b+c)^{k} t^{k}
$$

Since this is true for all real $t$, the coefficients of $t^{k}$ are equal, and we have, for all $k$,

$$
p_{k}\left((a-b)^{k}+(b-c)^{k}+(c-a)^{k}\right)=2 p_{k}(a+b+c)^{k}
$$

We are thus reduced to the case of the monomial $x^{k}$, which implies that $p_{k}=0$ for all $k$ different from 2 and 4, and $P$ is indeed of the form $\alpha x^{2}+\beta x^{4}$.

## Second Solution.

With $b=c=0$, we see that $P$ is even. Let $Q$ be the polynomial such that $P(x)=Q\left(x^{2}\right)$. The relation becomes

$$
Q\left((a-b)^{2}\right)+Q\left((b-c)^{2}\right)+Q\left((c-a)^{2}\right)=2 Q\left((a+b+c)^{2}\right)=2 Q\left(a^{2}+b^{2}+c^{2}\right)
$$

If we set $x=a-b, y=b-c$, we get

$$
Q\left(x^{2}\right)+Q\left(y^{2}\right)+Q\left(x^{2}+2 x y+y^{2}\right)=2 Q\left(x^{2}+x y+y^{2}\right)
$$

It is essential to note that this relation is satisfied for any pair of real numbers $(x, y)$. Indeed, if $x, y$ are any real numbers, finding real numbers $a, b, c$ such that $a b+b c+c a=0$ and $x=a-b, y=b-c$ amounts to solving the quadratic equation in $b$

$$
(b+x)(b-y)+b(b+x)+b(b-y)=0
$$

which is $3 b^{2}+2(x-y) b-x y=0$, with discriminant $4(x-y)^{2}+12 x y=4\left(x^{2}+x y+y^{2}\right)=(2 x+y)^{2}+3 y^{2}$, which is positive.

At this point, we have established that the functional equation in the statement is equivalent to the functional equation (IV.3) (for any real $(x, y)$) after the change of the unknown function given by $Q\left(x^{2}\right)=P(x)$. If we replace $y$ by $-y$ in (IV.3), we get

$$
Q\left(x^{2}\right)+Q\left(y^{2}\right)+Q\left(x^{2}-2 x y+y^{2}\right)=2 Q\left(x^{2}-x y+y^{2}\right)
$$

Subtracting the relations (IV.3) and (IV.4), and setting $p=x^{2}+y^{2}$ and $q=x y$, we get

$$
Q(p+2 q)-Q(p-2 q)=2 Q(p+q)-2 Q(p-q)
$$

for any pair of real numbers $(p, q)$ such that $2|q| \leqslant p$ (which is a necessary and sufficient condition for the existence of real numbers $x, y$ such that $p=x^{2}+y^{2}, q=x y$). Let $p>0$. Differentiating the relation (IV.5) three times with respect to $q$ in the interval $[-p / 2, p / 2]$, we get

$$
8 Q^{(3)}(p+2 q)+8 Q^{(3)}(p-2 q)=2 Q^{(3)}(p+q)+2 Q^{(3)}(p-q)
$$

With $q=0$, we get $Q^{(3)}(p)=0$ for all $p>0$; this implies that $Q$ is of degree 2. Since $Q(0)=0$ (with $x=y=0$), we get the necessary condition $Q(x)=\alpha x+\beta x^{2}$ and thus $P(x)=\alpha x^{2}+\beta x^{4}$.

Conversely, we see that these polynomials are valid by verifying that $Q(x)=x$ and $Q(x)=x^{2}$ satisfy (IV.3), which is an immediate calculation.

Remarks. Other computational solutions similar to the first are possible but offer little interest. One can even replace $c$ by $-\frac{a b}{a+b}$ and identify some of the coefficients of the $a^{i} b^{j}$ that appear after clearing the denominators.

The second solution allows generalizing the exercise to the case where we only assume that $P$ is a continuous function from $\mathbb{R}$ to $\mathbb{R}$ satisfying the functional equation. Having noted that $P$ is even, we set $Q(x)=P(\sqrt{x})$ for $x \geqslant 0$. The function $Q$ is continuous on $\mathbb{R}_{+}$ and satisfies, for positive $u, v$

$$
Q(u)-2 Q(u+v)+2 Q(u+3 v)-Q(u+4 v)=0
$$

(in the relation (IV.5), set $u=p-2 q, v=q$). A known argument then shows that $Q$ is $C^{\infty}$ on $\mathbb{R}_{+}$ (for fixed $u$, integrating this last relation from $v=0$ to $v=1$ yields a relation between $Q$ and one of its primitives, which implies that $Q$ is $C^{1}$ and, by induction, of class $C^{k}$ for all $k$). It then suffices to resume the differentiation method of the second solution to conclude that $Q^{(3)}=0$ on $\mathbb{R}_{+}$ and $P$ is still of the form $x \mapsto \alpha x^{2}+\beta x^{4}$.