Problem 1278

The given can be rewritten as saying that
$$\begin{aligned} Q(x, y, z) & :=x P(x)+y P(y)+z P(z) \\ & -x y z(P(x-y)+P(y-z)+P(z-x)) \end{aligned}$$
is a polynomial vanishing whenever $x y z \neq 0$ and $2 x y z=x+y+z$, for real numbers $x, y$, $z$.
Claim - This means $Q(x, y, z)$ vanishes also for any complex numbers $x, y, z$ obeying $2 x y z=x+y+z$.

Proof. Indeed, this means that the rational function
$$R(x, y):=Q\left(x, y, \frac{x+y}{2 x y-1}\right)$$
vanishes for any real numbers $x$ and $y$ such that $x y \neq \frac{1}{2}, x \neq 0, y \neq 0, x+y \neq 0$. This can only occur if $R$ is identically zero as a rational function with real coefficients. If we then regard $R$ as having complex coefficients, the conclusion then follows.
Remark (Algebraic geometry digression on real dimension). Note here we use in an essential way that $z$ can be solved for in terms of $x$ and $y$. If $s(x, y, z)=2 x y z-(x+y+z)$ is replaced with some general condition, the result may become false; e.g. we would certainly not expect the result to hold when $s(x, y, z)=x^{2}+y^{2}+z^{2}-(x y+y z+z x)$ since for real numbers $s=0$ only when $x=y=z$ !
The general condition we need here is that $s(x, y, z)=0$ should have "real dimension two". Here is a proof using this language, in our situation.
Let $M \subset \mathbb{R}^{3}$ be the surface $s=0$. We first contend $M$ is two-dimensional manifold. Indeed, the gradient $\nabla s=\langle 2 y z-1,2 z x-1,2 x y-1\rangle$ vanishes only at the points $( \pm 1 / \sqrt{2}, \pm 1 / \sqrt{2}, \pm 1 / \sqrt{2})$ where the $\pm$ signs are all taken to be the same. These points do not lie on $M$, so the result follows by the regular value theorem. In particular the topological closure of points on $M$ with $x y z \neq 0$ is all of $M$ itself; so $Q$ vanishes on all of $M$.
If we now identify $M$ with the semi-algebraic set consisting of maximal ideals $(x-a, y-$ $b, z-c)$ in Spec $\mathbb{R}[x, y, z]$ satisfying $2 a b c=a+b+c$, then we have real dimension two, and thus the Zariski closure of $M$ is a two-dimensional closed subset of Spec $\mathbb{R}[x, y, z]$. Thus it must be $Z=\mathcal{V}(2 x y z-(x+y+z))$, since this $Z$ is an irreducible two-dimensional closed subset (say, by Krull's principal ideal theorem) containing $M$. Now $Q$ is a global section vanishing on all of $Z$, therefore $Q$ is contained in the (radical, principal) ideal $(2 x y z-(x+y+z))$ as needed. So it is actually divisible by $2 x y z-(x+y+z)$ as desired.
Now we regard $P$ and $Q$ as complex polynomials instead. First, note that substituting $(x, y, z)=(t,-t, 0)$ implies $P$ is even. We then substitute
$$(x, y, z)=\left(x, \frac{i}{\sqrt{2}}, \frac{-i}{\sqrt{2}}\right)$$
to get
$$\begin{aligned} & x P(x)+\frac{i}{\sqrt{2}}\left(P\left(\frac{i}{\sqrt{2}}\right)-P\left(\frac{-i}{\sqrt{2}}\right)\right) \\ = & \frac{1}{2} x(P(x-i / \sqrt{2})+P(x+i / \sqrt{2})+P(\sqrt{2} i)) \end{aligned}$$
which in particular implies that
$$P\left(x+\frac{i}{\sqrt{2}}\right)+P\left(x-\frac{i}{\sqrt{2}}\right)-2 P(x) \equiv P(\sqrt{2} i)$$
identically in $x$. The left-hand side is a second-order finite difference in $x$ (up to scaling the argument), and the right-hand side is constant, so this implies $\operatorname{deg} P \leq 2$.

Since $P$ is even and $\operatorname{deg} P \leq 2$, we must have $P(x)=c x^{2}+d$ for some real numbers $c$ and $d$. A quick check now gives the answer $P(x)=c\left(x^{2}+3\right)$ which all work.