Problem: Find all polynomials $P$ with real coefficients such that
for all reals $a$, $b$, $c$ such that $ab+bc+ca = 0$, we have
\[ P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c). \]

Solution: The answer is \[ P(x) = \alpha x^4 + \beta x^2 \]
which can be checked to work, for any real numbers $\alpha$ and $\beta$.

It is easy to obtain that $P$ is even and $P(0) = 0$.
The trick is now to choose $(a,b,c) = (6x,3x,-2x)$
and then compare the leading coefficients to get
\[ 3^n + 5^n + 8^n = 2 \cdot 7^n \]
for $n = \deg f$ (which is even).
As $n \ge 7 \implies (8/7)^n > 2$, this means that we must have $n \le 6$.
Moreover, taking modulo $7$ we have $3^n + 5^n \equiv 6 \pmod 7$
which gives $n \equiv 2, 4 \pmod 6$.

Thus $\deg P \le 4$, which (combined with $P$ even) resolves the problem.