International Mathematical Olympiad

Let $P(x)$ satisfy the given equation. $$\text{If }a=b=c=0,\text{ then }P(0)=0.$$ $$\text{If }b=c=0,\text{ then }P(-a)=P(a)\text{ for all real }a.$$ Hence $P(x)$ is even. Without loss of generality, we may assume that $$P(x)=a_n{x}^{2n}+\cdots +a_1{x}^2,a_n\ne 0.$$ If $b=2a,c=-\frac{2}{3}a$, we have that $$P(-a)+P\left(\frac{8}{3}a\right)+P\left(-\frac{5}{3}a\right)=2P\left(\frac{7}{3}a\right),$$ or $$a_n\left[1+{\left(\frac{8}{3}\right)}^{2n}+{\left(\frac{5}{3}\right)}^{2n}-2{\left(\frac{7}{3}\right)}^{2n}\right]a^{2n}+\cdots +a_1\left[1+{\left(\frac{8}{3}\right)}^2+{\left(\frac{5}{3}\right)}^2-2{\left(\frac{7}{3}\right)}^2\right]a^2=0$$ for all $a\in \mathbb{R}$. Then all coefficients of the polynomial with variable $a$ are $0$. If $n\ge 3$, it follows from $8^6=262144>235298=2\times 7^6$ that ${\left(\frac{8}{7}\right)}^{2n}\ge {\left(\frac{8}{7}\right)}^6>2$. This implies $$1+{\left(\frac{8}{3}\right)}^{2n}+{\left(\frac{5}{3}\right)}^{2n}-2{\left(\frac{7}{3}\right)}^{2n}>0.$$ Hence $n\le 2$. Let $P(x)=a{x}^4+\beta x^2$, with $a,\beta \in \mathbb{R}$.

We now show that $P(x)=a{x}^4+\beta x^2$ satisfies the given equation. Let $a,b,c$ be real numbers satisfying $ab+bc+ca=0$. Then $$\begin{array}{rl}(a-b{)}^4+(b-c{)}^4+(c-a{)}^4-2(a+b+c{)}^4& =\sum (a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^4)-2(a^2+b^2+c^2{)}^2\\ & =\sum (a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^4)-\\ & 2a^4-2b^4-2c^4-4a^2{b}^2-4a^2{c}^2-4b^2{c}^2\\ & =\sum (-4a^{3}b+2a^2{b}^2-4a{b}^3)\\ & =-4a^2(ab+ca)-4b^2(bc+ab)-4c^2(ca+bc)+\\ & 2(a^2{b}^2+b^2{c}^2+c^2{a}^2)\\ & =4a^{2}bc+4b^{2}ca+4c^{2}ab+2a^2{b}^2+2b^2{c}^2+2c^2{a}^2\\ & =2(ab+bc+ca{)}^2=0,\\ (a-b{)}^2+(b-c{)}^2+(c-a{)}^2-2(a+b+c{)}^2& \\ & =\sum (a^2-2ab+b^2)-2\sum a^2-4\sum ab\\ & =0.\end{array}$$ Hence $P(x)=a{x}^4+\beta x^2$ satisfies the given equation.