Balkan Mathematical Olympiad Shortlisted Problems

Setting $y=0$ we get $$2P^2(x)+P^2(0)=2P(x^2)(1)$$ for every $x\in \mathbb{R}$. We claim that $P(x)$ is a monomial. Indeed, if this is not the case, let $a{x}^n$ and $b{x}^m$, with $n>m$ be the two non-zero terms with the largest powers of $x$. Comparing the coefficients of $x^{m+n}$ in (1) we get $2ab=0$, a contradiction. So $P(x)$ is a monomial. If $P(x)=c$, a constant, then substituting in the original equation we get $3c^2=2c$ giving $c=0$ or $c=\frac{2}{3}$. Otherwise $P(x)=a{x}^n$ for some $a\in \mathbb{R}\setminus \{0\}$ and $n\in \mathbb{N}$. Then $P(0)=0$ and substituting in (1) with $x=1$ we get $2a^2=2a$ giving $a=1$. Now for $x=y=1$ in the original equation we get $2+2^{2n}=2\cdot 3^n$. The cases $n=1,2$ are obvious solutions, while for $n\ge 3$ we have $$\frac{2+2^{2n}}{3^n}>{\left(\frac{4}{3}\right)}^n\ge \frac{64}{27}>2$$ showing that no other solutions exist. So the only possible solutions are $P(x)=0,\frac{2}{3},x,x^2$ which are easy to check that they satisfy the equation.

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Alternative solution.

If $P$ is constant, then we get $P(x)=0$ or $P(x)=\frac{2}{3}$ as in the first solution. So assume that the degree of $P$ is $n\ge 1$ and let $a_n$ be the leading coefficient. Equating the coefficients of $x^{2n}$ we $2a_n^2=2a_n$ giving $a_n=1$. Now letting $x=y$ we get $$2P^2(x)+P^2(2x)=2P(3x^2)$$ and equating the coefficients of $x^{2n}$ we get $2+2^{2n}=2\cdot 3^n$ which gives $n=1$ or $n=2$ as in the first solution. We can now try all polynomials of the form $P(x)=x+a$ and $P(x)=x^2+ax+b$ which leads to the same solutions as in the first solution.