3. (BUL 1) Find all polynomials $f(x)$ with real coefficients for which
$$
f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right) \text {. }
$$
3. An obvious solution is $f(x)=0$. We now look for nonzero solutions. We note that plugging in $x=0$ we get $f(0)^{2}=f(0)$; hence $f(0)=0$ or $f(0)=1$. If $f(0)=0$, then $f$ is of the form $f(x)=x^{k} g(x)$, where $g(0) \neq 0$. Plugging this formula into $f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right)$ we get $2^{k} x^{2 k} g(x) g\left(2 x^{2}\right)=\left(2 x^{2}+1\right)^{k} g\left(2 x^{3}+x\right)$. Plugging in $x=0$ gives us $g(0)=0$, which is a contradiction. Hence $f(0)=1$.
For an arbitrary root $\alpha$ of the polynomial $f, 2 \alpha^{3}+\alpha$ must also be a root. Let $\alpha$ be a root of the largest modulus. If $|\alpha|>1$ then $\left|2 \alpha^{3}+\alpha\right|>$ $2|\alpha|^{3}-|\alpha|>|\alpha|$, which is impossible. It follows that $|\alpha| \leq 1$ and hence all roots of $f$ have modules less than or equal to 1 . But the product of all roots of $f$ is $|f(0)|=1$, which implies that all the roots have modulus 1. Consequently, for a root $\alpha$ it holds that $|\alpha|=\left|2 \alpha^{3}-\alpha\right|=1$. This is possible only if $\alpha= \pm 1$. Since the coefficients of $f$ are real it follows that $f$ must be of the form $f(x)=\left(x^{2}+1\right)^{k}$ where $k \in \mathbb{N}_{0}$. These polynomials satisfy the original formula. Hence, the solutions for $f$ are $f(x)=0$ and $f(x)=\left(x^{2}+1\right)^{k}, k \in \mathbb{N}_{0}$