Croatian Mathematical Olympiad

Yes, there necessarily exists such polynomial $R(x)$.

Firstly, notice that we are allowed to assume, without loss of generality, that polynomials $P(x)$ and $Q(x)$ have leading coefficient equal to $1$. If the polynomial $Q(x)$ is constant, then the polynomial $P(x)$ is constant as well, and we have $P(x)=(Q(x){)}^2$.

Therefore, let us assume that the polynomial $Q(x)$ is not constant, and neither the polynomial $P(x)$. The following equality holds for degrees of polynomials $P(x)$ and $Q(x)$: $$(\mathrm{deg}P{)}^2=2\mathrm{deg}Q.$$ Thus, there exists a positive integer $n$ such that $\mathrm{deg}P=2n$ and $\mathrm{deg}Q=2n^2$. That means there exist polynomials $R(x)$ and $S(x)$ with real coefficients such that $$P(x)=(R(x){)}^2+S(x),\forall x\in \mathbb{R},$$ and their coefficients satisfy $\mathrm{deg}R=n$ and $\mathrm{deg}S<n$. Let us define $$A(x)=R(P(x))\text{and}B(x)=S(P(x)),\forall x\in \mathbb{R}.$$ Then we have $$B(x)=(Q(x){)}^2-(A(x){)}^2=(Q(x)-A(x))(Q(x)+A(x)),\forall x\in \mathbb{R}.$$ We also have $\mathrm{deg}A=2n^2=\mathrm{deg}Q$, so if the polynomial $B(x)$ is not the null polynomial, then its degree is greater than or equal to $2n^2$. However, it is impossible due to the fact that degree of the polynomial $B(x)$ is at most $(n-1)\cdot 2n=2n^2-2n<2n^2$.

Thus, polynomial $B(x)$ is the null polynomial, and since the polynomial $P(x)$ is not equal to $0$, we conclude that the polynomial $S(x)$ is not null polynomial. In other words, we have $P(x)=(R(x){)}^2$ for all real numbers $x$.