1. Determine all polynomials with integer coefficients $p(x)$ such that for every real number $x$ the following holds:

$$
16 p\left(x^{2}\right)=[p(2 x)]^{2}
$$
Solution. Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}, a_{n} \neq 0$ be the given polynomial. If we substitute $x=0$ in the given equation, we get $16 a_{0}=a_{0}^{2}$, from which it follows that $a_{0}=0$ or $a_{0}=16$. On the other hand, if we equate the coefficients of $x^{2 n}$ in the given equation, we obtain the equation $16 a_{n}=2^{2 n} a_{n}^{2}$ and since $a_{n} \neq 0$, we get $a_{n}=\frac{16}{4^{n}}$. However, $a_{n}$ is an integer, so $n=0$ or $n=1$ or $n=2$.

For $n=0$ the polynomials are $p(x)=0$ and $p(x)=16$.

For $n=1$ possible polynomials are $p(x)=4 x$ and $p(x)=4 x+16$. However, only $p(x)=4 x$ satisfies the conditions of the problem.

For $n=2$ possible polynomials are $p(x)=x^{2}+a_{1} x$ and $p(x)=x^{2}+a_{1} x+16$. By direct verification, we get $a_{1}=0$ and only $p(x)=x^{2}$ satisfies the conditions of the problem.