Let $p:\mathbb{R}\rightarrow \mathbb{R}$ be a polynomial with integer coefficients such that $p(n)$ is a perfect square for every $n\in\mathbb{N}^{*}$. Is it true that exists a polynomial $q$ with rational coefficients such that $p(x)=q(x)^2$?
If the roots of $p$ are integer we can prove it looking at the roots $x_1, x_2, \dots ,x_n$ and taking large prime numbers and positive integers $x$ such that $x-x_i=p$ for some $i\in\{1, 2, \dots, n\}$. Then we get that it must exist $j\neq i$ such that $x_i=x_j$.
Considering only integer coefficients I couldn't guess anything to prove or give a counterexample.
