It's essy to see that $\limsup_{n\to\infty}\frac{P(n)}{n}\le 1$ so $P(x) = \frac{a}{b}x+\frac{c}{d}$ for some constant $a,b,c,d$ or $P(x)$ is constant. If $P(x)=k$ then it satisfies the problem's condition iff $k$ is squarefree If the latter happens, we claim that $c=0$. Suppose that $c\ne 0$ take sufficiently large $n$ such that $P(n)=\mathrm{rad}(n)\mid n$. So $bdP(n)=adn+bc\mid bdn\mid abdn$. So $adn+bc\mid b^2c$ which fails for sufficiently large $n$. So $c=0$ which implies $a=1$. It's easy to check that this always work (i.e. take $n=b\mathrm{rad}(b)\cdot p$ for any prime $p\nmid b$) Hence the solution set is $P(x)=\frac{x}{b}$ or $P(x)=k$ for some squarefree $k$ and any positive integer $b$.

cjquines0 wrote: Let $n$ be a positive integer, with prime factorization $$n = p_1^{e_1}p_2^{e_2} \cdots p_r^{e_r}$$for distinct primes $p_1, \ldots, p_r$ and $e_i$ positive integers. Define $$rad(n) = p_1p_2\cdots p_r,$$the product of all distinct prime factors of $n$. Find all polynomials $P(x)$ with rational coefficients such that there exists infinitely many positive integers $n$ with $P(n) = rad(n)$. Did you mean $$\mathrm{rad}(n) = p_1p_2\cdots p_n$$

MarkBcc168 wrote: Hence the solution set is $P(x)=x$ or $P(x)=k$ for some squarefree $k$. What about $P(x)=\frac{x}{n}$ where $n$ is a positive integer???(coefficients of $P$ are rational)

Okay, I have edited the solution.