Real valued polynomial has real coefficients

Let $$P(x)=a_nx^n+...+a_1x+a_0 \\ Q(x)=\overline{a_n}x^n+....+\overline{a_1}x+\overline{a_0}$$

Now, for $x$ real you get $$P(x)=\overline{P(x)}=Q(x)$$

Therefore $P-Q$ is a polynomial which has infinitely may roots (all real numbers are roots).

P.S. The solution can be written alternatelly in this form: if you split the coefficients into real and imaginary part, your polynomial becomes $R_1(x)+iR_2(x)$ for some real polynomials $R_{1,2}$. Then your condition means $R_2$ has infinitely many roots.

It is the same solution...

Let be $P$ a counterexample of minimal ($>0$) degree. $a_0 = P(0)\in\Bbb R$ by hypothesis. Then $Q(x) = (P(x) - a_0)/x$ is real for $x\in\Bbb R$, so $Q$ has real coefficients. But then $P(x) = xQ(x) + a_0$ has real coefficients.

Below is a sketch of an alternative proof using strong induction on the degree of the polynomial.

A $0^{th}$ degree polynomial is a constant, so that constant must be real in order for the polynomial to take real values on $\mathbb{R}$.

Assume the statement holds true for polynomials of degree $\le n$ where $n \ge 0$ and let $P(x)$ be a polynomial of degree $n+1$. Then it is easy to prove that $Q(x) = P(x+1) - P(x)$ is a polynomial of degree $\le n$ which has real values for $\forall x \in \mathbb{R}$. By the hypothesis of the induction step, all coefficients of $Q(x)$ must be real. But the coefficients of $Q(x)$ are linear combinations with integer factors of the coefficients of $P(x)$ for powers $\le n$, therefore those must be real as well. Then the coefficient of $x^{n+1}$ must also be real since all others are, which concludes the induction step.