Roots of a polynomial must be real Say $P$ is a polynomial of degree $n$ with $n$ real roots. I have to prove that the polynomial $Q(z)=P(z+i)+P(z-i)$ has $n$ real roots.
First it happens that for a real number $t$ and because $P$ is real polynomial we have $$\overline{Q}(t)= \overline{Q(t)} =\overline{P(t+i)}+\overline{P(t-i)} \\
 = P(\overline{t+i})+P(\overline{t-i}) \\
= P(t-i)+P(t+i) = Q(t) $$
The polynomial $Q-\bar{Q}$ has infinite roots (namely all the real numbers) and thus $Q$ is a real polynomial
Truth is I don't know how to proceed further. This exercice was given at the beginning of a complex analysis course so I don't know if I must search for an analytical argument or a simpler one. If you can just give me a hint (I want to solve the problem by myself) that would be really helpful

Hint: If $t$ is real, what is the complex conjugate of $t+i$. What does that tell you about $Q$?
Hint 2: This is a heuristic hint. If you know, or assume, that a polynomial has only real roots, then you can assume the polynomial can be put in the for of a product of monomial factors, each of the form $(x-r_i)$, where r_i is a real root of the polynomial.
$$a_n\prod_{i=1}^n(x-r_i)$$
The heuristic hint is this. You will exclude all of the polynomials that have complex roots simply by using that form rather than the form:
$$a_0 +\sum_{i=1}^n a_nx^n$$
Hint 3: If $Q(z)= 0$, with $z\in\mathbb{C}$, what can one say about $|P(z+i)|$ as it relates to $|P(z-i)|$?
