Prove that if all of the roots of a polynomial $P(z)$ are real, then for any $b$ the roots of the polynomial $R(z)=P(z+ib)+P(z-ib)$ are also real.
I'm just trying take the polynomial $$P(z)=a_0+a_1z+\dotsb+a_nz^n$$ and look at $$P(z+bi)=a_0+a_1(z+bi)+\dotsb+a_n(z+bi)^n$$ and $$P(z-bi)=a_0+a_1(z-bi)+\dotsb+a_n(z-bi)^n$$ and then considering $$P(z+ib)+P(z-ib)=2a_0+2a_1z+2a_2(z^2-b^2)+\dotsb$$ we can see that every term hasn't the imaginary part, i.e all of $i$ are missing and we get new polynomial $G$ which have coefficients as in a polynomial $P(z)$ multiplying on $b$. But how to explain that all of the roots of $R(z)$ are real?
Here is a partial solution, based on an idea by Martin R. First of all, by rescaling we may assume without loss that $b=1$ ; we then have to deal with $Q=P(X-i)+P(X+i)$.
Martin's idea was to use a result by Saddeev & Sominski, that if $\sum_{k=0}^m\gamma_k X^k$ has only real roots then the same is true of $\sum_{k=0}^m\gamma_k P^{(k)}$ (see the proof here).
So we try to write $Q$ as a linear combination of derivatives of $P$. It turns out that
$$ Q=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^k}{(2k)!} P^{(2k)} \tag{1} $$
(the formula is easy to show by checking that it works for every monomial). So Martin's idea works iff the polynomial $C_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-x^2)^k}{(2k)!} $ has only real roots. This holds for $n\leq 5$, but unfortunately becomes false for $n=6$. Note that $C_n$ is the Taylor expansion of the cosine function.
So this method only shows the result when the degree of $P$ is $\leq 5$.