GCD between a polynomial with terms of even degree and a polynomial with terms of odd degree. We are given a polynomial $p(z)=a_0z^n+b_0z^{n-1}+a_1z^{n-2}+b_1z^{n-3}+\dots=P_1(z)+P_2(z)$, where $P_1(z)=a_0z^n+a_1z^{n-2}+\dots$, $P_2(z)=b_0z^{n-1}+b_1z^{n-3}+\dots$. Let $d(z)=\text{GCD}(P_1(z),P_2(z))$.
Observe that if $p(z_0)=p(-z_0)=0$ then
\begin{gather}
a_0z_0^n+b_0z_0^{n-1}+a_1z_0^{n-2}+b_1z_0^{n-3}+\dots=0,\\
a_0z_0^n-b_0z_0^{n-1}+a_1z_0^{n-2}-b_1z_0^{n-3}+\dots=0,
\end{gather}
and so $P_1(z_0)=P_2(z_0)=0$. Thus $d(z_0)=0$. So, if $z_0$ and $-z_0$ are root of $p$ then $z_0$ is a root of $d$.
Now, we can write $p(z)=d(z)p^*(z)$. How can I conclude that $p^*(z)$ hasn't any pair of opposite roots? The problem is when $p$ has a pair of opposite roots with multiplicity greater than one.
 A: Hint.
Do you think $d(z)$ is even, odd, or neither?
Further Hint. 
What happens if you multiply an even polynomial by an odd polynomial?
[Added later in response to Federico's comment]
What I had in mind was this. For clarity, I prefer to rename the polynomials $p_1,p_2$ as $p_e,p_o$ where all powers of $z$ in $p_e$ are even, and all those in $p_o$ are odd.
Consider first the case where p has no "opposite" roots. Suppose $z_0\ne0$ is a root of $d$, the gcd of $p_e,p_o$. Then $p_e(z_0)=p_o(z_0)=0$, so $p_e(-z_0)=p_o(-z_0)=0$, and hence $-z_0$ is also a root of $d$. But $p=p_e+p_o$, so $\pm z_0$ are also roots of $p$. Contradiction. So we must have $d(z)=z^k$ for some $k$ (which is obviously the highest power of $z$ dividing $p(z)$, but that is of no consequence to the argument).
Now consider the general case. Suppose $d=e\cdot f$, where $e(z)=\prod(z^2-z_i^2)^{\alpha_i} $ and $f(z)=\prod(z+z_i)^{\beta_i}\prod(z-z_i)^{\gamma_i}$ and at most one of each pair $(\beta_i,\gamma_i)$ is non-zero. In other words, we have factored out all the pairs of opposite roots that we can and put them into $e(z)$.
Obviously $d$ divides $p$, so $e$ divides $p$. Let $p=e\cdot q$. By the first case, the gcd of $q_e,q_o$ is 1, because $q$ has no opposite roots. 
So we are done. 
Since $e$ is even, and $p=e(q_e+q_o)=e\cdot q_e+e\cdot q_o$ we must have $p_e=e\cdot q_e,p_o=e\cdot q_o$ and hence $d$ is the gcd of $e\cdot q_e$ and $e\cdot q_o$, which is $e$. So $p=e\cdot q$, and since gcd of $q_e,q_o$ is 1, $q$ has no opposite roots. (In Federico's notation we have $q=p^*$).
