Do even degree polynomials have even degree factors (no conjugates)? An even degree polynomial is a polynomial which has terms of only even degree, for example $3x^6 + x^4 + 2x^2 + 5$.
Let $p$ and $q$ be two non-zero polynomials such that both don't have a term with the same degree (for example, $p = x^3 + x + 1$, $q = x^4 + 3x^2 + 5$ is not allowed since both have a term of degree 0). A conjugate factor pair is a pair of form $p + q$ and $p - q$.
Assume all factorisation is over integers.
Now suppose there is a even polynomial which doesn't have a conjugate factor pair in it's list of factors. Will all factors be even degree polynomials too?

I really have no idea how to even approach this problem. Only thing I tried was taking examples. But I wasn't successful in finding a counter example.
My inspiration behind asking this was that, if you remove the conjugate pair restriction, there there are examples like $(x-1)(x+1) = x^2 - 1$ or $(x^2 + 1 + x)(x^2 + 1 - x) = x^4 + x^2 + 1$.
 A: Let $R:=\Bbb{Z}[X^2]\subset\Bbb{Z}[X]$ be the subring of 'even degree' polynomials.  Let $P\in R$ factor in $\Bbb{Z}[X]$ as
$$P=\prod_{i=1}^nQ_i,$$
where the $Q_i$ are irreducible. Because $P\in R$ we have $P(X)=P(-X)$ and so
$$\prod_{i=1}^nQ_i(-X)=P(-X)=P(X)=\prod_{i=1}^nQ_i(X),$$
so by unique factorization, for each $i$ there exists some $j$ such that $Q_i(-X)=Q_j(X)$. In particular the coefficients of $Q_i$ and $Q_j$ are congruent mod $2$, and so for the polynomials
$$F:=\frac{Q_i+Q_j}{2}\qquad\text{ and }\qquad G:=\frac{Q_i-Q_j}{2},$$
we have $F,G\in\Bbb{Z}[X]$ and $Q_i=F+G$ and $Q_j=F-G$. So $Q_i$ and $Q_j$ are a conjugate pair, unless $G=0$ which is the case if and only if $Q_i=Q_j$.
So if $P$ does not have any conjugate pairs we must have $Q_i(X)=Q_i(-X)$ for all $i$, meaning that $Q_i\in R$ for all $i$, so indeed all factors of $P$ are 'even degree' as well.
A: Suppose $f(x)$ has even degree and let $f(x)=p(x)q(x)$ where $p(x)$ is an irreducible factor of $f(x)$. Then $f(-x)=p(-x)q(-x)$ and so $p(-x)$ is a factor of $f(x)$ as well. If $p(-x) \neq p(x)$, then $p(-x)$ divides $q(x)$ and so $f(x)=p(x)p(-x)g(x)$. Clearly $p(x)$ and $p(-x)$ are conjugate pairs. 
A: $$ { x ^ 2 } = { x \cdot x } $$
It does not look like $x$ and $x$ is a "conjugate pair" according to your definition, because it explicitly demands that $q$ be nonzero.
