If $p(x) \in \mathbb{Q}[X]$ is irreducible and has two roots $r,s$ such that $rs = 1$, then $p(x)$ is of even degree.
I'm not sure how to solve this problem. My initial idea was to consider the field tower $\mathbb{Q}(r, r+s) \supseteq \mathbb{Q}(r+s) \supseteq \mathbb{Q}$. We know that $\mathbb{Q}(r, r+s) = \mathbb{Q}(r)$, since $r+s \in \mathbb{Q}(r)$. Then, $$\operatorname{deg}p(x) = [\mathbb{Q}(r): \mathbb{Q}] = [\mathbb{Q}(r+s)(r): \mathbb{Q}(r+s)][\mathbb{Q}(r+s): \mathbb{Q}].$$ Note that $x^2 - (r+s)x + 1 \in \mathbb{Q}(r+s)[X]$ and has roots $r$ and $s$. If we could show that this polynomial is irreducible in $\mathbb{Q}(r+s)[X]$ or, equivalently, that $r \notin \mathbb{Q}(r+s)$, the extension $\mathbb{Q}(r, r+s)/\mathbb{Q}(r+s)$ would have degree $2$ and, therefore, $\operatorname{deg}p(x) = 2k$ for some $k \in \mathbb{Z}_{+}$. However, I wasn't able to come up with a proof of this. Is this the correct approach?
To expand on Bill Dubuque's comment: let deg$(p(x)) = d$. Then the reciprocal polynomial of $p(x)$ is given by $x^{d}p(x^{-1})$ and has root $s = r^{-1}$.
Note that $p(x)$ and its reciprocal share a root and are both irreducible, hence $x^{d}p(x^{-1})$ is a non-zero constant multiple of $p(x)$:
$$p(x) = cx^{d}p(x^{-1})$$
for some $c \in \mathbb{Q}$. Furthermore, $p(x)$ is irreducible, hence $1$ and $-1$ are not roots of $p(x)$.
If we insert $1$ into $p(x)$ we get $p(1) = c(1)^{d}p(1) \neq 0$, so $c = 1$. Now inserting $-1$ into $p(x)$ gives $p(-1) = (-1)^{d}p(-1)$ which implies $(-1)^{d} = 1$. Thus, $d$ must be even.