It is well-known that the number of irreducible monic polynomials of degree $n$ over the finite field of $q$ elements is given by the formula $$\frac{1}{n}\sum_{d\mid n}\mu\left(\frac{n}{d}\right)q^{d}.$$ Here $\mu(\cdot)$ is the Moebius function.
A polynomial $f(t)\in k[t]$ ($k$ any field) is called even if $f(-t)=f(t)$. I was wondering if the number of monic irreducible even polynomials of a given degree $n=2m$ is known.
What I have managed to show is that within the field extension ${\bf F}_{q^{2m}}\mid {\bf F}_q$, assuming $q$ is odd, it holds that $x\in {\bf F}_{q^{2m}}$ has an even minimal polynomial over ${\bf F}_q$ iff there exists $\sigma\in{\mathrm{ Gal}}({\bf F}_{q^{2m}}\mid {\bf F}_q)$ such that $\sigma(x)=-x$.
Proof- Let $m(t)\in{\bf F}_q[t]$ be the minimal polynomial of $x$. If $m(-x)=0$ then, by the fact that ${\bf F}_{q^{2m}}\mid{\bf F}_q$ is Galois, such an automorphism $\sigma$ exists.
Conversely, suppose $\sigma(x)=-x$, and define $f(t)=\frac{1}{2}(m(t)-m(-t))\in {\bf F}_q[t]$. Then $f(x)=\frac{1}{2}(m(x)-m(\sigma(x)))=0$. Since $\deg(f)\le \deg(m)$ this implies that either $f(t)=0$ or $f(t)=m(t)$. But note that $f(t)=-f(-t)$, and hence $m(t)=-m(-t)$. In particular, $m(0)=-m(-0)=-m(0)$, which means that $m(0)=0$, a contradiction to $m$ being irreducible.$\square$
Now, I'm pretty sure that this statement and proof are valid, and I'm pretty sure that they have some value in the enumeration I'm looking for, but I'm not sure how to apply it.
Would appreciate any help.
Thank you :)
Building on the good foundation laid out by the OP (and promoting my comments to an answer).
Indeed, the minimal polynomial of an element $x$ in some extension field $L$ of $K=\Bbb{F}_q$ is even, iff $x$ and $-x$ are conjugates. When we are interested in irreducible even polynomials of degree $n=2m$, we want $L=K(x)$. If $\sigma(x)=-x$, then $$\sigma^2(x)=\sigma(-x)=x,$$ and $x$ is thus in the fixed field of $\sigma^2$, and $\sigma^2$ is the identity mapping on $K(x)$. Let $F:z\mapsto z^q$ be the Frobenius automorphism of $L/K$. Basic Galois theory of finite fields then tells us that we must have $\sigma=F^m$. In other words we get the necessary condition $$ -x=\sigma(x)=x^{q^m}\Leftrightarrow x^{q^m}+x=0.\qquad $$
Lemma 1. If $z\in L, z\neq0,$ is a zero of $P(T)=T^{q^m}+T$, then $[K(z):K]=2d$ such that $d\mid m$ and $m/d$ is an odd integer. Furthermore, $z$ is then a zero of the polynomial $R(T)=T^{q^d}+T$.
Proof. Because $-z=-z+P(z)=z^{q^m}$, the elements $z$ and $-z$ are $K$-conjugates. Therefore the minimal polynomial of $z$ is even (by the OP's result), so its degree is $2d$ for some integer $d$. Because $z\in L$ we also have $2d\mid n=2m$, and $d\mid m$. By our earlier argument we have both $F^m(z)=-z$ and $F^d(z)=-z$. If $\ell=m/d$ we thus get $$ -z=F^m(z)=(F^d)^\ell(z)=(-1)^\ell z. $$ Therefore $\ell$ must be odd. QED.
Denote by $S_m$ the set $$ S_m:=\{z\in\overline{K}\mid z^{q^m}+z=0, z\neq0\}. $$
Lemma 2. If $d\mid m$ and $m/d=\ell$ is an odd integer, then $S_d\subseteq S_m$.
Proof. If $z\in S_d$, then $F^d(z)=-z$. Repeating the above calculation we see that $F^m(z)=-z$, and hence $z\in S_m$. QED.
Because an irreducible polynomial from $K[T]$ of degree $n=2m$ has $2m$ distinct zeros in $\overline{K}$, the following result is now immediate.
Proposition. Let $N_m$ be the number of even irreducible polynomials of degree $2m$ in $K[T]$. Then we have $$ 2m N_m=|S(m)\setminus\bigcup_{d\mid m, 2\nmid (m/d)}S(d)|. $$
Lemma 3. The polynomial $P(T)=T^{q^m}+T$ has $q^m$ distinct zeros in $L$.
Proof. Because $P'(T)=1$, the zeros of $P(T)$ in $\overline{K}$ are all simple. Because $P(T)^{q^m}-P(T)=T^{q^{2m}}-T$, and the zeros of the r.h.s. polynomial are exactly the elements of $L$, we see that all the zeros of $P(T)$ are in $L$. QED.
The inclusion-exclusion formula then gives us the following formula (Möbius inversion)
Corollary. $$ N_m=\frac1{2m}\sum_{d\mid m, 2\nmid (m/d)}\mu(\frac m d)(q^d-1). $$