It seems that usually smooth structures on compact 4-manifolds are distinguished by Seiberg-Witten/Donaldson invariants. And I don't know another direct way to do that. But in the case when $b\_2^+$ is even the invariants vanish identically. 


Does there exist an example of smooth compact exotic 4-manifold with $b\_2^+$ even?



Exotic 4-manifolds with even positive partial betti number
An often-overlooked invariant of a closed oriented 4-manifold $X$ is the closed oriented 4-manifold $\overline{X}$: the orientation-reversed manifold. Taking the Seiberg-Witten invariants of $\overline{X}$, one can distinguish smooth structures on manifolds $X$ with $b^->0$ odd, regardless of the parity of $b^+$. For instance, $\mathbb{C}P^2\# \overline{K3}$ has $b^+ = 20$ and infinitely many smooth structures (perform knot surgery on a K3 surface, blow up once, reverse orientation). 


The real problem, then, is to understand what happens when both $b^+$ and $b^-$ are even. 


I think one can distinguish infinitely many smooth structures on the connected sum of two K3 surfaces (so $b^+=6$, $b^-=38$) along the lines Tom Leness suggests in his answer.


The non-equivariant Bauer-Furuta invariant of a $\mathrm{Spin}^c$-structure $\mathfrak{s}$ on a closed oriented 4-manifold $X$ is a stable map $S^{(c\_1(\mathfrak{s})^2−σ)/4}\to S^{b^+}$. When $X$ is a homotopy K3 surface, and $c\_1(\mathfrak{s})^2=0$, we get the stable homotopy class of a map $S^4\to S^3$; this lives in $\mathbb{Z}/2$, and is actually in this case the mod 2 Seiberg-Witten invariant (I'm going on memory here, so this is a point to check). By Bauer's theorem, the invariant for a connected sum is the smash product of the invariants of the summands. Now, the smash-square of the Hopf map is the generator for $π^s\_2(S^0)=\mathbb{Z}/2$. Hence, if $X\_1$ and $X\_2$ are homotopy K3's, and $s(X\_i)$ is the number of $\mathrm{Spin}^c$-structures with $c\_1(\mathfrak{s})^2=0$ and odd SW invariant, the product $s(X\_1)s(X\_2)$ is an invariant of $X\_1\#X\_2$. Via Fintushel-Stern knot surgery along on K3 along knots with suitable Alexander polynomials, one can make $s(X\_i)$ arbitrarily large, and thereby distinguish infinitely many smooth structures on 
$K3 \# K3$. 


As so often in this subject, one has the feeling that information is being lost: maybe $X\_1\#X\_2$ actually knows $X\_1$ and $X\_2$.


