Simply-connected 4-manifolds can be blown up and down to complex projective planes. How about non-simply-connected ones? There is a nice theorem that for every simply connected, closed, smooth, oriented manifold $M$, we have for some $m \in \mathbb{N}$:
$$M \# \left(\mathop{\#}^m \left(\mathbb{C}\mathbb{P}^2 \# \overline{\mathbb{C}\mathbb{P}^2} \right) \right) \cong \left(\mathop{\#}^{m+b^+_2(M)} \mathbb{C}\mathbb{P}^2 \right) \# \left(\mathop{\#}^{m+b^-_2(M)} \overline{\mathbb{C}\mathbb{P}^2} \right)$$
The $b^\pm_2(M)$ are the positive and negative eigenvalues of the intersection form.
This has far-reaching consequences, for example any 4-manifold invariant that is multiplicative under direct sum $\#$ will only measure the intersection form on simply connected manifolds.
Is there a similar statement for non-simply-connected 4-manifolds? What is the closest we can get?
 A: Two smooth oriented 4-manifolds $M,N$ with fundamental group $\pi$ admit maps to $B\pi$, the classifying maps of their universal covers.  The manifolds $M$ and $N$ become diffeomorphic after connect summing with copies of complex projective space if and only if the images of the fundamental classes $[M],[N]$ in $H_4(B\pi)$ under the classifying maps are equal, for some choice of maps to $B\pi$, i.e. if we mod out $H_4(B\pi)$ by the action of outer automorphisms of $\pi$.  This follows from Kreck's surgery theory with normal 1-type $B\pi \times BSO$. This theory really tells us about connect summing with $S^2 \times S^2$s, but $\#^2(\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}) \cong S^2 \times S^2 \# \mathbb{CP}^2 \# \overline{\mathbb{CP}^2}$. It turns out that the class in $H_4$ is determined by the stable isomorphism classes of the $\mathbb{Z}\pi$ modules $\pi_2(M)$, $\pi_2(N)$.  
Thus for general groups you need many minimal models. For groups with vanishing $H_4$ the situation is rather similar to the trivial group situation; you just need to construct some 4-manifold with that group as its fundamental group, for the base manifold.  (Analogous to a hidden $S^4$ on the right hand side of your equation.)  
