Show that if $m \neq n$, then $\mathbb{R}P^m \times S^n$ and $\mathbb{R}P^n \times S^m$ have isomorphic homotopy groups but non-isomorphic integral homology groups.
As for isomorphism of homotopy groups I can say that it follows from the fact that $S^m \times S^n$ is $2$-sheeted covering space for $\mathbb{R}P^m \times S^n$ and similarly $S^n \times S^m$ for $\mathbb{R}P^n \times S^m$, but $S^m \times S^n$ and $S^n \times S^m$ obviously have similar homotopy groups.
I think that difference between integral homology groups may be deduced from the Kunneth formula, but unfortunately can't do it. Can anyone help me and give an example where and why these groups differ?
For simplicity let $m=2$ and $n=1$. Then for instance for the first homology group $H_1$ we have by Kunneth a short exact sequence $$0\to \oplus_{i+j=1}H_i(\mathbb{R}P^2)\otimes H_j(S^1)\to H_1(\mathbb{R}P^2\times S^1) \to \text{Tor}_1^{\mathbb{Z}}(H_0(\mathbb{R}P^2),H_0(S^1))\to 0$$ and a short exact sequence
$$0\to \oplus_{i+j=1}H_i(\mathbb{R}P^1)\otimes H_j(S^2)\to H_1(\mathbb{R}P^1\times S^2) \to \text{Tor}_1^{\mathbb{Z}}(H_0(\mathbb{R}P^1),H_0(S^2))\to 0.$$
Since $H_0$ is $\mathbb{Z}$ for any of the above spaces and $H_1(S^1)=\mathbb{Z}$, $H_1(S^2)=0$, $H_1(\mathbb{R}P^2)=\mathbb{Z}_2$, and $\mathbb{R}P^1 = S^1$, and $\text{Tor}_1^{\mathbb{Z}}(\mathbb{Z},\mathbb{Z})=0$ we have from the first sequence $$\mathbb{Z}_2\oplus \mathbb{Z} \cong H_1(\mathbb{R}P^2\times S^1),$$ and from the second sequence $$\mathbb{Z} \cong H_1(\mathbb{R}P^1\times S^2).$$