I am looking for a relatively simple way to see that $\Sigma \mathbb{C}P^2$ is not homotopy equivalent to $S^3 \vee S^5$.
Both the Euler characteristic of a space and its homology groups are invariant under homotopy equivalence, so I tried to consider these concepts. However, the Euler characteristic of both spaces is equal to $-1$, and the homology groups of both spaces are also isomorphic.
In the answer posted here, the author considers Steenrod squares to show that the homotopy equivalence can't exist, but this is a topic that I'm not quite familiar with.
Should I try to use the Freudenthal suspension theorem or something related to homotopy groups? Or, perhaps, cohomology rings and the cup product? I do know that the cup product of $H^{\ast}(\Sigma \mathbb{C}P^2)$ is trivial (see the result here); can I show that the cup product in $H^{\ast}(S^3 \vee S^5)$ is nontrivial?
Edit: As commented below, the cup product in $H^{\ast}(S^3 \vee S^5)$ is also trivial, so this won't help here, either.
Thanks!