= Solution
The standard <CW complex> structure on <Complex projective space> $\mathbb{CP}^3$ has one cell in each dimension $0,2,4,6$ and no other cells. Its <cellular chain complex> therefore has zero differentials: each boundary has an odd-dimensional target group, which is zero. The product of the usual two-cell structures on $S^2$ and $S^4$ likewise has exactly one cell in dimensions $0,2,4,6$. Thus <cellular homology> gives, for both spaces,
$$
\boxed{H_i(-;\mathbb Z)\cong\begin{cases}\mathbb Z,&i=0,2,4,6,\\0,&\text{otherwise}.\end{cases}}
$$
The <cohomology rings> detect more than these additive groups. Let $c\in H^2(\mathbb{CP}^3;\mathbb Z)$ be the hyperplane class. The <cohomology ring of complex projective space> is
$$
H^*(\mathbb{CP}^3;\mathbb Z)=\mathbb Z[c]/(c^4),\qquad |c|=2.
$$
For the normalization, a hyperplane is <Poincare dual> to $c$; intersecting three generic hyperplanes gives one point with positive complex <orientation>, so $\langle c^3,[\mathbb{CP}^3]\rangle=1$. Together with the computed additive groups this makes $c,c^2,c^3$ generators in their respective degrees.
Let $u$ and $v$ be the pullbacks of the <orientation> classes of $S^2$ and $S^4$. The <Künneth theorem> and naturality of the <cup product> give
$$
H^*(S^2\times S^4;\mathbb Z)=\mathbb Z[u,v]/(u^2,v^2),\qquad |u|=2,\quad |v|=4,
$$
where $uv$ generates degree six. In particular, every degree-two class on the product has square zero, whereas $c^2\ne0$. Any graded ring isomorphism would send $c$ to $\pm u$, contradicting this product relation. A <homotopy equivalence> induces an isomorphism of <cohomology rings>, so \b[the spaces have isomorphic integral <homology (mathematics)> but are not <homotopy> equivalent].
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