It is requested to build an example of two spaces $ X $ and $ Y $ CW-complexes 4-dimensional such that $ \pi_1 (X) = \pi_1 (Y) $, $ H_\ast (X) = H_\ast ( Y) $ but $ X $ and $ Y $ are not homotopically equivalent
I believe that this is not possible because of Whitehead's Theorem
Any advice to argue because they do not exist? or are the examples known?
Consider the spaces $\mathbb{R}P^4$ and $S^2 (\mathbb{R}P^2) \vee \mathbb{R}P^2$ where $S$ denotes suspension. These have the same homology groups because suspension shifts reduced homology groups while wedging sums them. These both have fundamental group $\mathbb{Z}/2$, for the second use van Kampen's theorem plus the fact that suspending makes connectivity go up 1.
Why are these not homotopy equivalent? Well the cohomology rings distinguish them. The cohomology ring of a wedge sum basically splits up into a direct product of the cohomology rings of the summands. In the cohomology ring of $\mathbb{R}P^4$ the nontrivial element in the second cohomology squares to a nontrivial element. However, the multiplication in any suspension is trivial, and $\mathbb{R}P^2$ has no cohomology in degree 4, so $H^2 (S^2 (\mathbb{R}P^2) \vee \mathbb{R}P^2)$ no such element. This means they are not homotopy equivalent.