The problem of recognizing the standard $S^n$ is the following: Given some simplicial complex $M$ with rational vertices representing a closed manifold, can one decide (in finite time) if $M$ is homeomorphic to $S^n$.

For $n=1$, this is obvious, and for $n=2$, one can solve it by computing $\chi(M)$. A solution for $n=3$ is due to

J.H. Rubinstein. An algorithm to recognize the 3-sphere. In Pro- ceedings of the International Congress of Mathematicians, vol- ume 1, 2, pages pp. 601–611, Basel, 1995. Birkhäuser.

By a theorem of S.P. Novikov, the problem is unsolvable if $n\geq 5$. The idea is the following: By the Adjan--Rabin theorem, there is a sequence of super-perfect groups $\pi_i$ for which the triviality problem is unsolvable. Now construct homology spheres $\Sigma_i$ with fundamental groups $\pi_i$. If one can decide which of the $\Sigma_i$ are standard spheres, then one can solve the triviality problem for the fundamental groups.

Question: Is the recognition problem for $S^4$ solvable?

The problem with this proof of S.P. Novikov's theorem is that there is no result that asserts that for any given super-perfect group $\pi$ there is a homology $4$-sphere satisfying $\pi_1(\Sigma) = \pi$. However, Kervaire has proved that every perfect group with the same amount of generators and relators may be realized as the fundamental group of a homology $4$-sphere.

Thus the question: Is there an improved Adjan--Rabin theorem that asserts the existence of a sequence of perfect groups $\pi_i$ with the same amount of generators and relators, the triviality problem of which is unsolvable?