TITLE: If a group $G$ has decidable word problem, must it have a decidable square problem?
QUESTION [9 upvotes]: My question is a refinement of this one about 'efficient' construction of square elements: If the word problem for a (finitely generated, finitely presented) group is decidable, must the 'square problem' (given an element $g$ of $G$, is there an element $h$ with $g=h^2$?) also be decidable? If not, how 'nice' can $G$ be while still having an undecidable square problem? For instance, can $G$ be automatic?  (It feels like there should be an argument based on the Dehn function that precludes this, but I'm not immediately seeing it.)  Could it even be hyperbolic?

REPLY [2 votes]: (This is not really an answer, rather a suggestion that the answer is probably negative.) 
It is known that for linear groups over integers $GL(n,\mathbb{Z})$
starting from $n=4$ the membership problem is undecidable. (It is decidable for $n=2$. The case $n=3$ is an open problem for what I know.)
Consider a finitely generated subgroup $G\subset GL(n,\mathbb{Z})$ with undecidable membership. It is, of course, very much decidable whether $g\in G$ 
is a square in $GL(n,\mathbb{Z})$ (using a Jordan normal form). There may be several square roots $h_1,h_2,\dots,h_k$, and the problem is to find out if any of them belong to $G$. I suspect that this restricted version of the membership problem is still undecidable in general, although I do not have a proof of this.