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The finite simple group $\operatorname{PSp}(4,7)$ has order $138297600 = 11760^2$.

There also seems to be a description of the $q$ such that $\operatorname{PSp}(4,q)$ has square order, see for example here .

Some natural questions:

Which finite simple groups have order $n^2$?

Are there any finite simple groups order $n^k$ for $k \geq 3$?

Is there a classification of finite simple groups of order $n^k$ ($k > 1$)?

For a sporadic simple group $G$, one can check there always exists a prime $p$ such that $p \mid |G|$ and $p^2 \nmid |G|$, so $|G| \neq n^k$ for $k > 1$. The same is also true for $G = \operatorname{Alt}(n)$ by Bertrand's postulate.

The difficult case is then that of the finite simple groups of Lie type.

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    $\begingroup$ On a vaguely related question, it has been proved that the only examples in which $|G|^m =|H|^n$ for distinct finite simple groups $G,H$ and $m,n \ge 1$ are the known examples when $m=n$. This is important because it means that finite characteristically simple groups are (almost) characterized by their orders. $\endgroup$
    – Derek Holt
    Commented Jul 19, 2019 at 11:12

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Not anything resembling an answer, but too long for a comment. This question seems very hard to me, from a purely diophantine perspective. We have formulas for the orders of all the finite simple groups. Let's take just the example of $PSp(4,q)$, where $q$ is an odd (for convenience) prime power. I believe the order of this group is $$f(q) = \frac{1}{2}q^4(q^4-1)(q^2-1).$$ If we fix some $k$ such as 2 or 3 and ask for the order of the group to equal $n^k$, then a group in this family of order $n^k$ corresponds to an integer point on the curve $$n^k = f(q)$$... and you impose the additional constraint that $q$ is a positive prime power. It sounds like the authors of the paper linked in the MSE post you reference have found some solutions to this equation (and I'm guessing this curve has genus zero if they are conjecturing it has infinitely many), but proving you've found them all sounds challenging. And then to do this for even all groups of the form $PSp(2m,q)$, let alone all FSG's, seems very hard to me. It wouldn't surprise me if you ran across a family of curves which was quite intractable.

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    $\begingroup$ With the Erdos-Selfridge theorem on products of consecutive integers, one can show that for any $k \geq 2$, the product $(q-1)q(q+1)$ has a prime divisor $p$ such that the largest power of $p$ dividing $(q-1)q(q+1)$ is $p^l$ for $l \not\equiv 0 \mod{k}$. So that would show that $|PSL(2,q)| \neq m^k$. (Maybe there is also a more elementary way to see this). For $k \geq 3$ I think this would show $f(q) \neq m^k$ as well in many cases, since $f(q) = \frac{1}{2}(q-1)^2q^4(q+1)^2(q^2+1)$. $\endgroup$
    – spin
    Commented Aug 24, 2019 at 10:37

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