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Mar 21, 2018 at 3:12 comment added GH from MO There is a more elegant way to finish. In your first display, each $d$ is odd, hence in fact we get $\tau(m^2)\leq m$. Therefore, we have equality in this bound, so each odd $d<m$ divides $m^2$. For $m>1$, this implies that $m-2\mid m^2$, hence also $m-2\mid 4$. As $m$ is odd, $m-2=1$, whence $m=3$.
Mar 20, 2018 at 17:48 history answered Alexander Kalmynin CC BY-SA 3.0