Prove the digital root of a square can be anything other than $2, 3, 5, 6, 8$? The digital root is the sum of the digits, unless that has more than one digit, so then you add up the digits again, until arriving at a single digit, e.g., $28$ -> $2 + 8 = 10$ -> $1 + 0 = 1$.
For what the digital root of a square can be, I just need to give an example of it, right? $0, 1, 4, 9$ are easy enough, and for $7$ I can just do $16$ or $25$ or $169$ or $196$ or $484$ or $529$ etc. (they seem to follow each other like that for some reason).
I'm not a professional mathematician, so please forgive me if what I'm saying is too obvious or too easy. I have looked among the first thousand squares and failed to find a single one with digital root $2, 3, 5, 6$ or $8$, but that doesn't prove anything, because maybe $1001^2$ does something contrary to expectation. I have this feeling that the explanation is in plain sight but for some reason I'm failing to see it.

Think about these $\bmod 3$.


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*If $n = 3k$, then $n^2 = 9k^2$.

*If $n = 3k + 1$, then $n^2 = 9k^2 + 6k + 1$.

*If $n = 3k$, then $n^2 = 9k^2 + 12k + 4 = 9k^2 + 12k + 3 + 1$.


Therefore $n^2 \equiv 2 \pmod 3$ is impossible if $n$ is an integer. What dose this have to do with digital roots? Well, if $n^2 \equiv 2, 5, \textrm{ or } 8 \pmod 9$, then $n^2 \equiv 2 \pmod 3$. But as I just demonstrated above, that congruence has no solutions.
To prove neither $3$ nor $6$ can be the digital root of a square, we have to broaden back out to $\bmod 9$. If $n = 9k + 3$, then $n^2 = 81k^2 + 54k + 9$, which is obviously divisible by $9$. Or if $n = 9k + 6$, then $n^2 = 81k^2 + 108k + 36$, which, again, is obviously divisible by $9$.
