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If we set $f(n)=$ the digital sum of $n$，for example, $f(2024)= 2+0+2+4= 8$.

Are there any $n>375501$ in solutions to the equation $f(n^2)=9,$ except $n=10k$, $n=10^a+10^b+1$, $n=5 \cdot 10^a+1$ or $n=10^a+5$? ($k,a,b$ are integers, $k>0$, $a \ge b \ge 0$)

or can it be proved that there are only limited nontrivial $n$ such $f(n^2)=9$ ?

And when $f(n^2)=m$, $m=4,7,10,11,\ldots$ whether such $n$ is still limited or not ?

(if $f(n^2)\equiv m \pmod 9$, then $n^2 \equiv m \pmod 9$, only when $m \equiv 0,1,4,7 \pmod 9$ there are integer solutions $n$)

trivial solution : if $f(n^2)=m$, then $f((10 n)^2)=m$，else if $n=A \times 10^k+B$ and $B^2<10^k$, $2AB<10^k$, $f(n)=m$, let $N= A \times 10^{k+1}+B$ then $f(N^2)=f(n^2)=m$

so these solutions like $10 n$ or $N$ are trivial to some degree.

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    $\begingroup$ Your question is hard to read. Is it possible to format it in Latex? And why do you work with integers as string. To me it seems better to use $n = \sum_{k=0}^m a_k \cdot 10^k$. $\endgroup$
    – Dieter Kadelka
    Commented Jan 4 at 15:32
  • $\begingroup$ I 'm sorry. The question is published with mobile phone, and I'm too careless to think about how it will be when reading this question. Thanks for editing, and I'm sorry. $\endgroup$
    – Tong Lingling
    Commented Jan 5 at 0:34
  • $\begingroup$ And there's mistake in the last two paragraph about trivial solution, it should be the digital sum of square of n , instead of n, equals m. I'm really careless. $\endgroup$
    – Tong Lingling
    Commented Jan 5 at 0:46
  • $\begingroup$ Just to clarify — you're asking about squares with digital sum 9? $\endgroup$
    – Daniel Weber
    Commented Jan 5 at 9:35
  • $\begingroup$ Maybe you can find some useful hints in this short preprint of mine about a related topic: arxiv.org/abs/2205.10163 Good luck! $\endgroup$
    – Marco Ripà
    Commented Jan 6 at 0:46

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