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.

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.

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(n²)=9 , except n=10k, n=10^a+10^b+1, n=5×10^a+1 or n=10^a+5 ? (k,a,b are integers, k>0, a≥b≥0)

or can it be proved that there are only limited nontrivial n such f(n²)=9 ?

And when f(n²)=m, m=4,7,10,11… whether such n is still limited or not ?

(if f(n²)≡m(mod 9), then n²≡m(mod 9), only when m≡0,1,4,7(mod 9) there are integer solutions n)

trivial solution : if f(n²)=m, then f((10×n)²)=m，else if n=A×10^k+B and B²<10^k, 2AB<10^k, f(n)=m, let N= A×10^(k+1)+B then f(N²)=f(n²)=m

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

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.