Interesting problem:

Find all positive integers such that they are
equal to the square of the sum of their
digits in base 10 representation.

Brute force for up to $n=25$ gives two solutions so far, $n=1$ and $n=9$, since
$$n=1~~\Rightarrow~~1^2=1=n\
n=9~~\Rightarrow~~9^2=81~~\Rightarrow~~8+1=9$$

Question

Interesting problem:

Find all positive integers such that they are 
equal to the square of the sum of their 
digits in base 10 representation.

Brute force for up to $n=25$ gives two solutions so far, $n=1$ and $n=9$, since
$$n=1~~\Rightarrow~~1^2=1=n\
n=9~~\Rightarrow~~9^2=81~~\Rightarrow~~8+1=9$$

anonymous · Answer

your first q and ur 99?????????????????????

anonymous · Answer

Third, actually.

anonymous · Answer

Just because he answers questions doesn't mean he can't ask them.

anonymous · Answer

im not sayin that ... like really

anonymous · Answer

I don't think your math matches your question description above.

anonymous · Answer

Suppose $f(x)$ is the sum of the digits of $x$ written in base $10$.
You found $f(n^2)=n$, but the question seems to ask $n=[f(n)]^2$.

anonymous · Answer

One thing that could help is $$
1\leq f(n)\leq 9(\lfloor\log_{10}(n)\rfloor+1)
$$

anonymous · Answer

It will help you figure out if there are any boundaries for solutions.

anonymous · Answer

In this case $\lfloor  \log_{10}(n)\rfloor + 1$ is the number of digits that $n$ has.  The highest it can be is for all of them to be $9$.

anonymous · Answer

Oh I see, I misinterpreted the question... Thanks for clearing that up @wio