f(n)=0,1,4,7 (mod 9).

Yes, but how to prove that all these values can be attained?

Rust wrote: f(n)=0,1,4,7 (mod 9). http://www.mathlinks.ro/Forum/viewtopic.php?t=86823: Rust wrote: It is equavalent $x^8-y^8=10(z^4-y^8)$. It had onlu solutution x=y=z=1. Very good and very complete solutions indeed. Just for your (and all others, there are many people doing that way) information: The next post (in Number Theory) only containing the trivial answers - or even more badly: only one special answer out of many - without any further statement or a proof (or a reasoning) or at least anything usefull will be deleted without any more comment. @all: at this case there is at least a small bit of information given, but answering to "find all solutions of $x^{1337}+y^{1337}=z^{1337}$" with "one answer is $x=y=z=0$" is pure spamming! [sarcasm] Note: such a very ingenious remark is surely much too complicated to be found out by others themself... [/sarcasm] It's understandable that there are users new to the forum not knowing about how,where and what to post. But from someone having more than 400 posts, I think one can expect to know that (and it was told to you before; I would not complain if this would be the first time). I think you can solve it, but either post a solution/a sketch of it or don't post. A remark to the problem: showing that $451132$ and $451332$ are in the image was the hardest problem in the third round of the german mathematical olympiad, grades 11/12/13.

It is obviosly $s(n)=n (mod \ 9)$, therefore ${s(n^2}=0,1,4.7 (moa \ 9)$. Let $n=10^k-i,i=1,2,3,5$. We have $n^2=99..9 (k-1 \ times)(10-2i)0000(k-1\ or \ k-2 \ times)i^2, s(n^2)=9k-2i+1+s(i^2)$ It give $s(n^2)=9k,9k+1,9k+4, 9k-2 (k>0)$. s(0^2)=0,s(1^2)=1,s(2^2)=4.

we can also prove it by using some well-known result . we know that $11\cdots 1122\cdots 225$ ($n$ times $1$ ,$n+1$ times $2$) is a perfect square . And the sum of digit is $n+2(n+1)+5=3n+7$ . Also from $(10^n-1)^2=100^n-2\cdot 10^n+1=99\cdots 9800\cdots 001$ ($n-1$ times $9$ ,$n-1$ times $0$) , we know that the sum of digit is $9(n-1)+8+1=9n$ . Hence $f(n)=0,1,4,7\mod 9$