BTW, check out this test, problem 3 It seems that this is an ISL 2004 problem. All the questions of the test No that Pierre has solved this not so long time ago. THis test seemed to be a great joke.

Upps, sorry, I did not see you precised the link.

Can someone post a link with Pierre's solution? It seems the Romanian students are not paying attention to the site (dzeta included!). I can count at least 3 problems from these TSTs that are posted with full solutions on the site

Valentin, on the other hand I do not think that Romanian TSTs should show who is paying attention to the site, although they seem to...

I have proved that $ f(p-1)=p-1 $ for all prime number $p$ and I d'ont know if it can help.

Look some three pages before in the proposed problems section at one of piglfly's posts with a very suggestive title $ f^2(m)+f(n)| (m^2+n)^2$ and you will see how close you are!

Valentine,let's have a look here: http://mathlinks.ro/Forum/topic-27188.html

I'm really sorry for I don't know how to creat a link

nalpaction wrote: I'm really sorry for I don't know how to creat a link Just write it down. It will be transformed automatically in a link. Or read the FAQ of the BBCode

This problem is from Mohsen Jamali designer of problem 6 of IMO 2004.

My solution. $f(1)^2+f(1)\mid 4\Rightarrow f(1)=1$ .let $p$ be a prime and large enough. we have $f(1)^2+f(p-1)\mid p^2\Rightarrow f(p-1)=p-1$($f(p-1)$ can not be $p^2$). we have $\left.\begin{matrix} (p-1)^2+f(m)\mid ((p-1)^2+m)^2\\ (p-1)^2+f(m)\mid (p-1)^4+f(m)(p-1)^2\\ (p-1)^2+f(m)\mid (f(m)-2m)(p-1)^2-m^2\\ (p-1)^2+f(m)\mid (f(m)-2m)(p-1)^2+(f(m)-2m)f(m) \end{matrix}\right\}\Rightarrow (p-1)^2+f(m)\mid(f(m)-2m)f(m)+m^2$ so $(f(m)-2m)f(m)+m^2=0$ because $p$ is large enough.so we have $(f(m)-2m)f(m)+m^2=0$ and $f(m)=m$