26th Balkan Mathematical Olympiad

First we prove that $f$ is injective. In fact, for any fixed $n$, if $f(m_1)=f(m_2)$, then: $m_1^2+2n^2=f(f(m_1{)}^2+2f(n{)}^2)=f(f(m_2{)}^2+2f(n{)}^2)=m_2^2+2n^2$, whence $m_1^2=m_2^2$ and $m_1=m_2$.

Since $f$ is injective we have: $$f(m{)}^2+2f(n{)}^2=f(p{)}^2+2f(q{)}^2\iff m^2+2n^2=p^2+2q^2(1)$$ Putting $f(1)=a$, we find for $m=n=1$, $f(3a^2)=3$. Then from (1) we get $$f(5a^2{)}^2+2f(a^2{)}^2=f(3a^2{)}^2+2f(3a^2{)}^2=3f(3a^2{)}^2=27.$$ Since the solution of the equation $x^2+2y^2=27$ in positive integers is $(x,y)=(3,3)$ and $(x,y)=(5,1)$, it follows that $f(a^2)=1$ and $f(5a^2)=5$.

Also, from (1) we have $$2f(4a^2{)}^2-2f(2a^2{)}^2=f(5a^2{)}^2-f(a^2{)}^2=24.$$ Since the unique solution in positive integers of the equation $x^2-y^2=12$ is $(x,y)=(4,2)$, it follows that $f(2a^2)=2$ and $f(4a^2)=4$.

Using (1) again we can have $$f((k+4)a^2{)}^2=2f((k+3)a^2{)}^2-2f((k+1)a^2{)}^2+f(k{a}^2{)}^2,$$ as it arises easily by the identity $(k+4{)}^2+2(k+1{)}^2=k^2+2(k+3{)}^2$, and therefore $f(k{a}^2)=k$, by induction on $k$. Then $f(a^2)=a=f(1)$ and thus $a=1$. Hence $f(k)=k$, for any $k\in {\mathbb{N}}^{\ast }$. Finally it is easy to verify that $f(k)=k$, is a solution of the problem.