K 11

Peter

Find all functions $f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $m,n\in \mathbb{N}_{0}$ : $mf(n)+nf(m)=(m+n)f(m^{2}+n^{2}).$

**Posted:** Thu May 24, 2007 4:25 pm

rem

Solution: $f(x)=c$ where $c\in\mathbb{N}_{0}$ for any $x\in\mathbb{N}_{0}$

First if we plug in $m=0$ and $n=1$ we get $f(0)=f(1)$

Now, assume there exist numbers $a,b\in\mathbb{N}_{0}$ not equal and neither equal to $0$ such that $f(a)$ is not equal to $f(b)$ wolog $f(a)>f(b)$ .
If we plug in $a=m$ and $b=n$ in the original equation we get: $af(b)+bf(a)=(a+b)f(a^{2}+b^{2})$ .
However, as $f(a)>f(b)$ we have $(a+b)f(b)<af(b)+bf(a)<(a+b)f(a)$ .
Then, $(a+b)f(b)<(a+b)f(a^{2}+b^{2})<(a+b)f(a)$ , and diving through by $a+b$ we get:
$f(b)<f(a^{2}+b^{2})<f(a)$ but then we can just repeat this procedure with $a$ and $a^{2}+b^{2}$ to get another number $c=(a^{2}+b^{2})^{2}+a^{2}$ so that $f(b)<f(x)<f(a^{2}+b^{2})$ and we can repeat this forever and because the function is $\mathbb{N}_{0}\to\mathbb{N}_{0}$ then we get infinitely many natural numbers in the range $(f(b),f(a)$ which is impossible. Therefore we can never have two numbers $a,b\in\mathbb{N}_{0}$ not equal and neither equal to $0$ so that $f(a)$ is not equal to $f(b)$ . So, $f(x)$ is constant, QED.

**Posted:** Thu Sep 20, 2007 4:17 pm

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Alexander Remorov