F.E (Canada-2002)

For discussing Olympiad Level Number Theory problems
mutasimmim
Posts:107
Joined:Sun Dec 12, 2010 10:46 am
F.E (Canada-2002)

Unread post by mutasimmim » Tue Aug 26, 2014 9:53 pm

Find all functions $ f:N_0\rightarrow N_0 $ such that for all nonnegative integers $m,n$ the following holds.
$mf(n)+nf(m)=(m+n)f(m^2+n^2)$

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SANZEED
Posts:550
Joined:Wed Dec 28, 2011 6:45 pm
Location:Mymensingh, Bangladesh

Re: F.E (Canada-2002)

Unread post by SANZEED » Tue Aug 26, 2014 11:04 pm

Let us denote the statement $mf(n)+nf(m)=(m+n)f(m^{2}+n^{2})$ by $P(m,n)$. Then,
$P(0,n)\Rightarrow nf(0)=nf(n^{2})$. So $f(n^{2})=f(0) \forall n\in\mathbb{N}_{0}$.
Now let $n_{0}$ be any non-negative integer. Let us take $m_{0}$ such that $(m_{0},n_{0})=1$. Then clearly $(n_{0},m_{0}^{2}+n_{0})=1$.
Now $P(m_{0}^{2},n_{0})\Rightarrow m_{0}^{2}f(n)+nf(0)=(m_{0}^{2}+n_{0})f(m_{0}^{4}+n_{0}^{2})$.
This means $m_{0}^{2}f(n)+nf(0)\equiv 0(mod (m_{0}^{2}+n_{0}))$. Thus,
$nf(0)-nf(n)=n(f(0)-f(n))\equiv 0(mod (m_{0}^{2}+n_{0}))$. But $(n_{0},m_{0}^{2}+n_{0})=1$. So,
$m_{0}^{2}+n_{0}|f(0)-f(n)$. Now taking $m_{0}$ as large as needed, we get that,
$f(n)=f(0)\forall n\in\mathbb{N}_{0}$.
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Nirjhor
Posts:136
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Location:Varies.

Re: F.E (Canada-2002)

Unread post by Nirjhor » Tue Aug 26, 2014 11:40 pm

We take the pair \((x,y)\in\mathbb{N}_0^2\) such that \(f(x)>f(y)\). Let \(f(x)-f(y)=d\) be the minimum possible. Now notice that
\[\begin{eqnarray}
(x+y)f(x) =xf(x)+yf(x)&>&xf(y)+yf(x) =(x+y)f\left(x^2+y^2\right)\\ &>& xf(y)+yf(y)=(x+y)f(y)
\end{eqnarray}\] implying that \(f(x)>f\left(x^2+y^2\right)>f(y)\) which is \(d=f(x)-f(y)>f\left(x^2+y^2\right)-f(y)>0\). Hence \(f\left(x^2+y^2\right)-f(y)<d\) leading to a contradiction. So the pair \((x,y)\in\mathbb{N}_0^2\) such that \(f(x)>f(y)\) doesn't exist. Hence \(f(x)=f(y)\) for all \((x,y)\in\mathbb{N}_0^2\) and verify that any constant function works.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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