ISL-2004N3

Discussion on International Mathematical Olympiad (IMO)
mutasimmim
Posts:107
Joined:Sun Dec 12, 2010 10:46 am
ISL-2004N3

Unread post by mutasimmim » Sun Sep 07, 2014 12:29 pm

Find all functions $f:N\to N$such that for any $m,n$ we have $([f(m)]^2+f(n))\mid(m^2+n)^2$.

Nirjhor
Posts:136
Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: ISL-2004N3

Unread post by Nirjhor » Sun Sep 07, 2014 1:12 pm

Let \(P(m,n)\Longrightarrow f(m)^2+f(n)\mid \left(m^2+n\right)^2\).

So \(P(1,1)\Longrightarrow f(1)^2+f(1)\in\{2,4\}\). These cases give only one natural solution \(f(1)=1\).

Now let \(p\in\mathbb{P}\). Then \(P(1,p-1)\Longrightarrow f(p-1)+1\mid p^2\Longrightarrow f(p-1)\in\{p-1,p^2-1\}\).

If \(f(p-1)=p^2-1\) then \(P(p-1,1)\Longrightarrow p^4-2p^2+2\mid p^4-4p^3+8p^2-8p+4\). This implies \[p^4-2p^2+2\le p^4-4p^3+8p^2-8p+4\Longrightarrow 10p^2-4p^3-8p+2\ge 0\] which is never true for any \(p>1\). Hence we deduce that \(f(p-1)=p-1~\forall~p\in\mathbb{P}\).

Now take any \(n\in\mathbb{N}\) and then
\[P(p-1,n)\Longrightarrow \left((p-1)^2+n\right)^2\equiv \left(f(n)-n\right)^2\equiv 0\pmod{(p-1)^2+f(n)}.\] Since \(p\) is unbounded whilst \(f(n)-n\) is bounded, we have \(f(n)-n=0\Longrightarrow f(n)=n ~\forall ~n\in\mathbb{N}\).
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Post Reply