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Difference between revisions of "2010 IMO Problems/Problem 3"

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Lemma 2) <math>|g(m)-g(m+1)| = 1</math> (we have show that it can't be 0)
 
Lemma 2) <math>|g(m)-g(m+1)| = 1</math> (we have show that it can't be 0)
  
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  Assume for contradiction, that <math>|g(m)-g(m+1)| > 1</math>. Then there must exist a prime number <math>p</math> such that <math>g(m)</math> and <math>g(m+1)</math> are in the same residue class modulo <math>p</math>.  
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  Assume for contradiction, that <math>|g(m)-g(m+1)| > 1</math>.  
 +
 
 +
Then there must exist a prime number <math>p</math> such that <math>g(m)</math> and <math>g(m+1)</math> are in the same residue class modulo <math>p</math>.  
  
 
If <math>|g(m)-g(m+1)| = p^aq</math> where <math>q</math> is not divisible by <math>p</math>.  
 
If <math>|g(m)-g(m+1)| = p^aq</math> where <math>q</math> is not divisible by <math>p</math>.  

Revision as of 22:40, 23 October 2010

Problem

Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $\left(g(m)+n\right)\left(g(n)+m\right)$ is a perfect square for all $m,n\in\mathbb{N}.$

Author: Gabriel Carroll, USA

Solution

Suppose such function $g$ exist then:

Lemma 1) $g(m) \ne g(m+1)$ 

Assume for contradiction that  $g(m) = g(m+1)$

$\left(g(m+1)+m\right)\left(g(m)+m+1\right)$ has to be a perfect square

but $\left(g(m)+m\right)^2<\left(g(m+1)+m\right)\left(g(m)+m+1\right)<\left(g(m)+m+1\right)^2$. 

A square cannot be between 2 consecutive squares. Contradiction. Thus,  $g(m) \ne g(m+1)$



Lemma 2) $|g(m)-g(m+1)| = 1$ (we have show that it can't be 0) 

Assume for contradiction, that $|g(m)-g(m+1)| > 1$.

Then there must exist a prime number $p$ such that $g(m)$ and $g(m+1)$ are in the same residue class modulo $p$.

If $|g(m)-g(m+1)| = p^aq$ where $q$ is not divisible by $p$. 


If $a=1$.

Consider an $n$ such that $g(m)+n =p^3$

$g(m+1)+n = p^3 \pm pq =p (r)$ , where $r$ is not divisible by $p$




If $a>1$.

Consider an $n$ such that $g(m)+n =p$

$g(m+1)+n = p \pm p^aq =p (r)$ , where $r$ is not divisible by $p$




At least one of $g(n)+m$ , $g(n)+m+1$ is not divisible by $p$. Hence,

At least one of $(g(m+1)+n )(g(n)+m +1)$,  $(g(m)+n )(g(n)+m)$ is divisible by an odd amount of $p$.

Hence, that number is not a perfect square. 



Thus, $g(x)=x+k$, $k\in\mathbb{N}$
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