# 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) | ||

− | 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>. | + | 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 23:40, 23 October 2010

## Problem

Find all functions such that is a perfect square for all

*Author: Gabriel Carroll, USA*

## Solution

Suppose such function exist then:

Lemma 1)

Assume for contradiction that

has to be a perfect square

but .

A square cannot be between 2 consecutive squares. Contradiction. Thus,

Lemma 2) (we have show that it can't be 0)

Assume for contradiction, that .

Then there must exist a prime number such that and are in the same residue class modulo .

If where is not divisible by .

If .

Consider an such that

, where is not divisible by

If .

Consider an such that

, where is not divisible by

At least one of , is not divisible by . Hence,

At least one of , is divisible by an odd amount of .

Hence, that number is not a perfect square.

Thus, ,