# Difference between revisions of "1994 IMO Problems/Problem 4"

(New page: Find all ordered pairs <math>(m,n)</math> where <math>m</math> and <math>n</math> are positive integers such that <math>\frac {n^3 + 1}{mn - 1}</math> is an integer. == Solution == Supp...) |
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Find all ordered pairs <math>(m,n)</math> where <math>m</math> and <math>n</math> are positive integers such that <math>\frac {n^3 + 1}{mn - 1}</math> is an integer. | Find all ordered pairs <math>(m,n)</math> where <math>m</math> and <math>n</math> are positive integers such that <math>\frac {n^3 + 1}{mn - 1}</math> is an integer. | ||

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== Solution == | == Solution == | ||

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From these, we have all solutions <math>(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)</math>. | From these, we have all solutions <math>(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)</math>. | ||

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+ | ==See also== |

## Revision as of 13:13, 4 October 2008

## Problem

Find all ordered pairs where and are positive integers such that is an integer.

## Solution

Suppose where is a positive integer. Then and so it is clear that . So, let where is a positive integer. Then we have which by cancelling out the s and dividing by yields . The equation is a quadratic. We are given that is one of the roots. Let be the other root. Notice that since we have that is an integer, and so from we have that is positive.

It is obvious that is a solution. Now, if not, and are all greater than , we have the inequalities and which contradicts the equations . Thus, at least one of is equal to .

If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .

If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .

From these, we have all solutions .