# Difference between revisions of "1977 Canadian MO Problems/Problem 1"

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In order for both <math>\displaystyle a</math> and <math>\displaystyle b</math> to be integers, the [[discriminant]] must be a [[perfect square]]. However, since <math>\displaystyle b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>\displaystyle b^2+b+1</math> cannot be a perfect square when <math>\displaystyle b</math> is an integer. Hence, when <math>\displaystyle b</math> is a positive integer, <math>\displaystyle a</math> cannot be. | In order for both <math>\displaystyle a</math> and <math>\displaystyle b</math> to be integers, the [[discriminant]] must be a [[perfect square]]. However, since <math>\displaystyle b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>\displaystyle b^2+b+1</math> cannot be a perfect square when <math>\displaystyle b</math> is an integer. Hence, when <math>\displaystyle b</math> is a positive integer, <math>\displaystyle a</math> cannot be. | ||

+ | {{alternate solutions}} | ||

== See also == | == See also == | ||

+ | |||

+ | |||

+ | [[Category:Olympiad Algebra Problems]] |

## Revision as of 22:07, 25 July 2006

## Problem

If prove that the equation has no solutions in positive integers and

## Solution

Directly plugging and into the function, We now have a quadratic in

Applying the quadratic formula,

In order for both and to be integers, the discriminant must be a perfect square. However, since the quantity cannot be a perfect square when is an integer. Hence, when is a positive integer, cannot be.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*