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

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Directly plugging <math>\displaystyle a</math> and <math>\displaystyle b</math> into the function, <math>\displaystyle 4a^2+4a=b^2+b.</math> We now have a quadratic in <math>\displaystyle a.</math> | Directly plugging <math>\displaystyle a</math> and <math>\displaystyle b</math> into the function, <math>\displaystyle 4a^2+4a=b^2+b.</math> We now have a quadratic in <math>\displaystyle a.</math> | ||

− | Applying the quadratic formula, | + | Applying the quadratic formula, <math> \displaystyle a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math> |

Because, <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. | Because, <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. | ||

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

## Revision as of 02:01, 22 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,

Because, the quantity cannot be a perfect square when is an integer. Hence, when is a positive integer, cannot be.