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

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If <math>\displaystyle f(x)=x^2+x,</math> prove that the equation <math>\displaystyle 4f(a)=f(b)</math> has no solutions in positive integers <math>\displaystyle a</math> and <math>\displaystyle b.</math> | If <math>\displaystyle f(x)=x^2+x,</math> prove that the equation <math>\displaystyle 4f(a)=f(b)</math> has no solutions in positive integers <math>\displaystyle a</math> and <math>\displaystyle b.</math> | ||

## Revision as of 15:51, 24 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.