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.
  
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{{alternate solutions}}
  
 
== See also ==
 
== See also ==
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[[Category:Olympiad Algebra Problems]]

Revision as of 22:07, 25 July 2006

Problem

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


Solution

Directly plugging $\displaystyle a$ and $\displaystyle b$ into the function, $\displaystyle 4a^2+4a=b^2+b.$ We now have a quadratic in $\displaystyle a.$

Applying the quadratic formula, $\displaystyle a=\frac{-1\pm \sqrt{b^2+b+1}}{2}.$

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

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

See also

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