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

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== Problem ==
 
== Problem ==
  
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>
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If <math>f(x)=x^2+x,</math> prove that the equation <math>4f(a)=f(b)</math> has no solutions in positive integers <math>a</math> and <math>b.</math>
  
  
 
== Solution ==
 
== Solution ==
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>
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Directly plugging <math>a</math> and <math>b</math> into the function, <math>4a^2+4a=b^2+b.</math> We now have a quadratic in <math>a.</math>
  
Applying the quadratic formula, <math> \displaystyle a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math>
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Applying the quadratic formula, <math>a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math>
  
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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In order for both <math>a</math> and <math>b</math> to be integers, the [[discriminant]] must be a [[perfect square]].  However, since <math>b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>b^2+b+1</math> cannot be a perfect square when <math>b</math> is an integer. Hence, when <math>b</math> is a positive integer, <math>a</math> cannot be.
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
== See also ==
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{{Old CanadaMO box|before=First question|num-a=2|year=1977}}
* [[1977 Canadian MO Problems]]
 
* [[1977 Canadian MO]]
 
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 22:49, 17 November 2007

Problem

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


Solution

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

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

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

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

1977 Canadian MO (Problems)
Preceded by
First question
1 2 3 4 5 6 7 8 Followed by
Problem 2
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