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

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== Problem ==
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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 14:51, 24 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.


See also

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