Difference between revisions of "1977 Canadian MO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | If <math> | + | 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> | + | 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> | + | Applying the quadratic formula, <math>a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math> |
− | In order for both <math> | + | 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}} | ||
− | == | + | {{Old CanadaMO box|before=First question|num-a=2|year=1977}} |
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 22:49, 17 November 2007
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.
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 |