Difference between revisions of "1977 Canadian MO Problems/Problem 1"
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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. | 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 Solution == | ||
+ | Write out the expanded form of <math>4f(a)=f(b)</math>: | ||
+ | |||
+ | <math>4a^2+4a=b^2+b</math> | ||
+ | |||
+ | Now simply factor it to get: <math>4a(a+1)=b(b+1)</math> | ||
+ | |||
+ | Subtract <math>b(b+1)</math> from both sides: <math>4a(a+1)-b(b+1)=0</math> | ||
+ | |||
+ | For the left side to equal <math>0</math>, <math>4a</math> or <math>a+1</math> must be <math>0</math> AND <math>b</math> or <math>b+1</math> must be <math>0</math>. | ||
+ | |||
+ | Set each one equal to <math>0</math> to find the possible solutions: | ||
+ | |||
+ | <math>4a=0</math> | ||
+ | |||
+ | <math>a+1=0</math> | ||
+ | |||
+ | <math>b=0</math> | ||
+ | |||
+ | <math>b+1=0</math> | ||
+ | |||
+ | Thus, <math>a</math> must be <math>0</math> or <math>-1</math>. The same applies to <math>b</math>. None of these solutions are greater than <math>0</math>. | ||
+ | |||
+ | <math>\therefore</math> There are no positive solutions for <math>a</math> or <math>b</math>, Q.E.D. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 18:46, 6 September 2008
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 Solution
Write out the expanded form of :
Now simply factor it to get:
Subtract from both sides:
For the left side to equal , or must be AND or must be .
Set each one equal to to find the possible solutions:
Thus, must be or . The same applies to . None of these solutions are greater than .
There are no positive solutions for or , Q.E.D.
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 |