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. | ||
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==Alternate Solutions?== | ==Alternate Solutions?== |
Revision as of 18:19, 3 January 2023
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?
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
1977 Canadian MO (Problems) | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 2 |