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. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Suppose there exist positive integral <math>a</math> and <math>b</math> such that <math>4f(a) = f(b)</math>. | ||
+ | |||
+ | Thus, <math>4a^2 + 4a = b^2 + b</math>, or <math>(2a+1)^2 = b^2 + b + 1</math>. Then in order for the original equation to be true, <math>b^{2} + b + 1</math> would have to be a perfect square. Completing the square of <math>b^{2} + b + 1</math> results in <math>(b+1/2)^{2} + 3/4</math>. Thus, <math>b^{2} + b + 1</math> is not a perfect square, and thus there is no <math>b</math> that satisfies <math>4f(a) = f(b)</math>. | ||
==Alternate Solutions?== | ==Alternate Solutions?== |
Revision as of 23:26, 18 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.
Solution 2
Suppose there exist positive integral and
such that
.
Thus, , or
. Then in order for the original equation to be true,
would have to be a perfect square. Completing the square of
results in
. Thus,
is not a perfect square, and thus there is no
that satisfies
.
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 |