Difference between revisions of "1977 Canadian MO Problems/Problem 1"
(Tag: Undo) |
(Fixed typo and added a third solution) |
||
Line 10: | Line 10: | ||
==Solution 2== | ==Solution 2== | ||
− | Suppose there exist positive | + | Suppose there exist positive integers <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>. | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | The given equation is <math>4f(a) = f(b)</math> or, <math>4a^2 + 4a = b^2 + b</math> or, <math>4a(a+1) = b(b+1)</math>. | ||
+ | |||
+ | For positive integers <math>a</math> and <math>b</math>, <math>4a(a+1)</math> is an even number because it is divisible by <math>2</math> whereas <math>b(b+1)</math> is an odd number (the product of two consecutive numbers is an odd number). Therefore, <math>a</math> and <math>b</math> both cannot be positive integers. | ||
+ | |||
+ | -ad1b314 | ||
==Alternate Solutions?== | ==Alternate Solutions?== |
Revision as of 08:47, 28 September 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 integers 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
.
Solution 3
The given equation is or,
or,
.
For positive integers and
,
is an even number because it is divisible by
whereas
is an odd number (the product of two consecutive numbers is an odd number). Therefore,
and
both cannot be positive integers.
-ad1b314
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 |