1977 Canadian MO Problems/Problem 1
If prove that the equation has no solutions in positive integers and
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.
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 .
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)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •||Followed by|