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.
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)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •||Followed by|