1977 Canadian MO Problems/Problem 1
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 Solution
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) | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 2 |