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.
Solution 2=
Suppose there exist positive integral and
such that
.
Thus, , or
. Then in order for the original equation to be true,
b^{2} + b + 1
(b+1/2)^{2} + 3/4
b^{2} + b + 1
b
4f(a) = f(b)$
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 |