1959 IMO Problems/Problem 1
Contents
Problem
Prove that the fraction is irreducible for every natural number
.
Solutions
First Solution
We observe that
Since a multiple of differs from a multiple of
by 1, we cannot have any postive integer greater than 1 simultaneously divide
and
. Hence the greatest common divisor of the fraction's numerator and denominator is 1, so the fraction is irreducible. Q.E.D.
Second Solution
Denoting the greatest common divisor of as
, we use the Euclidean algorithm as follows:
As in the first solution, it follows that is irreducible. Q.E.D.
Third Solution
Assume that is a reducible fraction where
is a divisor of both the numerator and the denominator:
Subtracting the second equation from the first equation we get which is clearly absurd.
Hence is irreducible. Q.E.D.
Fourth Solution
Let . Then
where
. Thus,
. Note: This solution, in hindsight, is just the first solution above in a slightly different notation.
Fifth Solution
We notice that:
So it follows that and
must be coprime for every natural number
for the fraction to be irreducible. Now the problem simplifies to proving
irreducible. We re-write this fraction as:
Since the denominator differs from a multiple of the numerator
by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that
is irreducible.
Q.E.D
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |