1959 IMO Problems/Problem 1
Contents
Problem
Prove that the fraction is irreducible for every natural number .
Solutions
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.
Solution 2
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.
Solution 3
Assume that is a reducible fraction.
If a certain fraction is reducible, then the fraction is reducible, too. In this case, .
This fraction consists of two consecutives numbers, which never share any factor. So in this case, is irreducible, which is absurd.
Hence is irreducible. Q.E.D.
Solution 4
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
Solution 5
By Bezout's Lemma, , so the GCD of the numerator and denominator is and the fraction is irreducible.
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 |