1959 IMO Problems/Problem 1

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

Proof by contradiction:

Let's 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, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that is irreducible.

Q.E.D

~ Solution by z6z61im4s99

For this fraction to be 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