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.

1959 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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