1959 IMO Problems/Problem 1
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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
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 cleary absurd.
Hence is irreducible. Q.E.D.
| 1959 IMO (Problems) | ||
| Preceded by First question | 1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
