Difference between revisions of "1959 IMO Problems/Problem 1"
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Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. | Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. |
Revision as of 02:39, 19 August 2013
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 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 |