Difference between revisions of "1959 IMO Problems/Problem 1"
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=== Solution === | === Solution === | ||
− | Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]] | + | Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]: |
− | < | + | <cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath> |
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− | + | It follows that <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | |
=== Solution 2 === | === Solution 2 === |
Revision as of 14:11, 15 December 2019
Contents
[hide]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:
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 |