Difference between revisions of "1959 IMO Problems/Problem 1"
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===Solution 4=== | ===Solution 4=== | ||
| − | By Bezout's | + | By [[Bezout's Lemma]], <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the GCD of the numerator and denominator is <math>1</math> and the fraction is irreducible. |
{{alternate solutions}} | {{alternate solutions}} | ||
| − | |||
== See Also == | == See Also == | ||
Revision as of 21:58, 31 December 2018
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 as follows:
As in the first solution, 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
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 4
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 | ||
