Difference between revisions of "1959 IMO Problems/Problem 1"
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(Added a new solution, similar to solution 2, but somewhat different. Feel free to comment on it! Love this page.) |
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| + | === Solution 3 === | ||
| + | |||
| + | [[Proof by contradiction]]: | ||
| + | |||
| + | If a certain fraction <math>\dfrac{a}{b}</math> is reducible, then the fraction <math>\dfrac{2a}{3b}</math> is reducible, too. | ||
| + | In this case, <math>\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}</math>. | ||
| − | === Solution | + | This fraction consists of two consecutives numbers, which never share any factor. So in this case, <math>\dfrac{2a}{3b}</math> is irreducible, which is absurd. |
| + | |||
| + | Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | ||
| + | <!--Solution by juanfkt93 - Juan Friss de Kereki--> | ||
| + | |||
| + | |||
| + | === Solution 4 === | ||
We notice that: | We notice that: | ||
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| − | ===Solution | + | ===Solution 5=== |
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. | 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. | ||
Revision as of 11:49, 14 October 2019
Contents
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
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 | ||
