Difference between revisions of "1959 IMO Problems/Problem 1"
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− | == Problem == | + | <math>Insert formula here</math>== Problem == |
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>. | ||
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Let <math>g = {\rm gcd}(21n + 4, 14n + 3)</math>. Then <math>g|h</math> where <math>h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1</math>. Thus, <math>g = h = 1</math>. ''Note: This solution, in hindsight, is just the first solution above in a slightly different notation.'' | Let <math>g = {\rm gcd}(21n + 4, 14n + 3)</math>. Then <math>g|h</math> where <math>h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1</math>. Thus, <math>g = h = 1</math>. ''Note: This solution, in hindsight, is just the first solution above in a slightly different notation.'' | ||
+ | === Fifth Solution === | ||
+ | We notice that: | ||
+ | |||
+ | <center> | ||
+ | <math>\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}</math> | ||
+ | </center> | ||
+ | |||
+ | So it follows that <math>7n+1</math> and <math>14n+3</math> must be coprime for every natural number <math>n</math> for the fraction to be irreducible. Now the problem simplifies to proving <math> \frac{7n+1}{14n+3} </math> irreducible. We re-write this fraction as: | ||
+ | |||
+ | <center> | ||
+ | <math> \frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}</math> | ||
+ | </center> | ||
+ | |||
+ | Since the denominator <math> 2(7n+1) + 1 </math> differs from a multiple of the numerator <math>7n+1</math> by 1, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is irreducible. | ||
+ | |||
+ | Q.E.D | ||
+ | |||
+ | ~ Solution by ''z6z61im4s99'' | ||
+ | |||
+ | |||
+ | |||
+ | For this fraction to be irreducible, | ||
{{alternate solutions}} | {{alternate solutions}} | ||
== See Also == | == See Also == |
Revision as of 01:38, 19 August 2013
== Problem ==
Prove that the fraction is irreducible for every natural number .
Contents
[hide]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 |