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Difference between revisions of "1959 IMO Problems/Problem 1"

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== Problem ==
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<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.''
  
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=== Fifth Solution ===
  
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We notice that:
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<center>
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<math>\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}</math>
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</center>
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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:
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<center>
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<math> \frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}</math>
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</center>
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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.
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Q.E.D
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~ Solution by ''z6z61im4s99''
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For this fraction to be irreducible,
 
{{alternate solutions}}
 
{{alternate solutions}}
 
== See Also ==
 
== See Also ==

Revision as of 01:38, 19 August 2013

$Insert formula here$== Problem ==

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.


Solutions

First Solution

We observe that

$3(14n+3) = 2(21n+4) + 1.$


Since a multiple of $14n+3$ differs from a multiple of $21n+4$ by 1, we cannot have any postive integer greater than 1 simultaneously divide $14n+3$ and $21n+4$. 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 $a, b$ as $(a,b)$, we use the Euclidean algorithm as follows:

$( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1$

As in the first solution, it follows that $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Third Solution

Proof by contradiction:

Let's assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is a divisor of both the numerator and the denominator:

$14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}$

$21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}$

Subtracting the second equation from the first equation we get $1\equiv 0\pmod{p}$ which is clearly absurd.

Hence $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.


Fourth Solution

Let $g = {\rm gcd}(21n + 4, 14n + 3)$. Then $g|h$ where $h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1$. Thus, $g = h = 1$. Note: This solution, in hindsight, is just the first solution above in a slightly different notation.

Fifth Solution

We notice that:

$\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}$

So it follows that $7n+1$ and $14n+3$ must be coprime for every natural number $n$ for the fraction to be irreducible. Now the problem simplifies to proving $\frac{7n+1}{14n+3}$ irreducible. We re-write this fraction as:

$\frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}$

Since the denominator $2(7n+1) + 1$ differs from a multiple of the numerator $7n+1$ by 1, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that $\frac{21n+4}{14n+3}$ 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
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