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Difference between revisions of "1959 IMO Problems/Problem 1"
(solution 1 incorrect, confusingly worded)
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== Solutions ==
 
== Solutions ==
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=== Solution ===
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=== First Solution ===
 
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For this fraction to be reducible there must be a number <math>x</math> such that <math>x \cdot (14n+3) = (21n+4)</math>, and a <math>1/x</math> such that <math>1/x \cdot (21n+4) = (14n+3)</math>. Since <math>x</math> can only be one number (<math>x</math> is a linear term) we only have to evaluate <math>x</math> for one of these equations. Using the first one, <math>x</math> would have to equal <math>3/2</math>. However, <math>3*3/2</math> results in <math>9/2</math>, and is not equal to our desired <math>4</math>. Since there is no <math>x</math> to make the numerator and denominator equal, we can conclude the fraction is irreducible.
 
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EDIT: It appears to me that this solution is incorrect because it assumes that if <math>ax + b = cx + d</math>, then <math>a = c</math> and <math>b = d</math> -- the problem says irreducible for ALL <math>n</math>. If you agree, please remove this solution. 12/17/2017
 
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EDIT 2: Disregard the first edit, because when comparing two polynomials (linear functions in this case), the coefficients for each term must be equal to eachother. For example, if <math>x</math> can be any number, then in <math>ax^2+4=6x^2+b</math>, we must have <math>(a,b)=(6,4)</math>. Thus, if <math>ax+b=cx+d</math> for all values <math>x</math>, then <math>a</math> must equal <math>c</math> and <math>b</math> must equal <math>d</math>.
 
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Clarification for edit 2: In this problem, we have <math>14nx+3x=21n+4</math>, with an unknown <math>n</math>, so we must have <math>14x=21</math> and <math>3x=4</math>, which cannot be true for any <math>1</math> value of <math>x</math>.
 
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=== Second Solution ===
 
  
 
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]] as follows:
 
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]] as follows:
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As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
 
As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
  
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=== Third Solution ===
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=== Solution 2 ===
  
 
[[Proof by contradiction]]:
 
[[Proof by contradiction]]:
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=== Fourth Solution ===
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=== Solution 3 ===
  
 
We notice that:
 
We notice that:
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===Fifth Solution===
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===Solution 4===
 
By Bezout's Theorem, <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the gcd of the numerator and denominator is 1 and the fraction is irreducible.
 
By Bezout's Theorem, <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the gcd of the numerator and denominator is 1 and the fraction is irreducible.
  

Revision as of 13:16, 25 May 2018

Problem

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


Solutions

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.

Solution 2

Proof by contradiction:

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.


Solution 3

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, 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 4

By Bezout's Theorem, $3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1$, so the gcd of the numerator and denominator is 1 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
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