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

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Q.E.D
 
Q.E.D
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===Fifth Solution===
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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 18:46, 14 April 2018

Problem

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


Solutions

First Solution

For this fraction to be reducible there must be a number $x$ such that $x \cdot (14n+3) = (21n+4)$, and a $1/x$ such that $1/x \cdot (21n+4) = (14n+3)$. Since $x$ can only be one number ($x$ is a linear term) we only have to evaluate $x$ for one of these equations. Using the first one, $x$ would have to equal $3/2$. However, $3*3/2$ results in $9/2$, and is not equal to our desired $4$. Since there is no $x$ to make the numerator and denominator equal, we can conclude the fraction is irreducible.

EDIT: It appears to me that this solution is incorrect because it assumes that if $ax + b = cx + d$, then $a = c$ and $b = d$ -- the problem says irreducible for ALL $n$. If you agree, please remove this solution. 12/17/2017

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 $x$ can be any number, then in $ax^2+4=6x^2+b$, we must have $(a,b)=(6,4)$. Thus, if $ax+b=cx+d$ for all values $x$, then $a$ must equal $c$ and $b$ must equal $d$.

Clarification for edit 2: In this problem, we have $14nx+3x=21n+4$, with an unknown $n$, so we must have $14x=21$ and $3x=4$, which cannot be true for any $1$ value of $x$.

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:

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

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


Fifth Solution

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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