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

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=== Solution 3 ===
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[[Proof by contradiction]]:
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If a certain fraction <math>\dfrac{a}{b}</math> is reducible, then the fraction <math>\dfrac{2a}{3b}</math> is reducible, too.
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In this case, <math>\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}</math>.
  
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=== Solution 3 ===
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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. 
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Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
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=== Solution 4 ===
  
 
We notice that:
 
We notice that:
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===Solution 4===
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===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

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

Proof by contradiction:

If a certain fraction $\dfrac{a}{b}$ is reducible, then the fraction $\dfrac{2a}{3b}$ is reducible, too. In this case, $\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}$.

This fraction consists of two consecutives numbers, which never share any factor. So in this case, $\dfrac{2a}{3b}$ is irreducible, which is absurd.

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


Solution 4

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 5

By Bezout's Lemma, $3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1$, 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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