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

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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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== Solution 1 ==
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== Video Solution ==
 
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https://youtu.be/zfChnbMGLVQ?t=1266
 
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~ pi_is_3.14
 
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=== Solution ===
 
  
 
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]:
 
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]:
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It follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
 
It follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
  
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=== Solution 2 ===
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== Solution 2 ==
  
 
[[Proof by contradiction]]:
 
[[Proof by contradiction]]:
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<!--Solution by tonypr-->
 
<!--Solution by tonypr-->
  
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=== Solution 3 ===
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== Solution 3 ==
  
 
[[Proof by contradiction]]:
 
[[Proof by contradiction]]:
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=== Solution 4 ===
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== Solution 4 ==
  
 
We notice that:
 
We notice that:
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===Solution 5===
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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.
  
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===Solution 6===
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== Solution 6 ==
 
To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with <math> \frac{21n}{14n} </math> + <math> \frac{4}{3} </math>.
 
To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with <math> \frac{21n}{14n} </math> + <math> \frac{4}{3} </math>.
 
Simplifying the fraction, we end up with.    <math> \frac{3}{2} </math>. Now combining these fractions with addition as shown before in the problem, we end up with <math> \frac{17}{6} </math>. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us
 
Simplifying the fraction, we end up with.    <math> \frac{3}{2} </math>. Now combining these fractions with addition as shown before in the problem, we end up with <math> \frac{17}{6} </math>. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us
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2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending up with results that at the end will only have a common divisor of 1.
 
2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending up with results that at the end will only have a common divisor of 1.
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== Video Solution ==
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https://youtu.be/zfChnbMGLVQ?t=1266
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~ pi_is_3.14
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 05:10, 21 July 2022

Problem

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

Solution 1

Denoting the greatest common divisor of $a, b$ as $(a,b)$, we use the Euclidean algorithm:

\[(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1\]

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

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:

Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction.

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.

Solution 6

To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with $\frac{21n}{14n}$ + $\frac{4}{3}$. Simplifying the fraction, we end up with. $\frac{3}{2}$. Now combining these fractions with addition as shown before in the problem, we end up with $\frac{17}{6}$. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us $\frac{25}{17}$. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). Trying 2, we end up with $\frac{46}{31}$. Again, both results are 1 off from being multiples of 15. Trying 3, we end up with $\frac{67}{45}$. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:

1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but

2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us $\frac{319}{199}$, we keep ending up with results that at the end will only have a common divisor of 1.


Video Solution

https://youtu.be/zfChnbMGLVQ?t=1266

~ pi_is_3.14

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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