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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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== Solutions ==
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Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]:
  
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=== First Solution ===
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<cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath>
  
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Their greatest common divisor is 1, so <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
  
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== Solution 2 ==
  
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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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[[Proof by contradiction]]:
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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 ===
 
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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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Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is the [[greatest common factor]] of <math>14n+3</math> and <math>21n + 4</math>. Thus,
 
<center>
 
<center>
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<math>( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1 </math>
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<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math>
 
</center>
 
</center>
  
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As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
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Subtracting the second [[equation]] from the first equation we get <math>1\equiv 0\pmod{p}</math> which is clearly absurd.
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Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
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<!--Solution by tonypr-->
  
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=== Third Solution ===
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== Solution 3 ==
  
 
[[Proof by contradiction]]:
 
[[Proof by contradiction]]:
  
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Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is a [[divisor]] of both the [[numerator]] and the [[denominator]]:
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Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]].
  
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<center>
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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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<math>14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}</math>
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In this case, <math>\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}</math>.
  
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<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math>
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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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</center>
 
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Subtracting the second [[equation]] from the first [[equation]] we get <math>1\equiv 0\pmod{p}</math> which is clearly absurd.  
 
  
 
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
 
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
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<!--Solution by tonypr-->
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<!--Solution by juanfkt93 - Juan Friss de Kereki-->
  
  
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=== Fourth Solution ===
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== Solution 4 ==
  
 
We notice that:
 
We notice that:
Line 64: Line 58:
 
Q.E.D
 
Q.E.D
  
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== Solution 5 ==
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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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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>.
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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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<math> \frac{25}{17} </math>. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16).
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Trying 2, we end up with <math> \frac{46}{31} </math>. Again, both results are 1 off from being multiples of 15.
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Trying 3, we end up with <math> \frac{67}{45} </math>. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:
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1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but
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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.
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== Solution 7 ==
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Let <math>\gcd (21n+4,14n+3)=a.</math> So for some co-prime positive integers <math>x,y</math> we have <cmath>21n+4 = ax \qquad (1)</cmath> <cmath>14n+3 = ay \qquad (2)</cmath> Multiplying <math>(1)</math> by <math>2</math> and <math>(2)</math> by <math>3</math> and then subtracting <math>(1)</math> from <math>(2)</math> we get <cmath>42n+9-(42n+8)=3ay-2ax</cmath> <cmath>\implies a(3y-2x)=1.</cmath> We must have <math>a=1</math>, since <math>a</math> is an positive integer. Thus, <math>\gcd(21n+4,14n+3)=1</math> which means the fraction is irreducible, as needed.
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== Video Solution  by OmegaLearn==
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https://youtu.be/zfChnbMGLVQ?t=1266
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 +
~ pi_is_3.14
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https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1
 +
- AMBRIGGS
 +
 +
https://youtu.be/Pfg4MVUML64
 +
 +
- little fermat
  
 
{{alternate solutions}}
 
{{alternate solutions}}
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== See Also ==
 
== See Also ==

Latest revision as of 09:59, 23 July 2023

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:

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

Their greatest common divisor is 1, so $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Solution 2

Proof by contradiction:

Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is the greatest common factor of $14n+3$ and $21n + 4$. Thus,

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

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.


Solution 7

Let $\gcd (21n+4,14n+3)=a.$ So for some co-prime positive integers $x,y$ we have \[21n+4 = ax \qquad (1)\] \[14n+3 = ay \qquad (2)\] Multiplying $(1)$ by $2$ and $(2)$ by $3$ and then subtracting $(1)$ from $(2)$ we get \[42n+9-(42n+8)=3ay-2ax\] \[\implies a(3y-2x)=1.\] We must have $a=1$, since $a$ is an positive integer. Thus, $\gcd(21n+4,14n+3)=1$ which means the fraction is irreducible, as needed.

Video Solution by OmegaLearn

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

~ pi_is_3.14

https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1 - AMBRIGGS

https://youtu.be/Pfg4MVUML64

- little fermat

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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