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

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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.
 
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 ==
+
== Video Solution by OmegaLearn==
 
https://youtu.be/zfChnbMGLVQ?t=1266
 
https://youtu.be/zfChnbMGLVQ?t=1266
  

Revision as of 03:12, 21 January 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:

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

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 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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