Prove that the fraction (21n + 4)/(14n + 3) is irreducible

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jlacosta

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Mathandski 3

N 21 minutes ago by ihategeo_1969

Source: Friend at school gave it to me

Given a continuous function $f$ such that $f(2x) = 3 f(x)$ and $\int_0^1 f(x) \, dx = 1$ , evaluate $\int_1^2 f(x) \, dx$ .

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ihategeo_1969

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KhuongTrang 4

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Source: own-CRUX

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KhuongTrang

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MellowMelon 14

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36 minutes ago

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reni_wee

an hour ago

ihategeo_1969

38 minutes ago

No more topics!

Prove that the fraction (21n + 4)/(14n + 3) is irreducible

DPopov 110

N Apr 22, 2025 by Shenhax

Source: IMO 1959 #1

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

110 replies

DPopov

Oct 5, 2005

Shenhax

Apr 22, 2025

Prove that the fraction (21n + 4)/(14n + 3) is irreducible

G H J

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Source: IMO 1959 #1

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DPopov

1398 posts

DPopov

#1 h Oct 5, 2005, 11:41 PM • 22 Y

Y by bowenying24, MathLuis, Adventure10, Adventure10, andyloo666, jhu08, centslordm, cttg8217, mathlearner2357, RedFireTruck, megarnie, rayfish, HWenslawski, ryanbear, OronSH, peelybonehead, Mango247, MarioLuigi8972, aidan0626, DensSv, ItsBesi, Sedro

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

This post has been edited 1 time. Last edited by DPopov, Oct 6, 2005, 12:13 AM

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MithsApprentice

2390 posts

MithsApprentice

#2 h Oct 5, 2005, 11:53 PM • 10 Y

Y by Kenkang1997, myh2910, Adventure10, jhu08, RedFireTruck, HWenslawski, ihatemath123, MarioLuigi8972, MathIQ., Ali_369

Let $k$ be a common factor, greater than zero, of both the numerator and the denominator. Let
$21n+4=Ak$
and
$13n+3=Bk$
Then
$k(3B-2A)=3Bk-2Ak=(42n+9)-(42n+8)=1$
The first expression is divisible by $k$ , so $k$ must be a divisor of 1. Thus the greatest common factor of the numerator and the denominator is one, and the fraction is not reducible.

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MithsApprentice

2390 posts

MithsApprentice

#3 h Oct 5, 2005, 11:53 PM • 6 Y

Y by Kenkang1997, nhusanboev, myh2910, Adventure10, jhu08, RedFireTruck

Using Euclid's algorithm for determining the greatest common divisor of the two numbers we obtain

$21n+4=1 \cdot (14n+3)+(7n+1)$
$14n+3=2 \cdot (7n+1)+1$
$7n+1=1 \cdot (2n+1)+0$

The greatest common divisor of the numerator and denominator is one. Therefore the given fraction is irreducible.

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bomb

365 posts

bomb

#4 h Oct 6, 2005, 6:49 AM • 6 Y

Y by artificial, Adventure10, jhu08, SanderDragon, RedFireTruck, DensSv

Easiest IMO prob. ever.

We have 2(21n+4)-3(14n+3)=-1, o we are done...

Bomb

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Valentin Vornicu

7301 posts

Valentin Vornicu

#5 h Oct 6, 2005, 6:55 AM • 24 Y

Y by ssk9208, RadioActive, claserken, imowinner, myh2910, heheXD1, Wizkid8402, OlympusHero, Adventure10, jhu08, centslordm, RedFireTruck, Toinfinity, Gnisrepfa2, michaelwenquan, shujingsheng, discretemathcombinatorics, judgefan99, cberybsb, aidan0626, DensSv, Sedro, Novmath, and 1 other user

bomb wrote:

Easiest IMO prob. ever.

Well it was the first imo problem .... ever

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bomb

365 posts

bomb

#6 h Oct 6, 2005, 12:33 PM • 4 Y

Y by Adventure10, jhu08, RedFireTruck, Mango247

It was

I never realised..

Bomb

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apiarist1

32 posts

apiarist1

#7 h Apr 22, 2008, 8:53 PM • 7 Y

Y by turko.arias, jhu08, RedFireTruck, hdnlz, Adventure10, Mango247, DensSv

Of course it seems easy to you since everyone knows how to use Euclid's algorithm, but 49 years ago it was a new method, probably.

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ZetaX

7579 posts

ZetaX

#8 h Apr 22, 2008, 9:03 PM • 37 Y

Y by Binomial-theorem, 62861, DominicanAOPSer, RadioActive, AKAL3, Python54, mathguy5041, claserken, Mudkipswims42, turko.arias, Kanep, OlympusHero, Adventure10, jhu08, math31415926535, centslordm, rayfish, RedFireTruck, Toinfinity, eagles2018, Lamboreghini, jrsbr, michaelwenquan, shujingsheng, judgefan99, OronSH, aidan0626, peelybonehead, Schur-Schwartz, hdnlz, Mango247, megarnie, Sedro, Andrew2019, and 3 other users

Probably, it was new 2300 years ago

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apiarist1

32 posts

apiarist1

#9 h Apr 22, 2008, 9:16 PM • 6 Y

Y by jhu08, RedFireTruck, hdnlz, Adventure10, Mango247, DensSv

I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece.

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darij grinberg

6555 posts

darij grinberg

#10 h Apr 23, 2008, 8:34 AM • 10 Y

Y by ssk9208, myh2910, Adventure10, jhu08, RedFireTruck, michaelwenquan, shujingsheng, judgefan99, Mango247, Andrew2019

apiarist1 wrote:

I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece.

1) Ordinary kids don't go to IMO's.
2) 49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...

dg

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test20

988 posts

test20

#11 h Apr 23, 2008, 9:02 AM • 3 Y

Y by Adventure10, jhu08, RedFireTruck

darij grinberg wrote:

49 years ago people DID learn the Euclidean algorithm at school; it's rather nowadays that they don't. You will be amazed at what people learnt at school 49 years ago; the textbooks contained things that are meanwhile forgotten enough that you can pose them as IMO problems...

Could you provide some examples for such IMO problems?

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sargeist

13 posts

sargeist

#12 h Jul 13, 2008, 8:42 PM • 5 Y

Y by jhu08, RedFireTruck, Adventure10, Mango247, and 1 other user

Hi, this is my first ever post here so please forgive me for being a little stupid (perhaps), but all I did was:

$\dfrac{21n + 4}{14n + 3} = \dfrac{14n + 3 + 7n + 1}{14n + 3} = 1 + \dfrac{7n + 1}{14n + 3} = 1 + \dfrac{7n + 1}{2(7n + 1) + 1}$

and so we see that the part by which the fraction exceeds 1 can never be reduced because the numerator does not divide the denominator.

EDIT: Sorry, what I mean is that nothing divides both the numerator and the denominator because of the extra "1" hanging around.

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yankeesrule007

123 posts

yankeesrule007

#13 h Mar 2, 2009, 7:58 AM • 3 Y

Y by jhu08, RedFireTruck, Adventure10

I'm not really sure if this is a solution. I'm not too smart so I might be wrong. What I did was assume that the fraction is reducible. Thus

(21n+4)/(14n+3)=k for some positive integer k.

Thus 21n+4=14nk+3k
7n(3-2k)=3k-4
7n =(3k-4)/(3-2k)
Since n is a natural number (3k-4)/(3-2k) must be a positive integer for some k. Unfortunately, (3k-4)/(3-2k)>0 only in the interval (4/3,3/2). We see that for (3k-4)/(3-2k) to be an integer k must be between 3/2 and 4/3. But then, k isn't an integer and we reach a contradiction. Thus (21n+4)/(14n+3) is irreducible.
I apologize for the lack of LaTeX in advance. I still have to learn.

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Yoni2b

503 posts

Yoni2b

#14 h Feb 13, 2011, 1:57 AM • 4 Y

Y by jhu08, RedFireTruck, Adventure10, Mango247

darij grinberg wrote:

apiarist1 wrote:

I mean that an ordinary kid didn't learn this method at school. I'm not ignorant - I know that Euclid lived in Ancient Greece.

People do realize this was in the soviet Union, america didn't enter till 1970's, with Paul Zeitz being first IMO member for the US (author of "Arts and Crafts of Problem Solving").

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al-mahed

58 posts

al-mahed

#15 h Feb 13, 2011, 2:14 AM • 3 Y

Y by jhu08, RedFireTruck, Adventure10

[moderator says: do not quote the entire post before yours]

... and your point is?__________________________

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EaZ_Shadow

1235 posts

EaZ_Shadow

#115 h Dec 2, 2024, 12:36 AM • 1 Y

Y by WHO_LET_ME_COOK

InftyByond wrote:

EaZ_Shadow wrote:

Problem just more easy than 2019 imo problem 1

im 75% sure ur sol is wrong on the wiki
please correct because the final step claims that all linear functions work which is incorrect

I’m pretty sure it is right because that the common difference won’t stay the same if the leading degree of the polynomial is at least 2

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polishedhardwoodtable

130 posts

polishedhardwoodtable

#116 h Dec 2, 2024, 1:25 AM

Y by

Assume otherwise. Then there is some prime $p$ such that $p|21n+4$ and $p|14n+3$ . Clearly $14n+3$ is never a multiple of $2$ or a multiple of $7$ , and clearly $21n+4$ is never a multiple of $3$ , so $p$ does not equal $2$ , $3$ , or $7$ . Now observe that $21n+4 \equiv 0 \pmod{p}$ implies that $n\equiv -\frac{4}{21} \pmod{p}$ , and $14n+3 \equiv 0 \pmod{p}$ implies that $n\equiv -\frac{3}{14}\pmod{p}$ , so we have $-\frac{4}{21}=-\frac{3}{14}\pmod{p}$ , cross-multiplying we get $56\equiv 63 \pmod{p}$ , dividing by $7$ gives $8\equiv 9 \pmod{p}$ , so $p|(9-8)=1$ , contradiction. Thus no prime exists and the fraction is irreducible.

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megahertz13

3183 posts

megahertz13

#117 h Dec 2, 2024, 3:09 AM

Y by

EaZ_Shadow wrote:

InftyByond wrote:

EaZ_Shadow wrote:

Problem just more easy than 2019 imo problem 1

im 75% sure ur sol is wrong on the wiki
please correct because the final step claims that all linear functions work which is incorrect

I’m pretty sure it is right because that the common difference won’t stay the same if the leading degree of the polynomial is at least 2

The function is not always a polynomial.

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EaZ_Shadow

1235 posts

EaZ_Shadow

#118 h Dec 12, 2024, 5:10 AM • 1 Y

Y by WHO_LET_ME_COOK

can you show it then?

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InftyByond

205 posts

InftyByond

#119 h Dec 12, 2024, 5:26 AM

Y by

EaZ_Shadow wrote:

can you show it then?

buh they're functions. not polys
which means that maybe something like $5x^\frac{1}{2}$ works. or something pathological like weirstrass.
anyways please find a different proof (maybe has to do with Cauchy? idk)

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kilobyte144

897 posts

kilobyte144

#120 h Jan 21, 2025, 6:59 PM

Y by

If $\dfrac{21n + 4}{14n + 3}$ is irreducible, we know that $\gcd(21n + 4, 14n + 3) = 1$ .

By the Euclidean Algorithm,
$\gcd(21n + 4, 14n + 3) = \gcd(7n + 1, 14n + 3) = \gcd(7n + 1, 1) = 1.$
Since $\gcd(21n, 14n + 3) = 1$ , we know that $\dfrac{21n + 4}{14n + 3}$ is irreducible for all integers $n$ . $\blacksquare$

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MathProGenius

10 posts

MathProGenius

#121 h Mar 5, 2025, 2:10 PM

Y by

It should be the easiest question. Using Euclid's Algorithm:
Gcd(21n+4;14n+3) = Gcd (7n+1;14n+3) = Gcd (7n+1;14n+3-(2*(7n+1)))= Gcd (1;7n+1) =1

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eric201291

208 posts

eric201291

#123 h Mar 5, 2025, 2:18 PM • 1 Y

Y by MathProGenius

（21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.

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MathProGenius

10 posts

MathProGenius

#124 h Mar 5, 2025, 2:20 PM

Y by

Bro did you copy my solution?

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ohiorizzler1434

759 posts

ohiorizzler1434

#125 h Mar 5, 2025, 2:26 PM • 2 Y

Y by MathProGenius, Sadigly

mathprogenius, stop the accusations. just because a solution uses the euclidean algorithm doesnt mean its copying you. this solution idea is not original and is well known.

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MathProGenius

10 posts

MathProGenius

#126 h Mar 5, 2025, 2:29 PM

Y by

Alright,my bad.

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ray66

34 posts

ray66

#127 h Mar 5, 2025, 6:03 PM

Y by

$\gcd(21n+4, 14n+3) = \gcd(7n+1, 14n+3) = \gcd(7n+1, 1) = 1$ , so it's irreductable

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eric201291

208 posts

eric201291

#128 h Mar 6, 2025, 1:29 PM

Y by

ohiorizzler1434 wrote:

mathprogenius, stop the accusations. just because a solution uses the euclidean algorithm doesnt mean its copying you. this solution idea is not original and is well known.

TYSM， I didn't copy.

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Levieee

215 posts

Levieee

#129 h Apr 1, 2025, 12:31 PM

Y by

$\textbf{Only IMO problem that i can do smh}$ :flex:

assume $\frac{21n+14}{14n+3}$ is reducible, then $gcd(21n+14,14n+3)=p$ $p>1, p \in \mathbb{N}$
$gcd(21n+4-(14n+3),14n+3)=p$
$gcd(7n+1,7n+2)=p$
$\rightarrow \leftarrow$
since two consecutive numbers cannot have $gcd \neq 1$
$\blacksquare$
:rotfl: