Question

Find the natural numbers a,b,c that satisfy
4/123 = 1/a + 1/b + 1/c

Answer 1

This problem turns out to be insanely difficult. There is an interesting open problem. We don't know yet whether below equation has solutions for every \(n\):
\[ 4/n = 1/a+1/b+1/c\]

Answer 2

dammit ganeshie!

Answer 3

i'm not even sure where to start..
multiplying 123abc through out feels like a bad idea!

wolfram is able to solve these for large n's quite comfortably
http://www.wolframalpha.com/input/?i=solve+4%2F123456789+%3D+1%2Fx%2B1%2Fy%2B1%2Fz,x%3E0,y%3E0,z%3E0+over+integers

Answer 4

Haha I sadly recognized this the second I saw it, I have tried to solve this one before. I'll give it another shot though. :D

Answer 5

yeah i do remember... there might exist nice ways to work few specific cases

here 3 must divide one of the numbers a, b, c
other than that i couldn't make any useful observations

Answer 6

\[(x+a)(x+b)(x+c)=x^3+(ab+bc+ac)x^2+(a+b+c)x+abc\]
Sorta starting at trying to use this somehow maybe

Answer 7

uhhh that's wrong haha

Answer 8

\[(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc\]

Answer 9

\[123(ab+bc+ca) = 4abc\]

WLOG we may let \(a = 41a'\) and get
\[3(ab+bc+ca) = 4a'bc\]

Answer 10

Nahh that just makes it more complicated as left hand side becomes worse after replacing a by 41a'

Answer 11

Haha reminds me of the arithmetic derivative if a,b, and c were primes \((abc)'=ab+bc+ca\) but I won't go there it's useless.

Answer 12

Here's an idea, not sure it's a large system of equations.
\[\small (+++) = (x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc\]\[\small (-++) = (x-a)(x+b)(x+c)=x^3+(-a+b+c)x^2+(-ab+bc-ac)x-abc\]\[\small (+-+) = (x+a)(x-b)(x+c)=x^3+(a-b+c)x^2+(-ab-bc+ac)x-abc\]\[\cdots\]\[\small (---) = (x-a)(x-b)(x-c)=x^3+(-a-b-c)x^2+(ab+bc+ac)x-abc\]
So using these 6 vectors with the basis \(\{1, x,x^2, x^3 \}\) I was thinking we could by some linear combination of these find some expressions for:

\[0=123(ab+bc+ac)-4abc\]

Answer 13

Ah I should have been more clear.
\[M: V \to U\]\(V=\{(+++),(-++),(+-+),(++-),(--+),(-+-),(+--),(---)\}\) \[U=\{1,x,x^2,x^3\}\]

So this means M is a 6x4 matrix mapping from a vector with coefficients on the basis vectors in V to U. We could write it out with \(v \in V\) and \(u \in U\) as:

\[Mv=u\]
and \[u=\begin{pmatrix} -4abc \\ 123(ab+bc+ac) \\ 0 \\ 0\end{pmatrix}\]

Answer 14

4x6 matrix sorry. since v is a 6x1 vector mapped to a 4x1 vector.

Answer 15

Wow, I still don't know how to count. There are clearly 8 vectors there I was thinking 3! = 1+3+3+1 but nevermind that. It should be 8x4... At any rate!

I'll admit this doesn't really turn the problem into something that looks nicer or easier or even integers necessarily so what good is it idk.

It might be better to combine the groups of 3 together so that it's a square matrix, since these coefficients will have to be the same anyways I think by symmetry... Phew...

So I'll go ahead and just define the new basis to be that way. So

\(\small W = \{ (+++), (++-)+(+-+)+(-++), (--+)+(-+-)+(+--), (---)\}\)
and then the same V as before,
\[M: W \to V\]
M is 4x4 matrix now, v is the same as before haha... Ok I am almost entirely certain I'm talking to myself now in my own world so maybe I should have left this on paper until I had actually gotten somewhere... Oh well just brainstorming for fun I just thought I'd share :X

Answer 16

Sorry was away... looks these transformations are giving us something concrete to play with..

Answer 17

I'm still going through your replies
i think the solution, if it exists would be some sort of an algorithm like fibonacci greedy algorithm.

using fibonacci greedy algorithm we can write 1/123 as 1/31 + 1/3813. But this is only sum of two fractions.. The number of fractions we get using this algorithm seems to be random

Answer 18

i was confused
you mean 4/123 as 1/31+1/3813
:p

Answer 19

typo.. yeah 4/123 = 1/31 + 1/3813

4/123 can be thought of as distributing 4 huge pizzas to 123 people

If we give them all maximum possible equal pieces, each piece must be of size 1/31. ( since 1/30 > 4/123)

Then we will be left with
4/123 - 1/31 = 1/3813

Answer 20

since the numerator is 1, we're done... we get only two fractions using this method for 4/123

Answer 21

hey so was the thing you somehow used:
\[\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a(a+1)}\]

Answer 22

I was trying to find the algorithm

Answer 23

\[\dfrac{4}{123} = \dfrac{1}{123/4} \lt \dfrac{1}{30}\]

clearly we can't distribute 1/30 of the pizza to the 123 ppl
so we do the next best thing, give 1/31 th of the pizza to each :

\[\dfrac{4}{123} = \dfrac{1}{31} + x\]

Answer 24

\[\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a(a+1)}\]

Answer 25

does that mean we can split an unit fraction like 1/31 into two unit fractions ? xD

Answer 26

yep i guess so! :p

Answer 27

\[\dfrac{4}{123} = \color{red}{\dfrac{1}{31}} + \dfrac{1}{3813} = \color{red}{\dfrac{1}{31+1} +\dfrac{1}{31(31+1)} } + \dfrac{1}{3813} \]

so \(a=32, b=992,c=3813\)

Simply awesome!

Answer 28

That is really a brilliant observation!

however, we're doomed if the fibonacci algorithm gives more than 3 fractions

Answer 29

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html?#section11.3
did you see this site @ganeshie8

Answer 30

They also talked about writing 4/n as a sum of 3 unit fractions

Answer 31

I guess there is like some sequence of solutions

Answer 32

http://oeis.org/A073101
how do I easily find the 123rd number in this sequence ?

Answer 33

That website on egyptian fractions looks very interesting...

Answer 34

4/72 has 191 different solutions !
I don't see an easy way to figure out this hmm

Answer 35

I've never heard of this fibonacci greedy algorithm thing before what is it and how can I learn more?

Answer 36

All I got is that it has something to do with pizza @Kainui :p

Answer 37

HAha ok wait there's actually a link to it on that cool site http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html?#section6

Answer 38

Consider the fraction 4/11

distributing 4 pizzas to 11 people`

Answer 39

we feed the people in installments giving them maximum possible equal amounts each time

Answer 40

\[\dfrac{4}{11} = \dfrac{1}{11/4} \lt \dfrac{1}{2}\]

Clearly we cannot give 1/2 pizza to each of them.
So we just give 1/3 pizza to each and save the remaining pizza :
\[\dfrac{4}{11} = \dfrac{1}{3} + x\]

Answer 41

Next we work with \(x\) :
\[x = \dfrac{4}{11} -\dfrac{1}{3} =\dfrac{1}{33} \]

Answer 42

Again the algorithm has ended in just 2 steps
i was hoping for it take more than 2 steps so that the structure of the algorithm becomes clear

Answer 43

It's kinda clear what's going on, sorta like how I might imagine one way of converting a number from base 10 to base 2 or something maybe... Well very loosely speaking lol

Answer 44

ok I did it
I just done something dumb earlier when I tried to do it on another problem
\[\frac{5}{158}=\frac{1}{\frac{158}{5}}< \frac{1}{31} \\ \frac{5}{158}=\frac{1}{32}+x \\ \implies x=\frac{1}{2528} \\ \frac{5}{158}=\frac{1}{32}+\frac{1}{2528}\]

Answer 45

Yeah similar to converting decimal to other bases
the numerators form a decreasing sequence and eventually terminate to 1

Answer 46

Nice!

Answer 47

Looks \(\dfrac{4}{13}\) takes more than 3 steps

Answer 48

so in general it goes like this
\[\frac{a}{b}=\frac{1}{\frac{b}{a}}<\frac{1}{[\frac{b}{a}]} \\ \frac{a}{b}=\frac{1}{[\frac{b}{a}]+1}+x \\ x=\frac{a}{b}-\frac{1}{[\frac{b}{a}]+1}\]
and I guess if we simplify x we will get 1/something
hopefully

Answer 49

[ ] I used this for floor function

Answer 50

Looks perfect!

`floor(x) + 1 = ceil(x)`

Answer 51

Is it easy to prove that the numerators eventually terminate to 1

Answer 52

\(\dfrac{\color{red}{4}}{13} = \dfrac{1}{4} + x_1 \implies x_1 = \dfrac{3}{13*4}\)

\(x_1= \dfrac{\color{red}{3}}{13*4} = \dfrac{1}{18}+x_2 \implies x_2 = \dfrac{2}{13*4*18} \)

\( x_2=\dfrac{\color{red}{2}}{13*4*18} = \dfrac{1}{13*4*9}+x_3 \implies x_3 = 0 \)

The red numbers seems to form a strictly decreasing sequence with the limit \(1\)..

Answer 53

Interesting, this seems this might be the same problem as using these functions:

\[f(x)=x+1\]\[f^{-1}(x)=x-1\]\[g(x)=\frac{-1}{x}\]

and representing every number as a composition of these 3 functions only. It's kind of like using the Euclidean algorithm or something. I don't know I'm only speaking intuitively and everything I've just said in this post could be 100% wrong.

Answer 54

Hey I think we have solved the conjecture !

Answer 55

( at least till one of us finds the flaw in my following reasoning )

Answer 56

Ahahaha I don't understand the reasoning yet I'm sorta scatterbrained thinking between reading what you wrote, reading what's on that egyptian fractions page, and my own separate work!

Answer 57

Step1 :
@myininaya has showed us how to split a unit fraction into two unit fractions using the identity :
\[\dfrac{1}{a} = \dfrac{1}{(a+1)} + \dfrac{1}{a(a+1)}\]

Answer 58

Step2 :
"Assuming" the numerators in the fibonacci greedy algorithm form a strictly decreasing sequence terminating at 1, we can conclude that the number of fractions given by the fibonacci algorithm for the input of form 4/n must be one of 2, 3, 4.

Answer 59

Step3 :
If the number of fractions is 3, we're done.
If the number of fractions is 2, we use myininaya's identity.
If the number of fractions is 4, then we're stuck >.<

Answer 60

so that is the flaw
i haven't considered the case of 4 fractions before