Hey all. Is there anyway to find ALL such integers a,b,c,d such that a<b<c<d and

1/a + 1/b + 1/c + 1/d = 1?

I know a = 2 since if a = 3 or more then the sum will be less than one. After that I'm stuck.

Any help is appreciated.

Thanks

Printable View

- Aug 28th 2010, 07:13 PMknockouwt4Sum of Unite fractions
Hey all. Is there anyway to find ALL such integers a,b,c,d such that a<b<c<d and

1/a + 1/b + 1/c + 1/d = 1?

I know a = 2 since if a = 3 or more then the sum will be less than one. After that I'm stuck.

Any help is appreciated.

Thanks - Aug 28th 2010, 08:37 PMundefined
- Aug 28th 2010, 10:35 PMSoroban
Hello, knockouwt4!

Quote:

$\displaystyle \text{Is there anyway to find ALL such integers }a,b,c,d$

$\displaystyle \text{such that }\,a<b<c<d\,\text{ and }\,\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} \:=\:1\,?$

I found four solution so far:

. . $\displaystyle \displaystyle{ \frac{1}{2} + \frac{1}{3} + \frac{1}{9}+ \frac{1}{18}\;=\;1\qquad\qquad\frac{1}{2}+\frac{1} {4}+\frac{1}{5}+\frac{1}{20} \;=\;1$

. . $\displaystyle \displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24} \;=\;1 \qquad\qquad \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42}\;=\;1 $

It can be done for three integers: .$\displaystyle \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} \;=\;1$

and five integers: .$\displaystyle \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{14} + \frac{1}{28}\;=\;1$

and nine integers: .$\displaystyle \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{31} + \frac{1}{62} + \frac{1}{124} + \frac{1}{248} + \frac{1}{496}\;=\;1$

I'll let*someone*else explain wheredenominators are coming from . . .*my*

- Aug 28th 2010, 10:57 PMundefined
- Aug 28th 2010, 11:12 PMIsomorphism
I have not found all possible solutions to this problem (Worried)

-----------EDIT--------------------

If a = 2 and b = 3 are fixed, the way to solve that part is to reduce it to:

6(a+b) - ab = 0

(a-6)(6-b)= 36

Now we can look at various factorizations of 36 and solve for (a,b) by a case by case analysis.

-----------EDIT-------------------------

However the previous posts (including the OP's post) are assuming that the integers are positive. The question allows negative integers too and here is a possible solution:

$\displaystyle \frac{1}{-6}+\frac{1}{-3}+\frac{1}{1}+\frac{1}{2} = 1$

We will have to include these solutions too . . - Aug 28th 2010, 11:13 PMundefined
Okay if I didn't make a mistake then this is the full list, gives b, c, d

Code:`3 7 42`

3 8 24

3 9 18

3 10 15

4 5 20

4 6 12

Code:`f()=for(b=3,5,for(c=b+1,11,d=1/2-1/b-1/c;if(d>0&&numerator(d)==1&&denominator(d)>c,print(b," ",c," ",denominator(d)))))`

- Aug 28th 2010, 11:42 PMundefined
Trying to move towards a complete solution: if not all of a,b,c,d are positive then either

(a,b,c,d) = (-12,2,3,4)

(a,b,c,d) = (-30,2,3,5)

or b = 1 or c = 1.

There is an infinite class of solutions

(a,b,c,d) = (-6k, -3k, 1, 2k)

where k ranges over the positive integers.

Edit: More solutions

(-2, 1, 3, 6)

(-3, 1, 4, 12) (-12, -4, 1, 3)

(-4, 1, 5, 20) (-20, -5, 1, 4)

(-4, 1, 6, 12)

(-5, 1, 6, 30) (-30, -6, 1, 5)

(-6, 1, 7, 42) (-42, -7, 1, 6)

(-6, 1, 8, 24) (-24, -8, 1, 6)

(-6, 1, 9, 18)

(-6, 1, 10, 15) (-15, -10, 1, 6)

(-7, 1, 8, 56) (-56, -8, 1, 7)

(-8, 1, 9, 72) (-72, -9, 1, 8)

(-8, 1, 10, 40) (-40, -10, 1, 8)

(-8, 1, 12, 24)

So the solution to this problem revolves around the question, find all pairs (u,v) of distinct positive integers such that 1/u + 1/v is a unit fraction; in other words, u+v divides uv.

There's a nice writeup by euler (post # 8) in the solution forum to this problem on Project Euler

http://projecteuler.net/index.php?se...roblems&id=108

but the solution forum is restricted to those who've solved that problem and entered the solution. Of course we would discard solutions of the form 1/8 + 1/8 = 1/4 but everything else would apply.

__Spoiler__: - Aug 29th 2010, 09:01 AMknockouwt4
Thanks everyone for the quick replies! theyve all been so helpful. I'll let you know of any other developments I come across...