# Sum of Unite fractions

• Aug 28th 2010, 07:13 PM
knockouwt4
Sum 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 PM
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Quote:

Originally Posted by knockouwt4
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

Maybe not elegant but you can continue with the same reasoning you just used.

1/b+1/c+1/d = 1/2

b must be greater than 2 and less than 6, and so on.

This makes exhaustive search pretty quick, for a computer at least.

Edit: Looks like there's only one solution.
• Aug 28th 2010, 10:35 PM
Soroban
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 where my denominators are coming from . . .

• Aug 28th 2010, 10:57 PM
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Quote:

Originally Posted by Soroban

I found four solution so far:

Ah yes I made the dumb mistake of restricting d without cause. The one I found was not listed by you,

1/2 + 1/4 + 1/6 + 1/12 = 1

Maybe I'll rewrite my PARI/GP one-liner in a while.
• Aug 28th 2010, 11:12 PM
Isomorphism

-----------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 PM
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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
PARI/GP

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)))))
Edit: Didn't think of negative integers.
• Aug 28th 2010, 11:42 PM
undefined
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:

PARI/GP

Code:

f(n)=local(t=divisors(n^2)); vector((#t-1)/2, i, [n+t[i], n+n^2/t[i]])
• Aug 29th 2010, 09:01 AM
knockouwt4
Thanks everyone for the quick replies! theyve all been so helpful. I'll let you know of any other developments I come across...