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Math Help - Sum of Unite fractions

  1. #1
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    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
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by knockouwt4 View Post
    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.
    Last edited by undefined; August 28th 2010 at 10:13 PM.
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  3. #3
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    Hello, knockouwt4!

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

    \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{ \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 \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: . \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} \;=\;1

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

    and nine integers: . \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 . . .

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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Soroban View Post

    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.
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  5. #5
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    I have not found all possible solutions to this problem

    -----------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:

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

    We will have to include these solutions too . .
    Last edited by Isomorphism; August 29th 2010 at 12:24 AM.
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  6. #6
    MHF Contributor undefined's Avatar
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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.
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  7. #7
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    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]])
    Last edited by undefined; August 29th 2010 at 02:10 AM.
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  8. #8
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    Thanks everyone for the quick replies! theyve all been so helpful. I'll let you know of any other developments I come across...
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