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Difficulty with the Erdos-Straus Conjecture

  1. Oct 4, 2003 #1
    The Erdos-Straus Conjecture proposes that:

    For n >= 1, 4/n = 1/a + 1/b + 1/c, has postive integer solutions.
    n = 2k (evens)
    n = 4k+3 (odds)
    n = 4k+1 (odds)

    For 2k I have found the pattern --> if k >= 1, 4/2k = 1/2k + 1/2k + 1/k. This is simple, I know.


    I am having difficulty finding the pattern for 4k+3 (no pattern has yet been found for 4k+1, I know).

    So what is the pattern for:

    4/4k+3 = 1/a + 1/b + 1/c

    ?

    Does anyone know this pattern? If you do, please reply with such pattern.


    Thank you
    J.E.H.
     
  2. jcsd
  3. Oct 5, 2003 #2

    MathematicalPhysicist

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    n>=1
    n=2k
    it should be 2k>=1 not k>=1
     
  4. Oct 5, 2003 #3

    MathematicalPhysicist

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    maybe: 4/4k+3=1/4k+3+1/4k+3+1/0.5*(4k+3)

    but im not sure this what you are after.
     
  5. Nov 19, 2004 #4
    The denominator of the last term in loop quantum gravity's equation [0.5*(4k+3)] will never be an integer, so it's not a valid unit fraction.

    The {a,b,c} triplet I came up with for 4/(4k+3) is:

    { 2(k+1), 2(k+1), (k+1)(4k+3) }

    This works too: { k+1, 2(k+1)(4k+3), 2(k+1)(4k+3) }.

    These two triplets have the same source: Substract 1/(k+1) from 4/(4k+3) and you get 1/(k+1)(4k+3). But since you now have an equation of the form

    4/(4k+3) = 1/a + 1/b

    you need to add a third term somehow to "fit" the conjecture. You can do so by splitting one of the right-hand terms as follows:

    1/a = 1/(2a) + 1/(2a).

    I've worked on this conjecture off and on the last 20 years but have never proved it. But I have found many results, including that the smallest {n} that is a counterexample--if one exists--is prime and equal to 1 mod 24.
     
    Last edited: Nov 19, 2004
  6. Nov 19, 2004 #5
    What a weird conjecture. Anyway 4k+1 with trivial solutions (making the substitution k = k - 0.5) is [tex]\frac {4}{4k+1} = \frac {1}{2(k+0.5)(4k+1)} + \frac {1}{2(k+0.5)(4k+1)} + \frac {1}{k+0.5}[/tex]. However, for n mod 4 = 1 (5, for example) neither this equation or the author above's works. Allan Swett (http://math.uindy.edu/swett/esc.htm) appears to be the most learned in this subject. Perhaps his page will give you some insights. Also, your solution to 1/2k I don't think is entirely correct because egyptian fractions are unique usually.
     
    Last edited by a moderator: Nov 19, 2004
  7. Aug 22, 2005 #6
    It's not that hard to find three different Egyptian fractions that add up to
    4/(4k+3).

    For any fraction of the form 4/(4k+3), compute the difference with the nearest integer fraction:

    4/(4k+3) - 1/(k+1) = 1/((k+1).(4k+3)).

    If you are satisfied with the absence of 1/c you are ready:
    4/(4k+3) = 1/(k+1) + 1/((k+1).(4k+3))

    Otherwise, the problem is to write any of the two fractions at the right hand side as the sum of two other - different - Egyptian fractions. You can use

    1/N = 1/(N+1) + 1/(N.(N+1)) for any N giving two schemes:

    for N = k+1 we get
    4/(4k+3) = 1/(k+2) + 1/((k+1).(k+2)) + 1/((k+1).(4k+3)) and

    for N=(k+1).(4k+3) we get
    4/(4k+3) = 1/(k+1) + 1/((k+1).(4k+3) + 1) + 1/((k+1).(4k+3).((k+1).(4k+3) + 1)).

    Since by the same method, we can easily find solutions for even n, the real problem is in the case that n = 1 mod 4. For all but some cases, the Erdös-Straus conjecture has been proved.

    Except for the case where n = 1^2, 11^2, 13^2, 17^2, 19^2, or 23^2 mod 840. Note that these are all subcases of the case n = 1 mod 4.

    See p.34 of http://www.kevingong.com/Math/EgyptianFractions.pdf for conclusion and more references.
     
  8. Aug 22, 2005 #7

    VietDao29

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    Please recheck the problem... For n = 1:
    [tex]4 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3[/tex]
    [tex]\Leftrightarrow 4 \leq 3[/tex]!!!!!
    So it's wrong for n = 1.
    I think it should reads:
    For n > 1, 4/n = 1/a + 1/b + 1/c, has postive integer solutions.
    Viet Dao,
     
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