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1/a + 1/b + 1/c = 1

  1. May 27, 2008 #1

    xax

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    Need to find all the posible solutions (a,b,c) for 1/a + 1/b + 1/c = 1, a,b,c integers.
     
  2. jcsd
  3. May 27, 2008 #2

    D H

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    xax, please show your work, and post your homework questions in the correct forum.
     
  4. May 27, 2008 #3

    xax

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    D H, I thought I was in the right forum. What I did so far: b+c = (a-1)*t and b*c = a*t and (c-1)*(b-1) = t+1. I'm stuck because I get too many posibilities.
     
  5. May 27, 2008 #4

    CRGreathouse

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    You can start by assuming, for the time being, that [itex]a\le b\le c[/itex] -- you can rearrange the terms later if you need to.

    If a = 1 then the sum is too large, so [itex]2\le a\le b\le c.[/itex] If a = 2, [itex]1/b+1/c=1/2[/itex] so clearly b needs to be no more than 4 (to get 1/4 + 1/4 = 1/2). Check the cases of b = 3 and b = 4.

    If a = 3 then b and c need to be small to make the sum 1, so check cases similarly.

    Finally, rearrange the solutions you have as needed (so a need not be the smallest).
     
  6. May 27, 2008 #5

    D H

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    That is true only if a, b, c are restricted to the positive integers. The OP just said integers.
     
  7. May 27, 2008 #6

    CRGreathouse

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    True, I missed that. There is one family of solutions which includes negative numbers.
     
  8. May 27, 2008 #7
    for example if a=-b an c is arbitrary big or similar we have an approximate solution in integers to your equation.

    or simply a=-b and c=1 b=-c and a=1 a=-c and b=1
     
    Last edited: May 27, 2008
  9. May 28, 2008 #8

    xax

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    Thanks alot to all of you and expecialy to CRGreathouse. I've proven that a<=3, b<=4 and found all the solutions(there are 3 in total).
    Edit: If one of them is negative the solution is (t,1,-t). Do you think there are more?
     
    Last edited: May 28, 2008
  10. May 28, 2008 #9

    CRGreathouse

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    If two are negative you have 1 - 1/a - 1/b < 1. If three are negative you have -1/a - 1/b - 1/c < 0.
     
  11. May 28, 2008 #10

    xax

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    Yes CRGreathouse, that's why I said that only one can be negative. My question was are there other solutions when one number is negative besides (t,1,-t)?
     
  12. May 28, 2008 #11

    CRGreathouse

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    If exactly one value is negative and another value is 1, the values are (1, v, -v). Otherwise, the sum of the reciprocals of the positive values is at most 1, so not a solution (since the negative will lower the result).
     
  13. Nov 27, 2009 #12
    try to prove if there is finite amount of solutions
     
  14. Mar 22, 2010 #13
    This is easy:
    2, 3, 6
    2, 6, 3
    3, 2, 6
    3, 6, 2
    6, 2, 3
    6, 3, 2

    6 solutions..
    1/2 + 1/3 + 1/6 = 1.

    Then i got a new question.
    1/a + 1/b + 1/c = 7.
    If a, b, c, are integers, how many solutions are there?
     
  15. Mar 22, 2010 #14
    Zero! Because 1/a, 1/b and 1/c are all <=1. Therefore their sum is <=3.
     
  16. Mar 22, 2010 #15
    I thought of it as well..
    Just needed a second opinion.
    Thanks! :)
     
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