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Number of positive integer solutions for given equation

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Let n be an odd integer ≥ 5. Find the number of triplets (x, y, z) of positive integers which satisfy the equation
    x + y + 2z = n

    2. Relevant equations
    Do not know

    3. The attempt at a solution
    Let n = 2k + 1, k ≥ 2

    x + y + 2z = n
    2z = 2k + 1 - (x + y) ≤ 2k + 1 - 2 (because x + y ≥ 2)
    2z ≤ 2k - 1
    Then stuck

    The answer is (n - 1)(n - 3)/4

    Thanks
     
  2. jcsd
  3. May 22, 2016 #2

    RUber

    User Avatar
    Homework Helper

    Sometimes these are easiest to look at with a few examples.
    If n = 5, then (1,2,1) and (2,1,1) are your only choices.
    If n = 7, you have more. For z = 1, you have (1,4,1), (2,3,1),(3,2,1),(4,1,1) and for z=2 you have (1,2,2),(2,1,2).
    What I see is that for any z, of which you are limited in choices, you will have an appropriate number of choices for (x+y)=n-2z.

    So how many options for z?
    For each of those options how many choices for y do you have? Once z and y and defined, you have only one choice for x.
     
  4. May 24, 2016 #3
    I think I get it. Thanks
     
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