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Homework Help: When will there exist solutions to ax + (a+2)y = c?

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data

    When will there exist solutions to ax + (a+2)y = c?

    2. Relevant equations


    3. The attempt at a solution

    Of course, this is a linear Diophantine equation. I know that a solution will exist when gcd(a, (a+2)) | c.

    gcd(a, (a+2)) = d ⇒ ∃ x,y ∈ ℤ ∋ xa + y(a+2) = d

    a = 0(a+2) + a
    a+2 = (a) + 2
    a = β(2) + r
  2. jcsd
  3. Jun 3, 2015 #2


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    Staff: Mentor

    gcd(a, (a+2)) has a shorter expression.
    Not sure what the last three lines are saying.
  4. Jun 3, 2015 #3
    I just started the algorithm for finding the GCD, but I don't think that I actually need to find an expression.

    Hm. Well, I know that a and a+2 are both odd or even depending on a.
  5. Jun 3, 2015 #4

    Ray Vickson

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    Homework Helper

    Have you forgotten to tell us whether or not a, c, x, y are positive (non-negative?) integers?
  6. Jun 3, 2015 #5
    Partially. a, b ∈ ℕ and x, y ∈ ℤ and 1 ≤ d ≤ min{a,b}.
  7. Jun 4, 2015 #6
    If you want to solve mu+nv = d when m and n are relatively prime, you should be able to find a specific solution with Bezout theorem. With that specific solution, you should be able to find a general solution (u,v) without too much difficulty.

    The thing is that ax +(a+2) y = c has not that form, or at least not always. Why ? Mfb gave you a good hint. Think about it
  8. Jun 4, 2015 #7


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    Staff: Mentor

    I think you should, as the final expression is very easy. It is a bit like leaving 4+5 in a final answer. It is not wrong in terms of mathematics, but you should replace it by 9.
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