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Math Help - elem.num.theo - diophantine equations

  1. #1
    Member cassiopeia1289's Avatar
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    elem.num.theo - diophantine equations

    define all solutions in integers to diophantine equations:

    56x+72y=40

    ok - I'll write my steps, my problem is that I end up with no solutions at all - maybe someone can find where I went wrong ... ?

    SO:
    72 = 56(1) + 16
    56 = 16(3) + 8
    16 = 8(2)
    so gcd(56,72) = 8

    next:
    8 = 56 - (16)(3)
    = 56 - 3(72-56)
    = 56(4) + 72(-3)

    40 = 56(20) + 72(-15)
    so x0=20 & y0=-15

    so:
    x = 20 + 9t, for some int. t
    y = -15 - 7t, for some int. t

    and I get:
    -(15/7) > t > -(20/9)
    but there are no integers that work.
    did I do something stupid in algebra or can there really be no solutions?
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  2. #2
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    Quote Originally Posted by cassiopeia1289 View Post
    x = 20 + 9t, for some int. t
    y = -15 - 7t, for some int. t
    Stop right there. Those are your solutions, for all integers t.
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  3. #3
    Member cassiopeia1289's Avatar
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    but normally we find the integers of t
    thats the what the question asks for: all solutions in the integers
    and then use that to find the x&y's for each of the t's
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  4. #4
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    t can be any integer and

    x = 20 + 9t
    y = -15 - 7t

    will be a solution.

    So that is the solution set, and it is defined for all integers t. If you want to know what possible values of t are, the answer is all integers.
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  5. #5
    Member cassiopeia1289's Avatar
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    well - ok - what I would normally do (what we always do in class and on homework) is what I did above:

    x = 20 + 9t, for some int. t
    y = -15 - 7t, for some int. t

    so you set 20 + 9t > 0
    where t > -(20/9)
    and then -15 - 7t > 0
    where t < -(15/7)

    and so you get
    -(15/7) > t > -(20/9)

    then (normally) from there we would take all the values of t in that restriction
    (say for instance you had .5>t>2.5 - so t=1&2, and then you would plug those t values back into your x=blah and y=blah - and those are your solutions)

    but with this problem - there are no integers in that restriction so that makes me wanna say no solutions ... ?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    well - ok - what I would normally do (what we always do in class and on homework) is what I did above:

    x = 20 + 9t, for some int. t
    y = -15 - 7t, for some int. t

    so you set 20 + 9t > 0
    where t > -(20/9)
    and then -15 - 7t > 0
    where t < -(15/7)

    and so you get
    -(15/7) > t > -(20/9)

    then (normally) from there we would take all the values of t in that restriction
    (say for instance you had .5>t>2.5 - so t=1&2, and then you would plug those t values back into your x=blah and y=blah - and those are your solutions)

    but with this problem - there are no integers in that restriction so that makes me wanna say no solutions ... ?
    no, note that what you were solving for were the positive integer solutions. you wanted both x and y to be greater than zero. so you set the expressions for x and y to be greater than zero respectively, and then solve for the range that t has to be within. no such restriction was given here. we accept negative as well as positive solutions as far as this problem is concerned
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  7. #7
    Member cassiopeia1289's Avatar
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    ah - ok that makes sense - thank you for being so patient!
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