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Homework Help: Simple Diophantine equation

  1. Feb 25, 2013 #1
    find integers x and y such that:

    3x - 98y = 12

    I understand how to solve for the values, but to make it a diophantine equation I need it to look like 3x + 98y = 12...........my question is what do I do with the negative sign? Carry it to the y? i.e: 3x + 98(-y) = 12, then deal with it after I find my values?
  2. jcsd
  3. Feb 25, 2013 #2


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    If you want positive coefficients, sure, do it that way.
  4. Feb 25, 2013 #3

    I had two other questions. One was quick the other not so much: Back to this equation above, I was wondering since the equation is 3x + 98(-y) = 12, how can I use the diophantine procedure if the values are not relatively prime? i.e don't I have to set it up in the form:

    3x + 98(-y) = 1............and then once I plow through with the euclidean algorithm Multiply through by 12 in this case. What happens if I don't get a gcd of 1 between my two values? Can I still use this?

    The other question had to do with squared terms:

    13x^2 - 23y^2 = 1...........how do I deal with the squares?
  5. Feb 25, 2013 #4


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    If the gcd isn't one, then you won't have a solution at all the if constant on the right side isn't divisible by the gcd. If it is, then divide the equation by the gcd. The second one I'd need to give some more thought to. Do you have any ideas from the course?
  6. Feb 26, 2013 #5
    I don't remember him talking about it specifically this time, but I sat in on some lectures during the last semester and what I remember is something along the lines of some sort of factoring technique......vague I know. I'm stumped and haven't found much online either.
  7. Feb 26, 2013 #6


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    How about considering ##13x^2 \equiv 1\,mod\,23?##?
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