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Math Help - Integers

  1. #1
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    Integers

    Let a, b be relatively prime positive integers and consider the equation ax+by = ab.
    (a) Show that the equation has no solution (x, y) belongs to N N.
    (b) Does it have a solution with (x, y) belongs to Z Z? Why/why not?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's consider the equation...

    a\cdot x + b\cdot y = a\cdot b (1)

    where a>0 and b> 0 are relatively prime integers. Deviding both terms of (1) by b [the same is if we devide by a...] we obtain the equation...

    \frac{a}{b}\cdot x + y = a (2)

    Since \frac{a}{b}\ni \mathbb{Z} and a \in \mathbb{Z}, no solution (x,y) \in \mathbb{Z}^{2} of (2) exist. Because \mathbb{N} \in \mathbb{Z}, the same holds for (x,y) \in \mathbb{N}^{2}...

    Kind regards

    \chi \sigma
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  3. #3
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    \chi \sigma,
    I am not sure what part of the question you thought you proved.
    However, consider 2x+3y=6 has many solutions in <br />
\mathbb{Z} \times \mathbb{Z} but no solution in <br />
\mathbb{N}^+ \times \mathbb{N}^+.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The obervation of Plato is correct, so that i have first to apologize ... and after to give the right solution...

    Let's consider the equation...

    a\cdot x + b\cdot y = a\cdot b

    ... where a>0 and b>0 are relatively prime integers. Deviding both terms of (1) by b [the same is if we devide by a...] we obtain the equation...

    \frac{a}{b}\cdot x + y = a \rightarrow y= a - \frac {a}{b}\cdot x (2)

    Now if a and b are relatively prime in order to have y \in \mathbb{Z} in necessary that x=k\cdot b with k integer, i.e. x devides b. In order to have also x\in \mathbb {N} and y\in \mathbb {N} however, must be k<1 and that is a contadiction. The conclusion is that no solution (x,y) \in \mathbb{N}^{2} exist and that infinite solutions (x,y) \in \mathbb{Z}^{2} exist under the condition  x= k\cdot b with k\in \mathbb{Z}...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 27th 2009 at 05:09 AM.
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