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Thread: 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...

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    $\displaystyle \chi \sigma$,
    I am not sure what part of the question you thought you proved.
    However, consider $\displaystyle 2x+3y=6$ has many solutions in $\displaystyle
    \mathbb{Z} \times \mathbb{Z}$ but no solution in $\displaystyle
    \mathbb{N}^+ \times \mathbb{N}^+$.
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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...

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

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

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

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

    Kind regards

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