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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- Apr 26th 2009, 09:54 PMsmithhallIntegers
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? - Apr 26th 2009, 11:25 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 27th 2009, 03:12 AMPlato
$\displaystyle \chi \sigma$,

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}^+$. - Apr 27th 2009, 04:20 AMchisigma
The obervation of Plato is correct, so that i have first to apologize (Headbang)... 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

$\displaystyle \chi$ $\displaystyle \sigma$