Integers

**smithhall** · April 26th 2009, 09:54 PM

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?

**chisigma** · April 26th 2009, 11:25 PM

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

$\chi$ $\sigma$

**Plato** · April 27th 2009, 03:12 AM

$\chi \sigma$ ,
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 $<br /> \mathbb{N}^+ \times \mathbb{N}^+$ .

**chisigma** · April 27th 2009, 04:20 AM

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

$\chi$ $\sigma$

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