# Find positive integers

• April 21st 2009, 04:15 AM
maria18
Find positive integers

Find all positive integers $x,y,z$ satisfying simultaneously two conditions.

i) $\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}$ is a rational number .

ii) $x^2 + y^2+z^2$ is a prime .
• April 21st 2009, 04:54 AM
Showcase_22
$

\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}=\frac{p}{q}
$

$q(x-y\sqrt{2009})=p(y-z \sqrt{2009})$

$qx-qy \sqrt{2009}=py-pz \sqrt{2009}$

$qx+pz \sqrt{2009}=py+qy \sqrt{2009}$

$qx+pz \sqrt{2009}=y(p+q \sqrt{2009})$

$y=\frac{qx+\sqrt{2009}}{p+q \sqrt{2009}}$

but since y has to be an integer, then $p=kqx$ and $q=k$ where $k \in \mathbb{N}$.

I'm not sure how this helps =S
• April 21st 2009, 06:15 AM
Soroban
Hello, maria18!

I found something, but I too haven't solved it yet.

Quote:

Find all positive integers $x,y,z$ satisfying simultaneously two conditions:

$(1)\;\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}$ is a rational number.

$(2)\;\;x^2 + y^2+z^2$ is a prime.

Rationalize: . $\frac{p}{q} \;=\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} \cdot \frac{y+z\sqrt{2009}}{y+z\sqrt{2009}} \;=\;$ . $\frac{xy + xz\sqrt{2009} - y^2\sqrt{2009} - 2009yz}{y^2-2009z^2}$

. . . . . . . . . $\frac{p}{q} \;=\;\frac{xy - 2009 + (xz - y^2)\sqrt{2009}}{y^2-2009z^2}$

Since $p$ is an integer: . $xz-y^2 \:=\:0 \quad\Rightarrow\quad xz \:=\:y^2$

Can we exploit that fact? . . . I don't know.

• April 21st 2009, 11:10 AM
Moo
Quote:

Originally Posted by Showcase_22
$

\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}=\frac{p}{q}
$

$q(x-y\sqrt{2009})=p(y-z \sqrt{2009})$

$qx-qy \sqrt{2009}=py-pz \sqrt{2009}$

$qx+pz \sqrt{2009}=py+qy \sqrt{2009}$

$qx+pz \sqrt{2009}=y(p+q \sqrt{2009})$

$y=\frac{qx+{\color{red}pz}\sqrt{2009}}{p+q \sqrt{2009}}$

Quote:

but since y has to be an integer, then $p=kqx$ and $q=k$ where $k \in \mathbb{N}$.

I'm not sure how this helps =S
I think it rather means that there exists k such that qx=pk and pz=qk
I'm not sure either how this helps lol, but it has a good look !
• April 22nd 2009, 07:09 AM
maria18
Quote:

Originally Posted by Soroban
Hello, maria18!

I found something, but I too haven't solved it yet.

Rationalize: . $\frac{p}{q} \;=\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} \cdot \frac{y+z\sqrt{2009}}{y+z\sqrt{2009}} \;=\;$ . $\frac{xy + xz\sqrt{2009} - y^2\sqrt{2009} - 2009yz}{y^2-2009z^2}$

. . . . . . . . . $\frac{p}{q} \;=\;\frac{xy - 2009 + (xz - y^2)\sqrt{2009}}{y^2-2009z^2}$

Since $p$ is an integer: . $xz-y^2 \:=\:0 \quad\Rightarrow\quad xz \:=\:y^2$

hello and thanks Soroban

I find==> with $xz \:=\:y^2$ we have $x^2+y^2+z^2 = (x+z)^2 - xz = (x+z)^2 - y^2 = (x+z-y)(x+z+y)$

so $(x+z-y)(x+z+y)$ is prime if ... (Worried)
• April 22nd 2009, 07:15 AM
Showcase_22
Thanks Moo for pointing that out.

This question is actually harder than it looks!

$

(x+z-y)(x+z+y)
$
can only be prime if only one of the expressions in the brackets is 1.

• April 22nd 2009, 02:25 PM
maria18
Quote:

Originally Posted by Showcase_22

$(x+z-y)(x+z+y)$ can only be prime if only one of the expressions in the brackets is 1.

we have obligatorily $x+z-y=1$ so $x+z=1+y$ and we have $xz= y^2$
so $x$ and $y$ are the roots of equation $X^2-(y+1)X+y^2=0$ so we get $y \in [-\frac{1}{3} ; 1]$ or $y=0$ or $y=1$
• April 22nd 2009, 02:37 PM
Showcase_22
You can use that to find all the other integers though can't you?

(I haven't tried it myself, but by substituting them into the equationswe can obtain values for x and z. They're going to be different depending on the y value).