• May 9th 2010, 09:19 AM
EinStone
Let $D \in \mathbb{Z}$ be square free. Consider $K = \{ x| ax^2+bx+c = 0, a,b,c \in \mathbb{Z} \}$.

For which conditions on $a,b,c\in \mathbb{Z}$ is $K= \{ u + v\sqrt{D}: u,v \in \mathbb{Q} \}$?
• May 9th 2010, 10:13 AM
Dinkydoe
With the abc-formula we have $K= \left\{\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\right\}$

Hence we must have $kD=b^2-4ac$ for some square $k$
( $\sqrt{k}\in \mathbb{Q}$)
• May 9th 2010, 10:51 AM
EinStone
Could you enlarge on that, what are the conditions on a,b,c?
• May 9th 2010, 11:26 AM
Dinkydoe
Ok.

So we can choose $k=p^2\in \mathbb{Z}$ free.

Then we choose $b\in \mathbb{Z}$ such that $ac=\frac{b^2-kD}{4}\in \mathbb{Z}$ is not prime.

Then $a,c$ divide this number.
• May 9th 2010, 11:46 AM
EinStone
So you say that we choose a and c arbitrarily, such that $ac$ is not a prime, and then b needs to be chosen such that $4ac = b^2 -kD$, for some $k=p^2$ $(p \in \mathbb{Z})$

Is it that what you mean? And why is this what we want??
• May 9th 2010, 12:12 PM
Dinkydoe
with the abc-formula we have that $K$ consists of the numbers $\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}$

To make sure $K$ has the desired property we must have $\sqrt{b^2-4ac}=p\sqrt{D}$ for some number $p\in \mathbb{Z}$.

That is, $b^2-4ac= p^2D$.

This number $p^2\in \mathbb{Z}$ can ofcourse be freely chosen.

Then we must choose $a,b,c\in\mathbb{Z}$ that satisfy this relation. Thus we must choose a $b\in \mathbb{Z}$ such that

there exists a composite number $ac\in\mathbb{Z}$ that satisfies $b^2 = p^2D+4ac$

wich is, the same as saying $\frac{b^2-p^2D}{4}$ must be a composite number. Only then we can choose $a,c\in \mathbb{Z}$ such that $ac=\frac{b^2-p^2D}{4}$.

And we've found $a,b,c$ such that $K$ has the desired property.
• May 9th 2010, 01:56 PM
EinStone
Why does ac need to be composite?
• May 9th 2010, 02:01 PM
Dinkydoe
Quote:

Why does ac need to be composite?
How else can we find $a,c\in \mathbb{Z}$?

If $ac=p$ prime then $a,c$ can not be integers.
• May 9th 2010, 02:06 PM
chiph588@
Quote:

Originally Posted by Dinkydoe
How else can we find $a,c\in \mathbb{Z}$?

If $ac=p$ prime then $a,c$ can not be integers.

$a=1,\; c=p$
• May 9th 2010, 02:09 PM
Dinkydoe
Ah shoot, big blunder here. Shameful moment here xp

Thanks chiph

Forget everything about ac being composite. I'm sorry for this error ;p

The rest you can take for granted.
• May 9th 2010, 02:11 PM
EinStone
So we just need to choose a,b,c such that $\sqrt{b^2-4ac}=p\sqrt{D}$ for some $p \in \mathbb{Z}$ and thats all?
• May 9th 2010, 02:15 PM
Dinkydoe
• May 9th 2010, 02:26 PM
EinStone
I don't see the difference. Which condition do you have except $b^2-4ac=p^2 D$ for $p \in \mathbb{Z}
$
.

I mean once this holds for integers a,b,c, $ac \in \mathbb{Z}$ trivially.
• May 9th 2010, 02:45 PM
Dinkydoe
No, it's not different. It's the same, but you wanted more details right ;p?