• May 9th 2010, 08:19 AM
EinStone
Let $\displaystyle D \in \mathbb{Z}$ be square free. Consider $\displaystyle K = \{ x| ax^2+bx+c = 0, a,b,c \in \mathbb{Z} \}$.

For which conditions on $\displaystyle a,b,c\in \mathbb{Z}$ is $\displaystyle K= \{ u + v\sqrt{D}: u,v \in \mathbb{Q} \}$?
• May 9th 2010, 09:13 AM
Dinkydoe
With the abc-formula we have $\displaystyle K= \left\{\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\right\}$

Hence we must have $\displaystyle kD=b^2-4ac$ for some square $\displaystyle k$
($\displaystyle \sqrt{k}\in \mathbb{Q}$)
• May 9th 2010, 09:51 AM
EinStone
Could you enlarge on that, what are the conditions on a,b,c?
• May 9th 2010, 10:26 AM
Dinkydoe
Ok.

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

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

Then $\displaystyle a,c$ divide this number.
• May 9th 2010, 10:46 AM
EinStone
So you say that we choose a and c arbitrarily, such that $\displaystyle ac$ is not a prime, and then b needs to be chosen such that $\displaystyle 4ac = b^2 -kD$, for some $\displaystyle k=p^2$ $\displaystyle (p \in \mathbb{Z})$

Is it that what you mean? And why is this what we want??
• May 9th 2010, 11:12 AM
Dinkydoe
with the abc-formula we have that $\displaystyle K$ consists of the numbers $\displaystyle \frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}$

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

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

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

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

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

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

And we've found $\displaystyle a,b,c$ such that $\displaystyle K$ has the desired property.
• May 9th 2010, 12:56 PM
EinStone
Why does ac need to be composite?
• May 9th 2010, 01:01 PM
Dinkydoe
Quote:

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

If $\displaystyle ac=p$ prime then $\displaystyle a,c$ can not be integers.
• May 9th 2010, 01:06 PM
chiph588@
Quote:

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

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

$\displaystyle a=1,\; c=p$
• May 9th 2010, 01:09 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, 01:11 PM
EinStone
So we just need to choose a,b,c such that $\displaystyle \sqrt{b^2-4ac}=p\sqrt{D}$ for some $\displaystyle p \in \mathbb{Z}$ and thats all?
• May 9th 2010, 01:15 PM
Dinkydoe
• May 9th 2010, 01:26 PM
EinStone
I don't see the difference. Which condition do you have except $\displaystyle b^2-4ac=p^2 D$ for $\displaystyle p \in \mathbb{Z}$.

I mean once this holds for integers a,b,c, $\displaystyle ac \in \mathbb{Z}$ trivially.
• May 9th 2010, 01:45 PM
Dinkydoe
No, it's not different. It's the same, but you wanted more details right ;p?