#### 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 $$\displaystyle K= \{ u + v\sqrt{D}: u,v \in \mathbb{Q} \}$$?

#### 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 $$\displaystyle k$$
($$\displaystyle \sqrt{k}\in \mathbb{Q}$$)

#### EinStone

Could you enlarge on that, what are the conditions on a,b,c?

#### 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.

#### 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??

#### 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.

#### EinStone

Why does ac need to be composite?

#### Dinkydoe

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.

#### chiph588@

MHF Hall of Honor
How else can we find $$\displaystyle a,c\in \mathbb{Z}$$?

If $$\displaystyle ac=p$$ prime then $$\displaystyle a,c$$ can not be integers.
$$\displaystyle a=1,\; c=p$$

#### 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.