# Rational points of conics over $\mathbb{Q}$

I am starting to read lecture notes on basics of arithmetic geometry by A. V. Sutherland.

In the second lecture, there is a procedure how to decide whether a conic over $\mathbb{Q}$ has a rational point (the lecture notes, p. 2, section 2.3 "Conics over $\mathbb{Q}$").

I am stuck on the part where this problem should be transformed into the problem of finding an element of a quadratic field with a given norm. Frankly, I don't think I understand the approach there at all:

1. One starts with a projective conic given by the equation

$$aX^2+bY^2+cZ^2=0, \;\; a, b, c \in \mathbb{Z},\; a>0,\; b, c<0,$$ multiply it by $a$ to get $$a^2X^2+abY^2+acZ^2=0$$ and then, suddenly, one has the equation

$$X^2+abY^2=(-ac)Z^2$$ instead of the (imho correct) equation $$a^2X^2+abY^2=(-ac)Z^2.$$

I don't see whether this is some kind of substitution trick or simply a mistake. What puzzles me the most is that it seems that this form is crucial to the following arguments.

1. Moreover, the author further claims that we can WLOG assume that the integers $-ab, -ac$ are square-free. Why should it be this case? What if one, for example, start with a curve given by the following equation: $$4X^2-9Y^2-25Z^2=0 \;.$$ Then how can I transform it into a form where the numbers $-ab, -ac$ are square-free?

Thank you in advance for any help.

• Points of the conic $a^2X^2+abY^2=(-ac)Z^2$ are in bijection with points of $X^2+abY^2=(-ac)Z^2$ simply mapping $(x,y,z)\mapsto (ax,y,z)$. For the same reason you can assume $-ab,-ac$ to be squarefree. – Ferra Oct 7 '15 at 10:38
• @Ferra Oh, I see. Thanks, I should have thought of that. I guess I can see now that one take care of the other problem similarly, right? – Pavel Čoupek Oct 7 '15 at 10:43
• precisely! your last curve is just $X^2-Y^2-Z^2=0$. – Ferra Oct 7 '15 at 10:55