Equivalent binary quadratic forms Given $f(x,y)=ax^2+bxy+cy^2$ we know that it is not necessarily equivalent to $f(-x,y)$ even if they represent the same numbers.
For example our textbook says that $x^2+xy+2y^2$ is equivalent to $x^2-xy+2y^2$, but $3x^2+xy+4y^2$ is not equivalent to $3x^2-xy+4y^2$. How can I see this?
Actually I thought $f(x,y)$ and $f(-x,y)$ cannot be equivalent, since the transformation has determinant $-1$.
I hope someone can clarify. 
 A: The point is that acting by the matrix
$$ \begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix} $$
or its negative is not the only way to move from the form $f(x,y)$ to the form $f(-x,y)$.
You can check by hand that acting on the form $f(x,y) = x^2 + xy + 2y^2$ by the matrix
$$ \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} $$
produces the desired form $f(-x,y) = x^2 - xy + 2y^2$, and this matrix has determinant $1$.
Proving that no matrix of determinant 1 will carry $3x^2 + xy + 4y^2$ to $3x^2 - xy + 4y^2$ is much harder.  The easiest way is to learn about reduction of forms and then see that these two forms are in different equivalence classes by reducing them and seeing that you get different reduced forms.
(It can be shown that in a class, either every form $f(x,y)$ has the property that it is equivalent to $f(-x,y)$ or no form does.  Later, when you learn about composition of forms, you discover Gauss's seminal result that the primitive classes of forms of discriminant $d$ form a group. The operation of going from $f(x,y)$ to $f(-x,y)$ is inversion in this group, so the classes of forms with $f(x,y)$ equivalent to $f(-x,y)$ are precisely the classes of order dividing $2$ in the class group.)
