Are reducible Integral Binary Quadratic Forms equivalent? By an integral binary quadratic form (IBQF for short) I mean an $$f(x,y) = ax^2 + bxy + cy^2$$ with $a,b,c \in \mathbb{Z}$. Note that I am not assuming that they are all coprime. 
Such an $f$ is said to be reducible if it factors as the product of two linear forms; equivalently, if its discriminant, defined as $\Delta(f) := b^2 - 4ac$, is a square in $\mathbb{Z}$. 
I would like to say that, if two IBQFs $f$ and $g$ are reducible and have the same discriminant, then they must be equivalent, by which I mean that there exists a matrix $\gamma \in SL_2(\mathbb{Z})$ such that $$\gamma f = g.$$
Is this true?
 A: That was fun. It turns out that, for discriminant $0$ and primitive, the two $SL_2 \mathbb Z$ classes are $f(x,y) = x^2$ and $g(x,y) = -x^2.$ For discriminant $1,$ the only class is $f(x,y) = xy.$
For discriminant $D = B^2$ with integer $B > 1,$ the count of classes is given by the Euler totient $\varphi(B),$ because all the distinct classes are given by taking $B > 0,$ also $\gcd(A,B) = 1$ and $1 \leq A < B,$ for each $A$ the class is given by
$$ f(x,y) = Ax^2 + B xy.   $$
Notice that there is no $y^2$ term. Also note that, with $B \geq 2,$ taking $A=0$ would give an imprimitive form, namely $Bxy,$ so we exclude that. Oh, it turns out that there is no need to have $B = - \sqrt D,$ requires proof but it is all elementary manipulations of how elements of $SL_2 \mathbb Z$ act on forms.  
GO FIGURE
A: $$ f(x,y) = x^2 - y^2  $$
$$ g(x,y) = 2 x y  $$
A different pair:
$$ f(x,y) = x^2 - 9y^2  $$
$$ g(x,y) = -x^2 + 9 y^2 $$
Here is a good one; 
$$ f(x,y) = x^2 - 36y^2  $$
$$ g(x,y) = 9x^2 - 4 y^2. $$
While $9-4 = 5,$ we can easily check $x^2 - 36 y^2 \neq 5,$ either because $5$ is not a quadratic residue $\pmod 3,$ or by solving $(x+6y)(x-6y) = 5$ where each factor needs to be $\pm 1, \pm 5,$ and we find, in four possible ways, $x= \pm 3, y= \pm \frac{1}{3}.$
