The principal form, up to equivalanc, is the only integral form of Discriminant $\mathbf{D}$, which represent one.

Question

Suppose that $g$ is an integral form of Discriminant $\mathbf{D}$, for which we have $g(u,v)=1$, can I deduce that it is equivalent to the principal form?

Answer 1

Yes, you can.

Since a common factor of $u$ and $v$ would end up as a squared factor of any number of the form $g(u,v),$ we know that $u$ and $v$ are relatively prime. Thus, we may find integers $\alpha$ and $\beta$ with $\beta u - \alpha v = 1$. The coefficient of $x^2$ in the form

$$ f(x,y) = g(ux + \alpha y, vx + \beta y) $$

is then $g(u,v)$. Thus, we have replaced $g$ by the equivalent form $f$, which has first coefficient $1$, just like the principal form.

Now we adjust the middle coefficient of $f$ so that it matches the middle coefficient of the principal form. Observe that the middle coefficient of a form is congruent to its discriminant modulo $2$. It follows that the middle coefficients of $f$ and the principal form have the same parity.

Suppose $f = ax^2 + bxy + cy^2$. For an arbitrary integer $n$, consider the form $f(x+ny,y)$. This will still have first coefficient $1$, and one checks that its middle coefficient is $b+2n$. By choosing $n$ appropriately, we can make this coefficient be any integer we wish with the same parity as $b$. In particular, we can replace $f$ by a form that has the same first and middle coefficients as the principal form. In particular, you can choose $n$ so that $f(x+ny,y)$ has the same first and middle coefficient as the principal form. But both forms will also have the same discriminant, and these conditions ensure that they will have the same third coefficient as well.