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I'm currently reading elementary number theory books, specifically in the chapter of binary quadratic forms. This problem got my attention.

Let $f(x,y)=ax^2+bxy+cy^2$ and $g(x,y)=rx^2$, where $r=gcd(a,b,c)$ and $f$ is a positive semidefinite quadratic form with discriminant $d=0$. Prove that $f$ is equivalent to $g$.

I know that $f$ can be expressed as $f(x,y)=r(hx+ky)^2$ since the dicriminant of $f$ is $d=0$. Then there exist integers $x_0$ and $y_0$ such that $f(x_0,y_0)=0$. Also $gcd(h,k)=1$ so there must be integers $u$ and $v$ with $hu+kv=1$. My guess is that the matrix $M=\begin{pmatrix}u&x_0\\v&y_0 \end{pmatrix}$ takes $f$ to $g$. Now, if $g(x,y)=Ax^2+Bxy+Cy^2$, then $A=f(u,v)=r$ and $C=f(x_0,y_0)=0$. But I'm not sure if this will make $B=0$ or if my assumption is right. Can someone help me to prove this? Thank you in advance.

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With integers $a,b,c$ and $b^2 = 4ac.$

take $$ g = \gcd(2a,b) $$ and solve Bezout equation $$ \frac{2a}{g} \; u + \frac{b}{g} v = 1 $$ in integers.

Now multiply out $$ \left( \begin{array}{cc} u & v \\ - \frac{b}{g} & \frac{2a}{g} \\ \end{array} \right) \left( \begin{array}{cc} 2a & b \\ b & 2c \\ \end{array} \right) \left( \begin{array}{cc} u & -\frac{b}{g} \\ v & \frac{2a}{g} \\ \end{array} \right) $$ In case of trouble, note that the determinant of the product is the same as the original determinant, as we arranged the left and right matrices to have determinant $1 \; . \; $

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  • $\begingroup$ Sir how does this product be equal to $\begin{pmatrix}2r & 0 \\0 & 0\end{matrixp}$ ? $\endgroup$ Commented May 1, 2020 at 6:17

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