Reducing Binary Quadratic Forms

Reducing Binary Quadratic Forms

Recall from the Reduced Binary Quadratic Forms page that a positive definite binary quadratic form $f(x, y) = ax^2 + bxy + cy^2$ is said to be reduced if either:

  • 1) $-a < b \leq a < c$.
  • 2) $0 < b < a = c$.

Also recall that $M_1 = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$ and $M_2 = M_2 = \begin{bmatrix} 1 & m\\ 0 & 1 \end{bmatrix}$ and if $a > c$ then applying $M_1$ to $f(x, y)$ gives us an equivalent binary quadratic form with $a < c$. If $b > a$ (equivalently $-a > b$) then applying $M_2$ to $f(x, y)$ gives us an equivalent binary quadratic form with $b < a$ (equivalently $-a < b$). We will now look at some examples of taking a binary quadratic form and finding an equivalent binary quadratic form that is reduced.

Example

Reduce the binary quadratic form $f(x, y) = 4x^2 - 24xy +39y^2$.

We have that $a = 4$, $b = -24$, and $c = 39$. So:

(1)
\begin{align} \quad -4 \not < -24 \leq 4 < 39 \end{align}

There exists an integer $m$ such that:

(2)
\begin{equation} -a < 2am + b < a \end{equation}

That is:

(3)
\begin{equation} -4 < 8m -24 < 4 \end{equation}

Observe that $m = 3$ works. So an equivalent binary quadratic form is:

(4)
\begin{align} \quad f(x + 3y, y) &= 4(x + 3y)^2 - 24(x + 3y)y +39y^2 \\ &= 4(x^2 + 6xy + 9y^2) - 24(xy + 3y^2) + 39y^2 \\ &= 4x^2 + 24xy + 36y^2 - 24xy - 72y^2 + 39y^2 \\ &= 4x^2 + 3y^2 \end{align}

Now we have:

(5)
\begin{align} -4 < 0 < 4 \not < 3 \end{align}

So applying $M_1$ to $f(x, y)$ gives us:

(6)
\begin{align} \quad f(-y, x) &= 4(-y)^2 + 3(x)^2 \\ &=3x^2 + 4y^2 \end{align}

So we have that $-3 < 0 < 3 < 4$, and so $f(x, y) = 3x^2 + 4y^2$ is a reduced binary quadratic form that is equivalent to $f(x, y) = 4x^2 - 24xy +39y^2$.

Example 2

Reduce the binary quadratic form $f(x, y) = 6x^2 - 5xy + 4y^2$.

We have that:

(7)
\begin{align} -6 < -5 \leq 6 \not < 4 \end{align}

Applying $M_1$ to $f(x, y)$ gives us:

(8)
\begin{align} \quad f(-y, x) &= 6(-y)^2 - 5(-y)(x) + 4(x)^2 \\ &= 6y^2 + 5xy + 4x^2 \\ &= 4x^2 + 5xy + 6y^2 \end{align}

Now we have that:

(9)
\begin{align} \quad -4 < 5 \not < 4 < 6 \end{align}

There exists an integer $m$ such that:

(10)
\begin{equation} -4 < 8m + 5 < 4 \end{equation}

So $m = -1$. Applying $M_2$ with $m = -1$ and we get:

(11)
\begin{align} \quad f(x - y, y) &= 4(x - y)^2 + 5(x - y)y + 6y^2 \\ &= 4(x^2 - 2xy + y^2) + 5(xy - y^2) + 6y^2 \\ &= 4x^2 - 8xy + 4y^2 + 5xy - 5y^2 + 6y^2 \\ &= 4x^2 -3xy + 6y^2 \end{align}

We have that $-4 < -3 \leq -4 < 6$, and so $f(x, y) = 4x^2 - 3xy + 6y^2$ is a reduced binary quadratic form that is equivalent to $6x^2 - 5xy + 4y^2$.

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