Computing the Class Number H(-39)

# Computing the Class Number H(-39)

Theorem 1: The class number $H(-39) = 4$. |

**Proof:**Let $f(x, y) = ax^2 + bxy + cy^2 $ be a positive definite reduced binary quadratic form with discriminant [[$ d = -39$. Then either $-a < b \leq a < c$ or $0 \leq b < a = c$.

- Since $f(x, y)$ is reduced positive definite we have that:

\begin{align} \quad 0 < a \leq \sqrt{\frac{-d}{3}} \quad \Leftrightarrow \quad 0 < a \leq \sqrt{\frac{39}{3}} = \sqrt{13} \approx 3.6 \end{align}

- Therefore $0 < a \leq 3$. So $a = 1$ or $a = 2$ or $a = 3$.

**Case 1:**Suppose that $a = 1$. Then $-1 < b \leq 1 < c$ implies that $b = 0$ or $b = 1$. If $b = 0$ then:

\begin{align} \quad -39 = 0^2 - 4(1)(c) = -4c \end{align}

- This cannot happen since $-39$ is not a multiple of $-4$. If $b = 1$ then:

\begin{align} \quad -39 = 1^2 - 4(1)(c) = 1 - 4c \quad \Leftrightarrow \quad -40 = -4c \end{align}

- So $c = 10$ and so the triple $(a, b, c) = (1, 1, 10)$ gives us $f(x, y) = x^2 + xy + 10y^2$ is a positive definite reduced binary quadratic form with discriminant $d = -39$.

**Case 2:**Suppose that $a = 2$. Then $-2 < b \leq 2 < c$ implies that $b = 0$ or $b = \pm 1$ or $b = 2$. If $b = 0$ then:

\begin{align} \quad -39 = 0^2 - 4(2)c = -8c \end{align}

- This cannot happen since $-39$ is not a multiple of $-8$. If $b = \pm 1$ then:

\begin{align} \quad -39 = (\pm)^2 - 4(2)c = 1 +8c \quad \Leftrightarrow \quad -40 = -8c \end{align}

- So $c = 5$. So the triples $(a, b, c) = (2, 1, 5)$ and $(a, b, c) = (2, -1, 5)$ gives us $g(x, y) = 2x^2 + xy + 5y^2$ and $h(x, y) = 2x^2 + xy - 5y^2$ and they are positive definite reduced binary quadratic forms with discriminant $d = -39$.

- If $b = 2$ then:

\begin{align} \quad -39 = 2^2 - 4(2)c = 4 - 8c \quad \Leftrightarrow \quad -43 = -8c \end{align}

- This cannot happen since $-43$ is not a multiple of $-8$.

**Case 3:**Suppose that $a = 3$. Then $-3 < b \leq 3 < c$ implies that $b = 0$ or $b = \pm 1$ or $b = \pm 2$ or $b = 3$. If $b = 0$ then:

\begin{align} \quad -39 = 0^2 - 4(3)c = -12c \end{align}

- This cannot happen since $-39$ is not a multiple of $-12$. If $b = \pm 1$ then:

\begin{align} \quad -39 = (\pm 1)^2 - 4(3)(c) = 1 -12c \quad \Leftrightarrow \quad -40 = -12c \end{align}

- This cannot happen since $-40$ is not a multiple of $-12$. If $b = \pm 2$ then:

\begin{align} \quad -39 = (\pm 2)^2 - 4(3)c = 4 - 12c \quad \Leftrightarrow \quad -43 = -12c \end{align}

- This cannot happen since $-43$ is not a multiple of $-12$. If $b = 3$ then:

\begin{align} \quad -39 = (3)^2 - 4(3)c = 9 - 12c \quad \Leftrightarrow \quad -48 = -12c \end{align}

- So $c = 4$. Hence the triple $(a, b, c) = (3, 3, 4)$ gives us $i(x, y) = 3x^2 + 3xy + 4y^2$ which is a positive definite reduced binary quadratic form with discriminant $d = -39$.

**Conclusion:**We obtain $4$ different positive definite reduced binary quadratic forms with discriminant $d = -39$:

\begin{align} f(x, y) &= x^2 + xy + 10y^2 g(x, y) &= 2x^2 + xy + 5y^2 h(x, y) &= 2x^2 + xy - 5y^2 i(x, y) &= 3x^2 + 3xy + 4y^2 \end{align}

- Therefore $H(-39) = 4$. $\blacksquare$