Exercise 2.9.10

Question

Determine the first three primes that are represented by $f = (1,1,1)$. For any of those primes $p$ determine a form $(p,b,c)$ that is equivalent to $(1,1,1)$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
If $p = f(x,y) = x^2 + xy + y^2$, then $4p = (2x +y)^2 + 3y^2$, therefore $p\equiv 1 \pmod 3$ or $p=3$.

Moreover $3 = f(1,1), 7 =f(2,1), 13 = f(3,1)$. Since $5 
ot \equiv 1$ and $11 
ot \equiv 1 \pmod 3$, the primes $3,7,13$ are the first three primes represented by the form $f = (1,1,1)$.

Take $U_3 = \begin{pmatrix} 1 & 1 \ 1 & 2 \end{pmatrix} \in \mathrm{SL}(2,\Z)$. Then 
\begin{align*}
(f\cdot U_3)(x,y) &= f(x+y, x +2y)\
&= (x+y)^2 + (x+y)(x+2y) + (x+2y)^2\
&= 3x^2 + 9xy + 7y^2
\end{align*}
Since $g = (3,3,1)$ is in the same $\Gamma$-orbit, because $(3,3,1) = (3, 9- 2 	imes 3, 3 - 9+7)$, we can take the more reduced form $g = (3,3,1)$, given by
$$g_3 = f\cdot U_3 T^{-1} = f \cdot \begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix} = (3,3,1).$$

\bigskip
Now take $U_7 = \begin{pmatrix} 2 & 1 \ 1 & 1 \end{pmatrix}$.
\begin{align*}
(f\cdot U_7)(x,y) &= f(2x+y, x +y)\
&= (2x+y)^2 + (2x+y)(x+y) + (x+y)^2\
&= 7x^2 +9xy +3 y^2.
\end{align*}

We can take also $g_7 = (7, 9 - 2	imes 7, 7-9 +3) = (7,-5,1) = f \cdot \begin{pmatrix} 2 & -1 \ 1 & 0 \end{pmatrix}$.

\bigskip
Now take $U_{13} = \begin{pmatrix} 3 & -1 \ 1 & 0 \end{pmatrix}$.
\begin{align*}
(f\cdot U_{13})(x,y) &= f(3x -y, x)\
&= (3x-y)^2 + (3x-y)x + x^2\
&= 13x^2 -7xy +y^2 .
\end{align*}

To conclude, the proper equivalences
$$(1,1,1) \overset{+}{\sim} (3,3,1) \overset{+}{\sim} (7, -5,1) \overset{+}{\sim} (13, -7, 1)$$
give  forms $(p,\cdot, \cdot)$ properly equivalent to $(1,1,1)$ for $p= 3,7,13$.
\end{proof}

Exercise 2.9.10

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Exercise 2.9.10

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