Exercise 3.7.6

Proof. Here $\mathrm{\Delta }=-3$ and $p=3$. Using Proposition 3.4.5, we obtain $R(\mathrm{\Delta },p)=R(-3,3)=-33+1=1$, and

From (3.10), we obtain $b(-3,3)=3$ and $c(-3,3)=1$, therefore

Write $f=(3,3,1)$. Then $3=f(1,0)$ gives a primitive representation of $3$ by $f$. Since $3$ is prime, all representations of $3$ by $f$ are primitive. By Proposition 2.6.2, there is only one equivalence class of proper representations of $3$ by $f$.

Here $\mathrm{Aut}(f)$ is the set of matrices $U(f,x,y)$ (see (2.20)), given by

where $x,y$ are the roots of $x^2+3y^2=4$, that is

By Exercise 2.9.9, these solutions form a cyclic group of order $6$ generated by $(1,1)$. Thus $\mathrm{Aut}(f)=⟨U⟩$, where

This gives

All (primitive) representations of $3$ by $f=(3,3,1)$ are given by the first column of these matrices. This gives

Verification: $f(-1,3)=3-9+8=3,f(-2,3)=12-18+9=3$.

Note: As $f=(3,3,1)\stackrel{+}{\sim }(1,1,1)$, we can determine all representations of $3$ by $(1,1,1)$. That is Exercise 2.9.11. □