Zagier's proofs are compact, so I don't think there's any way to reduce further. 4 pages can be a lengthy translation but here is enough to to get started.

**Theorem 2** Let $f(x,y) = ax^2 + bxy + cy^2$ be a primitive quadratic form of discriminant $D$ (squarefree). There is a bijection between solutions to Pell's equation and the automorphism group of $f$

$$ (t,u) \to \left( \begin{array}{cc} \frac{t - bu}{2} & - cu \\ au & \frac{t+bu}{2}\end{array} \right) $$

This bijection is a group isomorhism by the composition rule:
$$ (t_1, u_1) \circ(t_2, u_2) = \left( \frac{t_1 t_2 + D u_1 u_2}{2}, \frac{t_1 u_2 + t_2 u_1}{2}\right)$$
The group $U_f$ is finite if $D < 0$ and cyclic of order $w$

- $\mathbb{Z}/6\mathbb{Z}$ for $D = - 3$
- $\mathbb{Z}/4\mathbb{Z}$ for $D = - 4$
- $\mathbb{Z}/2\mathbb{Z}$ for $D < - 4$

For positive $D > 0$ the group of solutions $U_f \simeq \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

**Proof** Zagier says the positive discriminant case follows from Pell's equation (it does). If you have a solution to Pell's equation (for any $f(x,y) = n$) , both group laws are pretty clear and that they map to one another. There is a map (injective homomorphism) from $U_f$ either to $\mathbb{C}^\ast$ or $\mathbb{R}^\ast$ given by
$$ (t,u) \mapsto \frac{t + u \sqrt{D} }{2} $$
the case $D = - 3$ corresonds to $\mathbb{Z}[e^{2\pi i / 3}]$ and $D = - 4$ matches $\mathbb{Z}[i]$, with units $e^{2\pi i /3}$ or $i$. In all other cases $D < -4$ there are only the trivial units $\pm 1$. $\hspace{1.4in}\square$

**Theorem 3** Let $D$ be a "fundamental discriminant" $n \neq 0$ be a whole number. Then the total number $R(n)$ of representation of $n$ by primitive forms of discriminant $D$ is given by
$$ R(n) = \sum_{m|n} \chi_D(m) $$
where $m$ is a divisor of $n$ and $\chi_D(m)$ is a character. In particlar, $R(n)$ and all $R(n,f)$ are finite.

**Proof** There are no imprimitive forms, so we can leave outthe word "primitive" in the theorem. Let $R^\ast(n)$ be the number of inequivalent *primitive* representations of $n$ by (any) forms of discriminant $D$. Obviously
$$ R(n) = \sum_{g \geq 1, g^2 | n} R^\ast (n/g^2) $$
since every representation is a multiple of a primitive one. The main step in the proof is the formula
$$ R^\ast(n) = \{ b \pmod {2n} \big| b^2 \equiv D \pmod{4n}\} $$
Then he proceeds to explain the notion of a group action and variants of the **orbit stabilizer formula**

Let $G$ be a group $X$ and $Y$ be two sets and $S \subset X \times Y$. Then we can study the diagonal group action $(x,y) \mapsto (gx,gy)$ and observe the quotient sets $X/G$ as well as $S/G$.

- $Y_x = \{ y \in Y | (x,y) \in S \} $
- $G_x = \{ g \in G | gx = x \}$
- $|S/G| = \sum_{x \in X/G} |Y_x/G_x| $
- $|S/G| = \sum_{y \in Y/G} |X_y/G_y| $

The choice of group action by Zagier is set of invertible $2 \times 2$ matrices acting on pairs $X \times Y = $
$$ \big\{\text{quadratic forms }ax^2 + bxy+cy^2 \text{ with }b^2 - 4ac = D \} \times \{ \text{pairs of integers }x,y \in \mathbb{Z}\big\} $$

and $G = SL(2,\mathbb{Z})$ which is solutions to $ad-bc = 1$ for $a,b,c,d \in \mathbb{Z}$. Lastly the special set

- $S = S(n) =\{ (f,z) \in X \times Y : f(z) = n \} $

Then $X/G$ is the set of equivalence classes of discriminant $D$ , and by the group action formula:
$$ |S/G| = \sum_{[f]} R^\ast (n,f) = R^\ast(n) $$

Every element of $Y$ is equivalent to $z = (1,0)$ (Exercise) for this element:

$G_z = \left\{ \left(\begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right) : r \in \mathbb{Z}\right\} $

$X_z = \{ nx^2 + bxy + \frac{b^2 - D}{4n} y^2: b \in \mathbb{Z}, b^2 \equiv D \pmod {4n} \}$

these are related to the substitions $b \mapsto b + 2nr$

Zagier observes that $R^\ast(n)$ is multiplicative so that enough we solve for the primes
$$ R^\ast(p^r) = \# \big\{ b \pmod{p^r} \big| b^2 \equiv D \pmod{p^r} \big\} $$

Then we can reconstruct $R(n)$ from the count of "primitive" representations of $n$, $R^\ast(n)$

$R(p^r) = \sum_{0 \leq s < r/2}2 + \sum_{s=r/2} 1 = r+1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=+1$

$R(p^r) = \sum_{0 \leq s < r/2}0 + \sum_{s=r/2} 1 = 0 \text{ or }1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=-1$

$R(p^r) = \sum_{0 \leq s < (r-1)/2}0 + \sum_{(r-1)/2 \leq s \leq r/2} 1 = 1 = \sum_{0 \leq i \leq r} \chi_D(p^i) $ if $(\frac{D}{p})=+1$

and Theorem 3 checks in all cases. $\hspace{2in}\square$

**Corollary** The average number of representations of $n$ is exactly the value of the L-function
$$ \lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n) \right) = L(1, \chi_D) $$
The proof involves counting lattice points under a hyperbola. For a specific primitive $f$ the formula involves the fundamental unit.

**Theorem 4** Let $\epsilon_0$ be fundamental unit of $f$.

$$ \lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n,f) \right) =$$
$$\left\{ \begin{array}{cc}
\frac{2\pi}{w\sqrt{|D|}} & \text{ if }D < 0 \\
\frac{\log \epsilon_0 }{w\sqrt{D}} & \text{ if }D > 0 \\
\end{array}\right.$$