12. If the three sides of a right triangle are pairwise coprime positive integers, then such a right triangle is called a primitive right triangle.
(1) Find the number of primitive right triangles in which the area is exactly $n$ times the perimeter, and denote this number as $a_{n}$;
(2) Find $\max _{1 \leqslant n \leqslant 2023} a_{n}$, and list all primitive right triangles whose area is exactly 2023 times the perimeter.
(1) Let the sides of a primitive right-angled triangle be $x, y, z$, then $x, y$ are one odd and one even. Without loss of generality, assume $2 \mid y$.
Lemma: The indeterminate equation $x^{2}+y^{2}=z^{2}$, satisfying $(x, z)=1, x>0, y>0, z>0, 2 \mid y$, has all positive integer solutions $(x, y, z)$ that can be expressed as $x=a^{2}-b^{2}, y=2 a b, z=a^{2}+b^{2}$, where $a>b>0, a, b$ are one odd and one even, and $(a, b)=1$ for any integers.
Proof: $y^{2}=z^{2}-x^{2}=(z+x)(z-x)$, noting that $z-x, z+x$ have the same parity, and $2 \mid y$, then $x, z$ are both odd. Thus $\left(\frac{y}{2}\right)^{2}=\left(\frac{z+x}{2}\right)\left(\frac{z-x}{2}\right)$, by $\frac{z+x}{2}+\frac{z-x}{2}=x$ $\left.\left(\frac{z+x}{2}, \frac{z-x}{2}\right) \right\rvert\, x$, similarly we get $\left(\frac{z+x}{2}, \frac{z-x}{2}\right)\left|z \Rightarrow\left(\frac{z+x}{2}, \frac{z-x}{2}\right)\right|(x, z)=1$. Let $\left\{\begin{array}{l}\frac{z+x}{2}=a^{2}, \\ \frac{z-x}{2}=b^{2}\end{array} \Rightarrow\left\{\begin{array}{l}x=a^{2}-b^{2}, \\ z=a^{2}+b^{2},\end{array}\right.\right.$ thus $y=2 a b$, where $a, b$ are one odd and one even, $(a, b)=1$. By the condition, we have $\frac{1}{2} x y=n(x+y+z) \Rightarrow a b\left(a^{2}-b^{2}\right)=n\left(2 a^{2}+2 a b\right) \Rightarrow b(a-b)=2 n$, and $a, b$ are one odd and one even $\Rightarrow a-b$ is odd, then $b$ is even.
When $n=1$, $b(a-b)=2 \Rightarrow b=2, a-b=1 \Rightarrow a=3, b=2$, then there exists a unique $(x, y, z)=(5,12,13)$, so $a_{1}=1$; when $n=2$, $b(a-b)=4 \Rightarrow b=4, a-b=1 \Rightarrow a=5, b=4$, then there exists a unique $(x, y, z)=(9,40,41)$, so $a_{2}=1$. Let $n=2^{r} p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are the odd prime factors of $n$ greater than 1. Since $a, b$ are one odd and one even, then $a-b$ is odd, and $b$ is even. Also, $(a, b)=1 \Rightarrow(b, a-b)=1$, thus $b(a-b)=2 n=2^{r+1} p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}} \Rightarrow 2^{r+1} \mid b$. Let $b=2^{r+1} A, a-b=B$, then $p_{i}^{r_{i}} \mid A$ or $p_{i}^{r_{i}} \mid B(i=1,2, \cdots, k)$, and only one of these cases occurs, so there are exactly $2^{k}$ ways to choose $b$ and $a-b$.
Therefore, $a_{n}=2^{k}(k$ is the number of odd prime factors of $n$ greater than 1 $)$.
(2) $2023=7 \times 17^{2} \Rightarrow a_{2023}=2^{2}=4$. From $b(a-b)=2 \times 7 \times 17^{2}$,

If $b=2, a-b=7 \cdot 17^{2} \Rightarrow a=2025, b=2$, then $(x, y, z)=(4100621,8100,4100629)$;
If $b=2 \cdot 7, a-b=17^{2} \Rightarrow a=303, b=14$, then $(x, y, z)=(91613,8484,92005)$;
If $b=2 \cdot 17^{2}, a-b=7 \Rightarrow a=585, b=578$, then $(x, y, z)=(8141,676260,676309)$;
If $b=2 \cdot 7 \cdot 17^{2}, a-b=1 \Rightarrow a=4047, b=4046$, then $(x, y, z)=(8093,32748324,32748325)$.
Note that $3 \times 5 \times 7 \times 11=1155,3 \times 5 \times 7 \times 11 \times 13=15015$, i.e., the number of odd prime factors of 2023 greater than 1 is at most 4, so $\max _{1 \leqslant n \leqslant 2023} a_{n}=2^{4}=16$.