Generating Pythagorean Triples Using a New Method? Using a right triangle with side lengths $(a,b,c)$ where $a , b < c$,
I was thinking about how the area of a Pythagorean triple can be found using the Pythagorean triple right before it and I came across something that worked for a large number of Pythagorean triples,  $12r^2 + a_{k- 1}b_{k - 1} = a_kb_k$, a recursive formula where $k$ represents the $k$th term in a sequence. This seemingly generates a sequence of Pythagorean triples that I could not find used in any other formula. Its important to note that $12r^2$ is  twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$. Using this formula we can find the $1st$ term of sets where the inradius of each Pythagorean triple is $r + r^2k$ and the relationship between the side lengths are still defined by our recursive formula.
These $1st$ terms are triplets with an even value of $a$ where $r$ increases by $2$:
$(8,15,17),(12,35,37),(16,63,65)...$
Note: We find this using $(8,15,17)$ as we have a recursive formula as well as the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$,  which lets us find the side lengths of each Pythagorean triple.
Here is a sample of what they generate:
$$\begin{array}{c|c|c|c} 
set_1&15,8,17&33,56,65&51,140,149&69,260,269 \\ \hline
set_2&35,12,37&85,132,157&135,352,377&185,672,697 \\ \hline
set_3 &63,16,65&161,240,289&259,660,709&357,1276,1325& \\ \hline
set_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153  \\ \hline
\end{array}$$
I couldn't seem to find any similar formulas to this one or any method of generating Pythagorean triples that follow this sequence, I am looking for a proof.
 A: Your formula does generate Pythagorean triples but misses most of them and appears to require seeds to work.
I'm not sure what you are generating. You do generate triples where $C-A=2$ in the first column but that can be generated more easily by
$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\quad $ The rest of the table shows no pattern that I can see, like a consistent side difference within a set or consistent increment of side values within a set. The following formula generates all primitives and a few that are not but there is a consistent
$C-B=(2n-1)^2\quad$ and $\quad A_{n+1}-A_{n}=2(2n-1).\quad$ It is the formula derive when
$A=(2n-1+k)^2-k^2,\space
B=2(2n-1+k)k,\space
C=(2n-1+k)^2+k^2$
\begin{align*}
A=(2n-1)^2+ \quad &2(2n-1)k\\
B=\hspace{55pt} &2(2n-1)k\quad+2k^2\\ 
C=(2n-1)^2+ \quad &2(2n-1)k\quad +2k^2
\end{align*}
Here is a sample of what it generates
$$\begin{array}{c|c|c|c|c|} 
n & k=1 & k=2 & k=3 & k=4   \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41  \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline
\end{array}$$
Your formula generates the first column but nothing with a pattern I can see in the other cells. If you do want to work with areas, there is a list of them here.  If you can figure out how to generate this sequence, I can show you how to find all of the
$1,\space 2, \text{ or } 3\space $ triples
that correspond to each area.
A: We need to write generally speaking the more General equation:
$$aX^2+bXY+cY^2=jZ^2$$
Although I formula solutions recorded, but I see it is of interest expression solutions using any one of the known solution.
If we know what any one solution: $(x,y,z)$  - then you can write a formula for the solutions of this equation.
$$X=jxt^2-cxk^2+2(cyk-jzt)s+(by+ax)s^2$$
$$Y=jyt^2-2jztk+(cy+bx)k^2+2axks-ays^2$$
$$Z=jzt^2-(bx+2cy)kt+czk^2+(bzk-(2ax+by)t)s+azs^2$$
$k,t,s$ - any integer asked us.
