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.
Best Answer
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.