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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 solved for a large number of Pythagorean triples, $12r^2 + a_{r - 1}b_{r-1} = a_rb_r$, a recursive formula in terms of inradius($r$). Its important to note that $12r^2$ is twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$ although how this is exactly related, I am still unsure of. Using this formula we have two sequences of primitive Pythagorean triples.

Triplets with an odd value of $a$ where $r$ increases by $1$ :

$(3,4,5), (5,12,13), (7,24,25)...$

Triplets with an even value of $a$ where $r$ increases by $1$:

$(8,15,17),(12,35,37), (16,63,65)...$

Since we have a recursive formula, the area of each Pythagorean triplet can be found using the first term of each sequence and with the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$ we also can find their side lengths. My question is simple, can anyone prove or disprove this?

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  • $\begingroup$ What is meant by $b_r$? How does your recursive formula relate to the sequences of triples you've written? $\endgroup$ – kccu Feb 12 at 2:19
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    $\begingroup$ Thanks for the suggestion, I have fixed it $\endgroup$ – SpoonedBread Feb 12 at 2:27
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    $\begingroup$ Your triples are the upper and lower paths in the ternary tree of Pythagorean triples $\endgroup$ – Bill Dubuque Feb 12 at 2:28
  • $\begingroup$ Could you please explain why the recursive formula is related to this? $\endgroup$ – SpoonedBread Feb 12 at 14:33
  • $\begingroup$ To me, it looks like your formula is not for generating Pythagorean triples but rather for finding a multiplier given area. Also, for primitive triples, given Euclid's formula, et al, side-A is always odd, side-B is always even, and for half of all triples, $A>B$. There are ways of finding triples given an area if you are interested. $\endgroup$ – poetasis Feb 12 at 15:53
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Your use of $(8,15,17)$, etc. is not representative of any primitive triple generated by any formula except one I know of that generates triples with $B-A=1$ using a seed $T_0=(0,0,1)$. \begin{equation}T_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2 \qquad C_{n+1}=4A_n+3C_n+2\end{equation} $$T_1=(3,4,5)\quad T_2=(20,21,29)\quad T_3=(119,120,168)\quad\textbf{ ...}$$

I'm not sure what you are generating. Your first set of triples where $C-B=1$ is generated when $n=1$ and the second set where $C-A=2$ is generated when $k-1$ by this formula

\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 $$\begin{array}{c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline \end{array}$$ Perhaps your formula generates the first row and the first collumn in this sample but does it generate those where $n$ and $k$ are both greater than one?

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