# Generating Pythagorean Triples with 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 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?

• What is meant by $b_r$? How does your recursive formula relate to the sequences of triples you've written? – kccu Feb 12 at 2:19
• Thanks for the suggestion, I have fixed it – SpoonedBread Feb 12 at 2:27
• Your triples are the upper and lower paths in the ternary tree of Pythagorean triples – Bill Dubuque Feb 12 at 2:28
• Could you please explain why the recursive formula is related to this? – SpoonedBread Feb 12 at 14:33
• 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. – poetasis Feb 12 at 15:53

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 $$T_0=(0,0,1)$$. $$$$T_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2 \qquad C_{n+1}=4A_n+3C_n+2$$$$ $$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?