Generating Pythagorean Triples Using a New Method?

Question

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.

Answer 1

This formula will generate your table \begin{align*} A=(2n-1)^2+ & 2(2n-1)k \\ B= \qquad & 2(2n-1)k+ 2k^2 \\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2 \end{align*}

if provided with the following $(n,k)$ values $$\begin{array}{c|c|c|c|c|c|} X & Term_1 & Term_2 &Term_3 & Term_4 & Term_5 \\ \hline Set_1 & (1,1) & (1,2) & (1,3) & (1,4 & (1,5) \\ \hline Set_2 & (2,1) & (2,4) & (2,7) &( 2,10) & (2,13) \\ \hline Set_3 & (3,1) & (3,6) & (3,11) & (3,16) & (3,21) \\ \hline Set_4 & (4,1) & (4,8) & (4,15) & (4,22) & (4,29) \\ \hline Set_5 & (5.1) & (5,10) & (5,19) & (5,28) & (5,37) \\ \hline Set_6 & (6,1) & (6,12) & (6,23) & (6,34) & (6,45) \\ \hline \end{array}$$`

\begin{align*} \text{To achieve this we replace k by}\qquad \bigg((2n-1)(k-1)+1\bigg) \end{align*}

\begin{align*} A=(2n-1)^2+ & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg) \\ B= \qquad\qquad\quad & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg)+ 2 \bigg((2n+-)(k-1)+1\bigg)^2. \\ C=(2n-1)^2+ & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg)+ 2 \bigg((2n-1)(k-1)+1\bigg)^2 \end{align*}

and the result is

$$\begin{array}{c|c|c|c|c|c|} X & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & (3,4,5) & (5,12,13) & (7,24,25) & (9,40,41 ) & (11,60,61) \\ \hline Set_2 & (15,8,17) & (33,56,65) & (51,140,149) &( 69,260,269) & (87,416,425) \\ \hline Set_3 & (35,12,37) & (85,132,157) & (135,352,377) & (185,672,697) & (235,1092,1117) \\ \hline Set_4 & (63,16,65) & (161,240,289) & (259,660,709) & (357,1276,1325) & (413,1716,1765) \\ \hline Set_5 & (99,20,101) & (261,380,461) & (423,1064,1145) & (585,2072,2153) & (747,3404,3485) \\ \hline Set_6 & (143,24,145 ) & (385,552,673) & (627,1564,1685) & (869,3060,3181) & (1111,5040,5161) \\ \hline \end{array}$$`

$\textbf{Update}\qquad $. Strictly speaking, this formula is not recursive in that no triple is dependent on any other. It can be viewed as recursive, however, because $\quad\large{k_x = k_{x-1} + (2n-1)}$.

e.g. For $Set_3,\space$ where $(2n-1)=5,\space$ for column one, $\quad k_0=1+(1-1)(5)=1,\quad$ for column two $\quad k_1=k_0+(2-1)(5)=k_0+5=6,\quad$ for column three $\space k_2=k_0+(3-1)(5)=1+(5+5)=(1+5)+5=k_1+5=11,\space $ etc.

To implement this recursiveness, we let the first triple $T_1$ in each set be $$A=4n^2-1\qquad B=4n\qquad C=4n^2+1$$ and let all other $T_x$ be \begin{align*} A=(2n-1)^2+ & 2(2n-1)k_x \\ B= \qquad & 2(2n-1)k_x + 2k_x ^2 \\ C=(2n-1)^2+ & 2(2n-1)k_x + 2k_x ^2 \end{align*} where $\qquad k_x=k_{(x-1)}+(2n-1)$

Answer 2

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.

Answer 3

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.