I realised recently I could use this equation for the pythagorean triples.
As you can see the 345 triangle seems to be special? as we can create it using x=2 AND x=3.
I know the 345 triangle is related to this golden triangle.
Using the theorem of geometry called the Power of a Point.
why is there a square-root relationship here**?**
Is it correct to say that the angle of the 345 triangle is related to the Geometric Progression. 4, 2, 1, 1/2, 1/4 where x = (a+b) = 2 and 1/x = (a-b) = 1/2, x^2 = 4 and 1/x^2 = 1/4, x^4 = 16 and 1/x^4 = 1/16 and therefore the square as shown below**?**
We find the 345 triangle in the construction of the square as shown below.
This is Euclid's formula in another form shown here $$A=(m^2-k^2)\quad B=2mk\quad C=m^2+k^2)$$ I use $\space F(m,k)\space $ but others use $\quad F(m,n),\space F(p,q),\space F(s,t), \cdots$
In this case, $\space m=x,\space k=1 \quad A=\bigg(x-\frac{1}{x}\bigg) \quad B=2^2 \quad C= \bigg(x+\frac{1}{x}\bigg)^2\quad$ and we shall see how it transforms back to Euclid's formula by multiplying each of the three terms by $\space x^2.$
\begin{align*} 2^2 &= \bigg(x+\frac{1}{x}\bigg)^2 - \bigg(x-\frac{1}{x}\bigg)^2 \\ \bigg(x-\frac{1}{x}\bigg)^2 +2^2 &= \bigg(x+\frac{1}{x}\bigg)^2 \\ \big(x^2-1\big)^2+\big(2x(1)\big)^2 &= \big(x^2+1\big)^2 \\ A^2+B^2&=C^2 \end{align*}
What this formula $\space F(m,1)\space $ generates is \begin{align*} &F(2,1)=(3,4,5),\quad F(3,1)=(8,6,10),\\ &F(4,1)=(15,8,17),\quad F(5,1)=(24,10,26),\\ &F(6,1)=(35,12,37),\quad F(7,1)=(48,14,50),\\ &F(8,1)=(63,16,65),\quad F(9,1)=(80,18,82),\\ &F(10,1)=(99,20,101),\quad F(11,1)=(120,22,122),\space \cdots \end{align*} These are not your results and note that every other triple is imprimitive.
Is the golden ration related to $\space(3,4,5)?$
Given $\space\tan^{-1}(\phi) \approx \tan^{-1}(1.61803398874989) \approx 58.28^\circ \\\longrightarrow F(617,344)=(262353,424496,499025)\\ \dfrac{424496}{262353}\approx 1.61803371792966 $
$$ (3,4,5)\longrightarrow \tan^{-1} (1.333\space\cdots) \approx 53.13^\circ\\ \longrightarrow F(2,1)=(3,4,5)$$