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For our purpose,let $a,b,c$ and $x\gt2$ be natural numbers such that the positive integers $a,b$ and $c$ form a special pythagorean triple $(a,b,c)$,then it is conjectured that the following is true

$$G(x)=\cfrac{2}{3x+\cfrac{(-1)(7)} {9x+\cfrac{(2)(10)}{15x+\cfrac{(5)(13)}{21x+\cfrac{(8)(16)}{27x+\ddots}}}}}$$

$$G(x)=\frac{a}{2b}+\frac{c}{b}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c+a}{2c}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c-a}{2c}\right) \right]$$

where,

$$a=4x(x^2-1)\\b = x^4-6x^2+1\\c=(x^2+1)^2$$

such that,

$$a^2+b^2=c^2=(x^2+1)^4$$

enter image description here

Corollaries

Here are some of its closed forms with their related not necessarily primitive pythagorean triples next to them

$$G(3)=\frac{12}{7}+\frac{25}{7}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{50}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1}{50}\right) \right] ; (7,24,25)$$

$$G(4)=\frac{120}{161}+\frac{289}{161}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{529}{578}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{578}\right) \right] ;(161,240,289)$$

$$G(5)=\frac{60}{119}+\frac{169}{119}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{289}{338}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{289}\right) \right] ;(119,120,169)$$

$$G(6)=\frac{420}{1081}+\frac{1369}{1081}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2209}{2738}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{529}{2738}\right) \right] ;(840,1081,1369)$$

$$G(7)=\frac{168}{527}+\frac{625}{527}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{961}{1250}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{289}{1250}\right) \right] ;(336,527,625)$$

$$G(8)=\frac{1008}{3713}+\frac{4225}{3713}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{6241}{8450}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2209}{8450}\right) \right] ;(2016,3713,4225)$$

$$G(9)=\frac{360}{1519}+\frac{1681}{1519}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2401}{3362}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{961}{3362}\right) \right] ;(720,1519,1681)$$

$$G(10)=\frac{1980}{9401}+\frac{10201}{9401}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{14161}{20402}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{6241}{20402}\right) \right] ;(3960,9401,10201)$$

$$G(11)=\frac{660}{3479}+\frac{3721}{3479}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{5041}{7442}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2401}{7442}\right) \right] ;(1320,3479,3721)$$

$$G(12)=\frac{3432}{19873}+\frac{21025}{19873}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{27889}{42050}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{14161}{42050}\right) \right] ;(6864,19873,21025)$$

and so on for all values of $x\gt2$...

Question :Can the conjecture be rigorously proved,so that the connection between the continued fraction and the primitive pythagorean triples can be established?

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Nicco
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    I modified your post a bit to give a simpler and unified formula. – Tito Piezas III May 07 '16 at 16:20
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    It's a beautiful cfrac, by the way, and with an unexpected connection to a special kind of Pythagorean triples that involve squares of squares. – Tito Piezas III May 07 '16 at 16:23
  • @Tito Piezas III :thanks Tito,your modification did the post much justice. – Nicco May 07 '16 at 17:29
  • Hmm... the expression of hypergeometric functions looks familiar. For $\theta \in [0,\pi]$, we have $$ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1-\cos\theta}{2}\right)
    • {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1+\cos\theta}{2}\right)

    = \cos\frac{\theta+\pi}{3} $$ Not sure whether this helps.

     – achille hui May 09 '16 at 04:19
  • @achille hui :yes it does help,as it reveals that the closed form is actually a cubic root as observed by Tito ,but I avoided it in favour of the hypergeometric function at rational points for the sake of neatness. – Nicco May 09 '16 at 12:11
  • @Nicco: I found the general solution of the superfamily which this belongs to. (They are algebraic numbers of degree $m$.) However, it remains to be proven rigorously. Kindly see this post. – Tito Piezas III May 09 '16 at 21:06

1 Answers1

1

(Too long for a comment.)

I. Level $2$

From your other post, we have,

$$G_1(x,n)=\cfrac{1}{2x+\cfrac{(-1)(-1+n)} {6x+\cfrac{(1)(1+n)}{10x+\cfrac{(3)(3+n)}{14x+\cfrac{(5)(5+n)}{18x+\ddots}}}}}\tag1$$

with $2v+1 =-1,1,3,5,\dots$ The special case $n=6$,

$$G_1(x,\color{brown}6)=\frac{1}{3b}\left(a+\sqrt{c^3}\right)\tag2$$

where,

$$a=-x^3+3x,\;b=3x^2-1,\;c=x^2+1$$

So $G_1(x,\color{brown}6)$ is a root of a quadratic. Note that,

$$\color{blue}{a^2+b^2=c^3=(x^2+1)^3}\tag3$$

II. Level $3$

From this post, we have the analogous,

$$G_2(x,n)=\cfrac{1}{3x+\cfrac{(-1)(-1+n)} {9x+\cfrac{(2)(2+n)}{15x+\cfrac{(5)(5+n)}{21x+\cfrac{(8)(8+n)}{27x+\ddots}}}}}\tag4$$

with $3v+2 =-1,2,5,8,\dots$ The special case $n=8$,

$$G_2(x,\color{brown}8)=\frac{a}{4b}+\frac{c}{2b}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c+a}{2c}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c-a}{2c}\right) \right]\tag{5a}$$

Equivalently,

$$G_2(x,\color{brown}8)=\frac{1}{4b}\left(a+c\Big(u+\frac{1}{u}\Big)\right)\tag{5b}$$

where $\displaystyle u =e^{4\pi\, i /3}\left(\frac{a+b\,i}{c}\right)^{1/3}$ and,

$$a = 2m,\;b=m^2-1,\;c=m^2+1,\quad\text{with}\quad m=\frac{x^2-1}{2x}$$

So $G_2(x,\color{brown}8)$ is a root of a cubic. Note that,

$$\color{blue}{a^2+b^2=c^2=\left(\frac{x^2+1}{2x}\right)^4}\tag6$$

III. Level $4$

Another cfrac with a closed-form, presumably a root of a quartic?

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