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I am trying to find an expansion into an arithmetic continued fraction of $\sqrt{x}$, where $x=((4a^2+1)b+a)^2+4ab+1$, where $a,b \in \mathbb{N}$.

So far I have:

Clearly $x < ((4a^2+1)b+a+1)^2$, so the integral part of $\sqrt{x}$, $I = (4a^2 + 1)b + a$, which means $x - I^2 = 4ab + 1$ and $a_1 = \dfrac{1}{\sqrt{x}-I}=\dfrac{\sqrt{x}+I}{x-I}$. Is my work correct so far? and, How might I proceed?

Clyde Kertzer
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1 Answers1

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hard to imagine why you would do this with no actual numbers For $a=1, b=1$ I get $x=41$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 41} = 6 + \frac{ \sqrt {41} - 6 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {41} - 6 } = \frac{ \sqrt {41} + 6 }{5 } = 2 + \frac{ \sqrt {41} - 4 }{5 } $$ $$ \frac{ 5 }{ \sqrt {41} - 4 } = \frac{ \sqrt {41} + 4 }{5 } = 2 + \frac{ \sqrt {41} - 6 }{5 } $$ $$ \frac{ 5 }{ \sqrt {41} - 6 } = \frac{ \sqrt {41} + 6 }{1 } = 12 + \frac{ \sqrt {41} - 6 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccc} & & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 6 }{ 1 } & & \frac{ 13 }{ 2 } & & \frac{ 32 }{ 5 } & & \frac{ 397 }{ 62 } & & \frac{ 826 }{ 129 } & & \frac{ 2049 }{ 320 } \\ \\ & 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 41 \cdot 0^2 = 1 & \mbox{digit} & 6 \\ \frac{ 6 }{ 1 } & 6^2 - 41 \cdot 1^2 = -5 & \mbox{digit} & 2 \\ \frac{ 13 }{ 2 } & 13^2 - 41 \cdot 2^2 = 5 & \mbox{digit} & 2 \\ \frac{ 32 }{ 5 } & 32^2 - 41 \cdot 5^2 = -1 & \mbox{digit} & 12 \\ \frac{ 397 }{ 62 } & 397^2 - 41 \cdot 62^2 = 5 & \mbox{digit} & 2 \\ \frac{ 826 }{ 129 } & 826^2 - 41 \cdot 129^2 = -5 & \mbox{digit} & 2 \\ \frac{ 2049 }{ 320 } & 2049^2 - 41 \cdot 320^2 = 1 & \mbox{digit} & 12 \\ \end{array} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ \sqrt { 130} = 11 + \frac{ \sqrt {130} - 11 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {130} - 11 } = \frac{ \sqrt {130} + 11 }{9 } = 2 + \frac{ \sqrt {130} - 7 }{9 } $$ $$ \frac{ 9 }{ \sqrt {130} - 7 } = \frac{ \sqrt {130} + 7 }{9 } = 2 + \frac{ \sqrt {130} - 11 }{9 } $$ $$ \frac{ 9 }{ \sqrt {130} - 11 } = \frac{ \sqrt {130} + 11 }{1 } = 22 + \frac{ \sqrt {130} - 11 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccc} & & 11 & & 2 & & 2 & & 22 & & 2 & & 2 & & 22 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 11 }{ 1 } & & \frac{ 23 }{ 2 } & & \frac{ 57 }{ 5 } & & \frac{ 1277 }{ 112 } & & \frac{ 2611 }{ 229 } & & \frac{ 6499 }{ 570 } \\ \\ & 1 & & -9 & & 9 & & -1 & & 9 & & -9 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 130 \cdot 0^2 = 1 & \mbox{digit} & 11 \\ \frac{ 11 }{ 1 } & 11^2 - 130 \cdot 1^2 = -9 & \mbox{digit} & 2 \\ \frac{ 23 }{ 2 } & 23^2 - 130 \cdot 2^2 = 9 & \mbox{digit} & 2 \\ \frac{ 57 }{ 5 } & 57^2 - 130 \cdot 5^2 = -1 & \mbox{digit} & 22 \\ \frac{ 1277 }{ 112 } & 1277^2 - 130 \cdot 112^2 = 9 & \mbox{digit} & 2 \\ \frac{ 2611 }{ 229 } & 2611^2 - 130 \cdot 229^2 = -9 & \mbox{digit} & 2 \\ \frac{ 6499 }{ 570 } & 6499^2 - 130 \cdot 570^2 = 1 & \mbox{digit} & 22 \\ \end{array} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ \sqrt { 370} = 19 + \frac{ \sqrt {370} - 19 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {370} - 19 } = \frac{ \sqrt {370} + 19 }{9 } = 4 + \frac{ \sqrt {370} - 17 }{9 } $$ $$ \frac{ 9 }{ \sqrt {370} - 17 } = \frac{ \sqrt {370} + 17 }{9 } = 4 + \frac{ \sqrt {370} - 19 }{9 } $$ $$ \frac{ 9 }{ \sqrt {370} - 19 } = \frac{ \sqrt {370} + 19 }{1 } = 38 + \frac{ \sqrt {370} - 19 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccc} & & 19 & & 4 & & 4 & & 38 & & 4 & & 4 & & 38 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 19 }{ 1 } & & \frac{ 77 }{ 4 } & & \frac{ 327 }{ 17 } & & \frac{ 12503 }{ 650 } & & \frac{ 50339 }{ 2617 } & & \frac{ 213859 }{ 11118 } \\ \\ & 1 & & -9 & & 9 & & -1 & & 9 & & -9 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 370 \cdot 0^2 = 1 & \mbox{digit} & 19 \\ \frac{ 19 }{ 1 } & 19^2 - 370 \cdot 1^2 = -9 & \mbox{digit} & 4 \\ \frac{ 77 }{ 4 } & 77^2 - 370 \cdot 4^2 = 9 & \mbox{digit} & 4 \\ \frac{ 327 }{ 17 } & 327^2 - 370 \cdot 17^2 = -1 & \mbox{digit} & 38 \\ \frac{ 12503 }{ 650 } & 12503^2 - 370 \cdot 650^2 = 9 & \mbox{digit} & 4 \\ \frac{ 50339 }{ 2617 } & 50339^2 - 370 \cdot 2617^2 = -9 & \mbox{digit} & 4 \\ \frac{ 213859 }{ 11118 } & 213859^2 - 370 \cdot 11118^2 = 1 & \mbox{digit} & 38 \\ \end{array} $$

Will Jagy
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    The reason is that in that particular form, there is a "nice" periodic expression for the continued fraction: $\sqrt{x} = [(4a^2+1)b+a;\overbrace{2a,2a,(8a^2+2)b+2a},\ldots]$ (I don't know how to write continued fraction, but that's what Sage gives). – Gareth Ma Apr 28 '22 at 22:24
  • @GarethMa yes. If you look at the ends of my three cf's, you see your "digits" on the right column. I like to display the digits until the Pell $p^2 - n y^2 = 1$ appears; this takes two of your repeated portion, as for these numbers $n,$ there are solutions to $p^2 - n q^2 = -1$ – Will Jagy Apr 28 '22 at 22:31
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    I see it now, thanks for the clarification ;) Let's see if OP will be satisfied with "observing pattern from small numbers then proving pattern is correct" – Gareth Ma Apr 28 '22 at 22:33
  • @GarethMa one can just manage the proof by concentrating on $ w^2 - n y^2 = -1,$ where the $n$ is his original $x,$ then $w = 16ba^4 + 4a^3 + 8ba^2 + 3a + b, $ and $y= 1 + 4 a^2 .$ Here the fraction $ \frac{w}{y}$ is the third convergent of the c.f. you specify. – Will Jagy Apr 29 '22 at 17:52