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How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$?

I've been reading the Plimpton 322 news, and this fits in the gap in the Babylonian table between 0.6996 ($a=1679,b=2400$) and 0.75 ($a=3,b=4$). The Babylonians apparently tabulated a lot of triples and then looked for the best one, but is there a more direct way?

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    $\begingroup$ This amounts to finding natural numbers $u>v$ such that $\frac{u^2-v^2}{2uv}\approx \frac57$ or $\frac75$. $\endgroup$ – Arthur Aug 29 '17 at 13:36
  • $\begingroup$ $400^2+561^2=689^2$, $561/400\approx 7/5$ $\endgroup$ – Professor Vector Aug 29 '17 at 13:40
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    $\begingroup$ @prof That triple wouldn't happen to satisfy the constraints of the Plimpton 322 tablet though, since the longer side is 561 which is not regular in base 60. They needed that length to be regular since they use it as a divisor to get other numbers in the table. $\endgroup$ – Joe Knapp Aug 31 '17 at 1:39
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Pythagorean triplets are characterized by being of the form $$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$ So you could look for two integers $m>n$ such that $$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$ or, equivalently, $$\frac{m}{n}- \frac{n}{m} \approx 2 \cdot \frac{5}{7}$$

Now, let $x$ be unique positive solution of $$x-x^{-1} = 2 \cdot \frac{5}{7}$$ one can easily compute that $x= \frac{5 + \sqrt{74}}{7} \approx 1.94$. The idea is to appximate $x= m/n$with a rational number. The continued fraction of $x$ is periodic and it is $[1; \overline{1, 16, 1,1,1}]$. This gives us the first convergents of $x$: $$\frac{1}{1} , \frac{2}{1} , \frac{33}{17} , \frac{35}{18} , \frac{68}{35} , \frac{103}{53} , \frac{171}{88},\dots$$ which are the best rational approximations of $x$. Taking $m= 33$ and $n= 17$ you get the triplet $$a= 800 ; \ \ b= 1122 ; \ \ c= 1378$$ which is not a primitive triplet. Dividing it by $2$ you get the primitive triplet $$a= 400 ; \ \ b= 561 ; \ \ c= 689$$ with $$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{400}{561} - \frac{5}{7} \right|= \frac{5}{3927} \approx 0.0013$$

Going on like this ($m= 35 , n=18$) you get the next solution $$a= 901 ; \ \ b= 1260 ; \ \ c= 1549$$ with $$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{901}{1260} - \frac{5}{7} \right|= \frac{1}{1260} \approx 0.0008$$ and so on.

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The parametrization $c=m^2+n^2, a=m^2-n^2, b=2mn$ comes from the fact that $-m/n$ is the slope of the line connecting points $Q=(1,0)$ and $P=(a/c,b/c)$ of the unit circle.

When you want $a:b=5:7$, you want to be near the point $P=(5/\sqrt{74},7/\sqrt{74})$. In other words, you want $$ \frac mn\approx\frac{7}{\sqrt{74}-5}. $$ A way to find progressively better rational approximations to a given real number such as $7/(\sqrt{74}-5)$ is to expand it as a continued fraction.

Some such approximations $$ \begin{array}{c|c|c|c|c} m&n&c&a&b\\ \hline 33&17&1378&800&1122\\ 68&35&5849&3399&4760 \end{array} $$

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  • $\begingroup$ Pointless for me to continue the table. JKabrg and Crostul (+1) have given the periodic continued fraction, and that is what we need here. $\endgroup$ – Jyrki Lahtonen Aug 29 '17 at 14:01
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$$\left(a \over b\right)^2 + 1 = \left(c \over b\right)^2 $$ $$\implies 1 = \left(c \over b\right)^2 - \left(a \over b\right)^2$$ $$\implies 1 = x^2 - y^2 \textrm{ where } x = {c \over b} \textrm{ and } y={a \over b}$$

So what you want to find is a rational point on the unit rectangular hyperbola for which $y \approx {5 \over 7}$.

The following is an exhaustive formula for the rational points on the unit hyperbola (derivable from Euclid's formula for the Pythagorean triples, and with a geometric meaning described by J. Lahtonen):

$$x = {1 + t^2 \over 1 - t^2}, y = {2t \over 1 - t^2}$$

Now we find the exact solution for ${2t \over 1 - t^2} = {5 \over 7}$, and then find rational approximations to the resulting $t$.

The exact value we get is $t = {\sqrt{74} \over 5} - {7 \over 5}$. We can get rational approximations using the continued fraction $[0; \overline{3, 8,3}]$. This gets you values of $t$ of $1/3, 8/25, 25/78...$

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Looks like you noticed the gap between row 11 and row 12 of Plimpton 322? The scribe left out a triple there for some reason--it would have been:

$a = 11529$, $b = 16000$, $c = 19721$

Or in the sexagesimal system:

$a=3.12.9$, $b=4.26.40$, $c=5.28.41$

${a\over{b}} = {5.044\over{7}}$

You could find this triple by tabulating all possible triples (using, say, the $m,n$ method) up to some maximum $c$, say 20,000, and selecting only those triples that have the longer non-hypotenuse side length as a regular number in the sexagesimal system, i.e., expressible as $2^x3^y5^z$, with $x,y,z \in \mathbb{N}$. That length had to be regular since they use it as a divisor to get $\delta^2$ in the first column.

If you sort that list by ${a\over{b}}$ it turns out that there is only a very limited set of possible ratios in the list and if you take the simplest triangle (smallest short side) for each distinct ratio value you get the rows of Plimpton 322, plus the ones the scribe left out.

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  • $\begingroup$ I think you mean $2^x 3^y 5^z$, without the +. $\endgroup$ – user210229 Aug 31 '17 at 3:44
  • $\begingroup$ @Matt: Thanks, corrected. $\endgroup$ – Joe Knapp Aug 31 '17 at 8:28
  • $\begingroup$ Thanks to you for answering the question I might have asked about the selection of entries! It would be an interesting addition to the recent article to fill in both missing columns and missing rows. $\endgroup$ – user210229 Aug 31 '17 at 14:21
  • $\begingroup$ @Matt -- Check the paper, Table 9 is an expanded table (38 entries), along with two columns that hypothetically were on the broken off piece. sciencedirect.com/science/article/pii/S0315086017300691 $\endgroup$ – Joe Knapp Aug 31 '17 at 17:47
  • $\begingroup$ Right, that's why it would be an interesting addition to fill in the missing rows too. $\endgroup$ – user210229 Aug 31 '17 at 18:36

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