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I am trying to find sequence of Pythagorean triples $(x_{n}, y_{n}, z_{n})\in\mathbb{Z}^{3}$ such that $x_{n}/y_{n}\rightarrow 1$. This way, both $z_{n}/x_{n}$ and $z_{n}/y_{n}$ would converge to $\sqrt{2}$ as $n\rightarrow\infty$.

I am aware there are various other ways of approximating $\sqrt{2}$ by rational numbers such as here, here, and here, but none of these give me a way to generate Pythagorean triples in the way I desire. I am wondering if there is an elegant way to do this.

Maximal Ideal
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  • Are you familiar with the (old, popular) technique for constructing pythagorean triples by intersecting a unit circle with a line at two places, (-1, 0) and anywhere else in the positive-positive quadrant? – DanielV Nov 06 '22 at 21:01
  • Just to clarify : You want to create large pythogearean triples $(a,b,c)$ with $a\approx b$ WITHOUT using the value of $\sqrt{2}$ or any other closely related value. Is this right ? – Peter Nov 06 '22 at 21:05
  • @Peter I suppose yes. Preferably it'd give an algorithm to approximate/find $\sqrt{2}$ without knowing the value of $\sqrt{2}$ ahead of time (if that's what you were asking). – Maximal Ideal Nov 06 '22 at 21:16
  • @DanielV No. (I may have encountered it before, but probably forgot.) – Maximal Ideal Nov 06 '22 at 21:17
  • @MaximalIdeal Well if you learn that technique for generating pythagorean triples (there's a million youtube videos explaining it), your question is equivalent to the line approaching the intersection at $(\sqrt 2, \sqrt 2)$ – DanielV Nov 06 '22 at 21:27
  • I am not sure whether the below answers are hidden uses of sequences converging to $\sqrt{2}$ (the one I commented is definitely circular). But considering that even the Heron method establishes excellent approximations of $\sqrt{2}$ , this new approach to approximate it has little merit anyway. Far more interesting is the other way round , namely to find triples $(a,b,c)$ with $a\approx b$ with the help of $\sqrt{2}$ – Peter Nov 07 '22 at 07:57
  • @Peter but it is not circular. The OP knows how to generate rational approximations for $\sqrt{2}$. All I did was explain how to turn one of them into another approximation which is also of the form the OP wanted. I then gave a concrete example where I generated a rational approximation from scratch. – Numerical Disintegration Nov 07 '22 at 19:40

3 Answers3

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We begin with Euclid's formula for generating Pythagorean triples, shown here as

$$A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2$$ The $\,mk\,$ values for a series of triples where $\,|B-A|=1\,$ are contiguous Pell numbers, i.e. $\,\big\{1,2,5,12,29,70,169,\cdots\big\}$ and may be generated by the following funcions: $$ m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}$$ For example \begin{align*} m_1&= \frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 & k_1= \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1\\ m_2&= \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 & k_2= \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2\\ m_3&= \frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 & k_3= \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5\\ m_4&= \frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 &k_4= \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12\\ \end{align*}

The values grow quickly and exceed $\,15\,$ digits when $\,n>19.\quad$ Here is a series sample \begin{align*} F(2,1)&=(3,4,5)\\ F(5,2)&=(21,20,29)\\ F(12,5)&=(119,120,169)\\ F(29,12)&=(697,696,985)\\ F(70,29)&=(4059,4060,5741)\\ F(169,70)&=(23661,23660,33461)\\ F(408,169)&=(137903,137904,195025)\\ F(985,408)&=(803761,803760,1136689)\\ \end{align*} With $\,n\,$ sufficiently high, we can generate a Pythagorean triple where $C/A\text{ and }C/B\,$ are both arbitrarily close to $\,\sqrt{2}.$

poetasis
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    Very interesting (+1). I looked at the first few iterates and it coincides with my example (now with $q_{1,0}=2$ and $q_{2,0}=3$) as well as with J. W. Tanner's example. I wonder if we could prove this. – Numerical Disintegration Nov 07 '22 at 00:59
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    @Numerical Disintegration Another way of generating these triples may be what you are suggesting and comes from Wacław Sierpiński: Pythagorean triangles, THE SCRPTA MATHEMATICA STUDIES Number NINE, GRADUATE SCHOOL OF SCIENCE YESHIVA UNIVERSITY, NEW YORK, 1962, pp. 17-22 with a starting "seed" of $T_0=(0,0,1)$. $$T_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2 \qquad C_{n+1}=4A_n+3C_n+2$$ $$T_1=(3,4,5)\quad T_2=(20,21,29)\quad T_3=(119,120,168)\quad T_4=(696,697,985)\quad ...$$ – poetasis Nov 07 '22 at 03:35
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I am trying to find sequence of Pythagorean triples $(x_{n}, y_{n}, z_{n})\in\mathbb{Z}^{3}$ such that $x_{n}/y_{n}\rightarrow 1$.

Start with $(x,x+1,z)=(3,4,5)$.

Given $x^2+(x+1)^2=z^2$, it follows that $(3x+2z+1)^2+(3x+2z+2)^2=(4x+3z+2)^2$.

So take $x_{n+1}=3x_n+2z_n+1$, $y_{n+1}=x_{n+1}+1$, and $z_{n+1}=4x_n+3z_n+2$

to get a sequence you desire.

J. W. Tanner
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You can generate Pythagorean triplets as follows:

Take two integers, $q_1$ and $q_2$. Then define $p_1 = q_1 +q_2$ and $p_2 = 2q_1 +q_2$. Taking $a = 2 q_1 p_1$, $b=q_2 p_2$, $c = q_1 p_2 + q_2 p_1$ it holds that $a^2 + b^2 = c^2$. See here.

For $a_n/b_n\to 1$ we require to find $q_{1,n}$, $q_{2,n}$ such that $2q_{1,n}^2/q_{2,n}^2\to 1$.

That is $q_{2,n}/q_{1,n}\to \sqrt{2}$. But we can take any rational sequence that tends to $\sqrt{2}$. Taking $q_{2,n}$ and $q_{1,n}$ to be the numerator and denominator of such a sequence yields the result.

Edit:

After the criticism from Peter and the down vote (both of which I think were a bit unfair) here is a concrete example.

Let us consider $q_{1,0}=2$, $q_{2,0}=3$ and take $$ q_{1,n+1} = q_{1,n} + q_{2,n},$$ $$ q_{2,n+1} = 2q_{1,n} + q_{2,n}.$$ This is a rational approximation of $\sqrt{2}$ taken from this example given by the OP.

Then we can take $$ a_n = 2 q_{1,n} q_{1,n+1}\quad;\quad b_n = q_{2,n}q_{2,n+1}\quad;\quad c_n = q_{1,n}q_{2,n+1} + q_{2,n}q_{1,n+1}.$$

Numerical Disintegration
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