Approximating $\sqrt{2}$ by Pythagorean triples?

Question

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.

Answer 1

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 $\,\sqrt{2}.$

Answer 2

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

Answer 3

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.