Approximating $\sqrt{2}$ by Pythagorean triples? 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.
 A: 
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.
A: 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
$$ a_n = 2 q_{1,n} q_{1,n+1}\quad;\quad b_n = q_{2,n}q_{2,n+1}\quad;\quad c_n = q_{1,n}q_{2,n+1} + q_{2,n}q_{1,n+1}.$$
