What are the new formulas for x and y that will converge to $\sqrt{k}$?

[ineedhelpnow]

I'm not sure which category to post this question under :)

I'm not sure if any of you are familiar with "Greek Ladders"
I have these two formulas:
${x}_{n+1}={x}_{n}+{y}_{n}$
${y}_{n+1}={x}_{n+1}+{x}_{n}$

x	y	$\frac{y}{x}$
1	1	1
2	3	1.5
5	7	~1.4
12	17	~1.4
29	41	~1.4
70	99	~1.4

As shown in the table, the higher n gets, the closer $\frac{{y}_{n}}{{x}_{n}}$ converges to $\sqrt{2}$ (which is approximately 1.4).
$\lim_{{n}\to{\infty}}\frac{{y}_{n}}{{x}_{n}}=L$

If n is large enough, then $\frac{{y}_{n}}{{x}_{n}}=\frac{{y}_{n+1}}{{x}_{n+1}}\approx L$

$\frac{{y}_{n}}{{x}_{n}}=\frac{{x}_{n}+{x}_{n+1}}{{x}_{n}+{y}_{n}}$

$\frac{{y}_{n}}{{x}_{n}}=\frac{{x}_{n}+{x}_{n}+{y}_{n}}{{x}_{n}+{y}_{n}}$

$\frac{{y}_{n}}{{x}_{n}}=\frac{{2x}_{n}+{y}_{n}}{{x}_{n}+{y}_{n}}$

${y}_{n}{x}_{n}+{{y}_{n}}^{2}={{2x}_{n}}^{2}+{x}_{n}{y}_{n}$

$\frac{{{y}_{n}}^{2}}{{{x}_{n}}^{2}}=2$

$L=\frac{{y}_{n}}{{x}_{n}}=\sqrt{2}$

SO, as shown, the two formulas at the top will always close into root 2 as n increases. My question is how many I alter those formulas so that they may converge to $\sqrt{k}$. So for example, not just 2. But 3. Or 5. Etc.

 

[MarkFL]

To extend the ladder for $\sqrt{k}$, you want:

\(\displaystyle x_{n+1}=x_n+y_n\)

\(\displaystyle y_{n+1}=x_{n+1}+(k-1)x_n\)