<s>[INST] $\sqrt{3}$ represented as continued fraction Am I incorrect in the assumption that I deal with $\sqrt{3}$ in the same way I would approach $\sqrt{2}$, by adding and subtracting $1$ such that:
$\sqrt{3}=1+\sqrt{3}-1=1+\cfrac{2}{1+\sqrt{3}}$  
The table representation is $[1; 1,2]$.  
This is what I have so far:  
$\sqrt{3}=1+\cfrac{2}{2+\cfrac{2}{1+\sqrt{3}}}$   
Which seems to lead me to $[1;2,2,...]$ any help would be appreciated, thanks
 [/INST] If you obtain a rational approximation of $\sqrt{3}$, for example $\frac{362}{209}$ you'll find its continued fraction is $$1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}}}}}$$ That 3 looks extraneous, so we might form the conjecture that $$\sqrt{3}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}$$
Then we can prove that by calling the RHS x
$$x=1+\frac{1}{1+\frac{1}{1+\color{red}{1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}}=1+\frac{1}{1+\frac{1}{1+x}}$$
Then by simple algebra we can verify that $x=\sqrt{3}$. 
On a small digression, these rational approximations might seem hard to obtain but I found that with the chakravala method in a few minutes.
</s>