Rationally Irrational

Let,

$a=x + \cfrac{1}{y + \cfrac{1}{x + \cfrac{1}{y+_\ddots} } }$

Then, $\sqrt{38}=x+\cfrac{1}{a}=x+\cfrac{1}{x+\cfrac{1}{y+\cfrac{1}{a}}}$

Solving for $a$ from the above equations, we get

$a=\cfrac{xy\pm\sqrt{x^2y^2+4xy}}{2y}$

$\implies x+\cfrac{2y}{xy\pm\sqrt{x^2y^2+4xy}}=\sqrt{38} \implies (x-\cfrac{y}{2})\pm\cfrac{1}{2x}\sqrt{x^2y^2+4xy}=\sqrt{38}$ [by rationalization of denominator]

Now, for any three integers $(p,q,r)$ it is not possible to write $\sqrt{p}=q+\sqrt{r}$ where $p$ or $r$ are not perfect squares. It is very easy to prove by taking squares on both sides.

So, $x=\cfrac{y}{2}$ and $\cfrac{1}{2x}\sqrt{x^2y^2+4xy}=\sqrt{38}$

$\implies x=6, y=12$ and hence, $\boxed{x+y=18}$

let $n=a^2+b$ where $a,b$ are integers, $a,b\neq0$

$n=a^2+b\\ n-a^2=b\\ (\sqrt{n}+a)((\sqrt{n}-a)=b\\ \sqrt{n}=a+\dfrac{b}{a+\sqrt{n}}$

substituting for $\sqrt{n}$ repeately, we get a contiued fraction

$\sqrt{n}=a+\dfrac{b}{2a+\dfrac{b}{2a+\dfrac{b}{2a+_\ddots}}}=a+\dfrac{1}{\dfrac{2a}{b}+\dfrac{1}{2a+\dfrac{1}{\dfrac{2a}{b}+_\ddots}}}$

in the given question $n=38$

also from the information that x and y are integers,

we get, $\dfrac{2a}{b}$ is an integer

so $a=6$ and $b=2$ is the only possibilty

thus we get $\sqrt{38}=6+\dfrac{1}{\dfrac{12}{2}+\dfrac{1}{12+\dfrac{1}{\dfrac{12}{2}+_\ddots}}}=6+\dfrac{1}{6+\dfrac{1}{12+\dfrac{1}{6+_\ddots}}}$

thus $x=6$ and $y=12$ , $x+y=12+6=\boxed{18}$