Solve for $x$ and $y$, $(\sqrt x+\sqrt 3)^2=y+6\sqrt 2$

Question

I have this simple looking problem, but I am stumped. It isn't for homework, it's just for fun.

Solve for $x$ and $y$, $(\sqrt x+\sqrt 3)^2=y+6\sqrt 2$, where $x$ and $y$ are positive integers.

I get $x=6$ and $y=9$ from Wolfram, but I'm not sure about steps. Can anyone help?

So far I have $x+2 \sqrt {3x} + 3 = y + 6\sqrt 2$

Answer 1

$x$ must have $\sqrt2\cdot\sqrt3$ as factor to produce $2\sqrt{3x}=6\sqrt2$

$\implies3x=18$

and we already have $x+3=y$

Answer 2

Hint:

$2\sqrt{3x}-6\sqrt2=y-x-3\in\mathbb{Z}_{\ge0}$ therefore there is one only possibility,since $x$ an integers, $2\sqrt{3x}-6\sqrt2=0$

Answer 3

From $x+2 \sqrt {3x} + 3 = y + 6\sqrt 2$ we get $$\underbrace{x-y+3\over 2}_{q\in \mathbb{Q}} = 3\sqrt 2-\sqrt{3x}\;\;\;(*)$$

Now we square it again $$q^2 = 18-6\sqrt{6x} -3x$$ so$$\underbrace{q^2 +3x-18\over -6}_{r\in \mathbb{Q}}= \sqrt{6x}$$ So $\sqrt{6x} = {a\over b}$ for some relatively prime integers $a,b$ (and we can assume $b>0$). From here we see that (after squaring and clearing the denominator) $$b^26x=a^2\implies b^2\mid a^2\implies b=1$$ So $x=6c^2$. If we plug this in equation $(*)$ we get, if $c\ne 1$ $$\sqrt{2}= {q\over 3-3c}\in \mathbb{Q}$$

which is nonsense. So $c=1$ and $x=6$ and $y=9$