A system of the form $a+b\sqrt{2}$.

Given $x$ and $y$ are of the form $a+b\sqrt{2}$ ($a$ and $b$ are both positive integers) that satisfy the equation

$2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$.

Find such $x$ and $y$.

Let [MATH]\color{yellow}\bbox[5px,purple]{x=a+b\sqrt{2}}[/MATH], [MATH]\color{yellow}\bbox[5px,green]{y=c+d\sqrt{2}}[/MATH], where $a,\,b,\,c,\,d\in \Bbb{N}$.

Squaring the given equality, and rearrange it, we have:

$x^2+(y-8-4\sqrt{2})x+(6+4\sqrt{2}-4y-2\sqrt{2}y)=0$

Replace $y$ by [MATH]\color{yellow}\bbox[5px,green]{y=c+d\sqrt{2}}[/MATH], and rearrange the terms we get:

$x^2+(c-8+(d-4)\sqrt{2})x+(6-4c-4d+(4-4d-2c)\sqrt{2})=0$

Assume the quadratic in $x$ above has roots $x_1=a+b\sqrt{2}$ and $x_2=a-b\sqrt{2}$, we find from the sum and product of roots information that $d=4,\,c=-6$ but we're told $c>0$, so our assumption is wrong.

Therefore, it's safe to assume that the quadratic in $x$ above:

1. is part of the quartic equation $(x^2+(c-8+(d-4)\sqrt{2})x+(6-4c-4d+(4-4d-2c)\sqrt{2}))(g(x))=0$

2. has roots $x_1=a+2\sqrt{2},\,x_2=a-2\sqrt{2}$ and $x_3=p-2\sqrt{2},\,x_4=p+2\sqrt{2}$, where $p$ is an integer, and $b=2$, and together they formed the quartic that looks like:

$(x^2+(c-8+(d-4)\sqrt{2})x+(6-4c-4d+(4-4d-2c)\sqrt{2}))(g(x))$

$=(x-(a+2\sqrt{2}))(x-(p-2\sqrt{2}))(x-(a-2\sqrt{2}))(x-(p+2\sqrt{2}))=0$

$=\underbrace{(x-(a+2\sqrt{2}))(x-(p-2\sqrt{2}))}_{(x^2+(c-8+(d-4)\sqrt{2})x+(6-4c-4d+(4-4d-2c)\sqrt{2}))}\underbrace{(x-(a-2\sqrt{2}))(x-(p+2\sqrt{2}))}_{g(x)}=0$


Again, using the information from sum and product of rules we find:

$d=4$, $ap=-2(1+2c),\,a+p=8-c,\,p-a=-6-c$.

If we let $p=-2$, then we get $c=3$ and therefore $a=7$.

Thus, one of the solutions for $(x,\,y)$ is $(7+2\sqrt{2},\,3+4\sqrt{2})$.

You might wonder if there exists other solution(s) if you make other assumption. Yes, this type of problem has more than one solution that fits the given condition, therefore if you end up with other pair of $(x,\,y)$ and you don't make any slip in your calculation, then you're most probably correct.