Find x and y of form a+b√2 satisfying 2x+y-√(3x²+3xy+y²)=2+√2

[anemone]

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 $y$.

 

[Greg]

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 $y$.

Spoiler

Rearranging and squaring, then rearranging again (with the quadratic formula in mind), we arrive at

$$x=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+12(2+\sqrt2)^2}}{2}=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+72+48\sqrt2}}{2}$$

We now seek positive integers $a,b,c,d$ such that

$$a^2+2b^2+72=c^2+2d^2$$

and

$$2ab+48=2cd$$.

By inspection, $a=b=1$ and $c=d=5$, so $x=6+4\sqrt2,y=1+\sqrt2$ are one such $x$ and $y$.

 

[anemone]

Spoiler

Rearranging and squaring, then rearranging again (with the quadratic formula in mind), we arrive at

$$x=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+12(2+\sqrt2)^2}}{2}=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+72+48\sqrt2}}{2}$$

We now seek positive integers $a,b,c,d$ such that

$$a^2+2b^2+72=c^2+2d^2$$

and

$$2ab+48=2cd$$.

By inspection, $a=b=1$ and $c=d=5$, so $x=6+4\sqrt2,y=1+\sqrt2$ are one such $x$ and $y$.

Thanks greg1313 for participating! Your solution is neat and elegant, good job!

My solution (which is clearly more convoluted than yours):

Spoiler

Let $x=a+b\sqrt{2},\,y=c+d\sqrt{2}$, where $a,\,b,\,c,\,d\in \Bbb{N}$.

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

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

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

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

Next, it's safe to assume that the quadratic in $x$ above has roots $x_1=a+2\sqrt{2}$ and $x_2=p-2\sqrt{2}$, where $p$ is an integer, and $b=2$, and comes from

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

Again, we find $d=4$. We find $ap=-2(1+2c),\,a+p=8-c,\,p-a=-6-c$. If we let $p=-2$, then we get $c=3$ and therefore $a=7$.

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