MHB What are the integer values for x, y, z, a, b, c in the equation?

[anemone]

Given that $1+\sqrt{2}+\sqrt{2(2+\sqrt{2})}=\sqrt{\dfrac{x+\sqrt{y+\sqrt{z}}}{a-\sqrt{b+\sqrt{c}}}}$, where $x,\,y,\,z,\,a,\,b,\,c$ are integers. Find the values for all of integers $x,\,y,\,z,\,a,\,b,\,c$.

 

[anemone]

Hint:

Spoiler

Note that for example if we have $36+20 \sqrt{2}+8 \sqrt{2+\sqrt{2}}+4 \sqrt{2 (2+\sqrt{2}})$, it can be factorized as $(4+2\sqrt{2})(8+\sqrt{2}+2\sqrt{2+\sqrt{2}})$.

Also:

Spoiler

$2=(2+\sqrt{2})(2-\sqrt{2})$.

 

[anemone]

My solution:

Spoiler

$1+\sqrt{2}+\sqrt{2(2+\sqrt{2})}=\sqrt{\dfrac{x+\sqrt{y+\sqrt{z}}}{a-\sqrt{b+\sqrt{c}}}}$

Squaring it gives

$7+4\sqrt{2}+2\sqrt{2(2+\sqrt{2})}+4\sqrt{2+\sqrt{2}}=\dfrac{x+\sqrt{y+\sqrt{z}}}{a-\sqrt{b+\sqrt{c}}}$

If we let

$\begin{align*}7+4\sqrt{2}+2\sqrt{2(2+\sqrt{2})}+4\sqrt{2+\sqrt{2}}&=(a(2+\sqrt{2})(b+c\sqrt{2}+d\sqrt{2+\sqrt{2}})\\&=2ab+2ac+(ab+2ac)\sqrt{2}+ad\sqrt{2(2+\sqrt{2})}+2ad\sqrt{2+\sqrt{2}}\end{align*}$

Equating the coefficients and constant give

$ad=2$, $ab=3$ and if we let $a=1$, it then suggests that $d=2$, $b=3$ and $c=\dfrac{1}{2}$

$\begin{align*}\therefore 7+4\sqrt{2}+2\sqrt{2(2+\sqrt{2})}+4\sqrt{2+\sqrt{2}}&=\left(\dfrac{1}{2}(2+\sqrt{2})(6+\sqrt{2}+4\sqrt{2+\sqrt{2}}\right)\\&= \dfrac{(2+\sqrt{2})(6+\sqrt{2}+4\sqrt{2+\sqrt{2}})}{2}\\&= \dfrac{(2+\sqrt{2})(6+\sqrt{2}+4\sqrt{2+\sqrt{2}})}{(2+\sqrt{2})(2-\sqrt{2})}\\&=\dfrac{6+\sqrt{2}+4\sqrt{2+\sqrt{2}}}{2-\sqrt{2}}\\&\end{align*}$

Now, the tricky part is to recognize that $2-\sqrt{2}=(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})$, and also, $6+\sqrt{2}+4\sqrt{2+\sqrt{2}}=2^2+2(2)\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2}}^2=(2+\sqrt{2+\sqrt{2}}) ^2$

$\begin{align*}\therefore 7+4\sqrt{2}+2\sqrt{2(2+\sqrt{2})}+4\sqrt{2+\sqrt{2}}&=\dfrac{(2+\sqrt{2+\sqrt{2}}) ^2}{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\\&=\dfrac{2+\sqrt{2+\sqrt{2}}}{2-\sqrt{2+\sqrt{2}}}\\&=\sqrt{\dfrac{x+\sqrt{y+\sqrt{z}}}{a-\sqrt{b+\sqrt{c}}}}\end{align*}$

So, $a=b=c=x=y=z=2$.