# Find {x,y} in NN such that (1-sqrt(2)+sqrt(3))/(1+sqrt(2)-sqrt(3))=(sqrt(x)+sqrt(y))/2?

Oct 3, 2016

$\left\{x , y\right\} = \left\{2 , 6\right\}$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this a couple of times to find:

(1-sqrt(2)+sqrt(3))/(1+sqrt(2)-sqrt(3)) = ((1-sqrt(2)+sqrt(3))(1+sqrt(2)+sqrt(3)))/((1+sqrt(2)-sqrt(3))(1+sqrt(2)+sqrt(3))

color(white)((1-sqrt(2)+sqrt(3))/(1+sqrt(2)-sqrt(3))) = (((1+sqrt(3))-sqrt(2))((1+sqrt(3))+sqrt(2)))/(((1+sqrt(2))-sqrt(3))((1+sqrt(2))+sqrt(3))

$\textcolor{w h i t e}{\frac{1 - \sqrt{2} + \sqrt{3}}{1 + \sqrt{2} - \sqrt{3}}} = \frac{{\left(1 + \sqrt{3}\right)}^{2} - {\left(\sqrt{2}\right)}^{2}}{{\left(1 + \sqrt{2}\right)}^{2} - {\left(\sqrt{3}\right)}^{2}}$

$\textcolor{w h i t e}{\frac{1 - \sqrt{2} + \sqrt{3}}{1 + \sqrt{2} - \sqrt{3}}} = \frac{1 + 2 \sqrt{3} + 3 - 2}{1 + 2 \sqrt{2} + 2 - 3}$

$\textcolor{w h i t e}{\frac{1 - \sqrt{2} + \sqrt{3}}{1 + \sqrt{2} - \sqrt{3}}} = \frac{2 + 2 \sqrt{3}}{2 \sqrt{2}}$

$\textcolor{w h i t e}{\frac{1 - \sqrt{2} + \sqrt{3}}{1 + \sqrt{2} - \sqrt{3}}} = \frac{\sqrt{2} + \sqrt{2} \sqrt{3}}{2}$

$\textcolor{w h i t e}{\frac{1 - \sqrt{2} + \sqrt{3}}{1 + \sqrt{2} - \sqrt{3}}} = \frac{\sqrt{2} + \sqrt{6}}{2}$

So $\left\{x , y\right\} = \left\{2 , 6\right\}$