Question

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

Answer 1

`{ x, y } = { 2, 6 }`

Explanation:

The difference of squares identity can be written:

`a^2-b^2=(a-b)(a+b)`

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))`

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

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

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

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

`color(white)((1-sqrt(2)+sqrt(3))/(1+sqrt(2)-sqrt(3))) = (sqrt(2)+sqrt(6))/2`

So `{ x, y } = { 2, 6 }`