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

1 Answer
George C.
Oct 3, 2016

#{ 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 }#

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