How to find x and y?

#(x^5+y^5)/"x+y"=1#

#x^2+y^2=2#

1 Answer

#(1,1) and (-1,-1)#

#(x_1, pm y_1)#

#(x_2, pm y_1)#

#(x_3, pm y_2)#

#(x_4, pm y_2)#

Total: 10 points

Explanation:

#y = epsilon sqrt {2 -x^2}#

#x^5 + y^5 = x + y#

#x^5 + epsilon (2 - x^2)^2 sqrt {2 -x^2} = x + epsilon sqrt {2 -x^2}#

#epsilon [(x^2 - 2)^2 - 1] sqrt {2 -x^2} = x -x^5#

#[x^4 - 4x^2 + 3]^2 * (2 - x^2) = x^2 (x^4 - 1)^2#

#(x^2 - 3)^2(x^2 - 1)^2 * (2 - x^2) = x^2 (x^2 - 1)^2 (x^2 + 1)^2#

#x^2 - 1 = 0# or #z = x^2# and #(z - 3)^2 * (2 -z) = z (z + 1)^2#

#x = pm 1# or #-z^3 + 6z^2 - 9z + 2z^2 - 12z + 18 = z^3 + 2z^2 + z#

#0 = 2z^3 - 6z^2 + 22z - 18#

#z^3 - 3z^2 + 11z - 9 = 0#

#(z - 1)(z^2 -2 z + 9) = 0#

#Delta = 4 - 4 * 9 = -32 => z = 1 pm 2 i sqrt 2 = x^2#

#x^2 = 3 (1/3 pm i sin arccos frac{1}{3})#

#x = pm sqrt 3 (cos t pm i sin t), t = 1/2 arccos frac{1}[3}#

#x_1 = sqrt 3 (cos t + i sin t)#

#x_2 = - sqrt 3 (cos t + i sin t)#

#=> y_1^2 = 2 - 3 (1/3 + i sin arccos frac{1}{3}) = 1 - 6i sqrt 2#

#x_3 = sqrt 3 (cos t - i sin t)#

#x_4 = - sqrt 3 (cos t - i sin t)#

#=> y_2^2 = 2 - 3 (1/3 - i sin arccos frac{1}{3}) = 1 + 6i sqrt 2#

#|y^2| = sqrt{1^2 + 36 * 2} = sqrt 73#

#y^2 = sqrt 73 (cos arctan 6 sqrt 2 + i sin arctan 6 sqrt 2)#

#y_1 = root{4} 73 (cos frac{1}{2} arctan 6 sqrt 2 - i sin frac{1}{2} arctan 6 sqrt 2)#