Suppose x, y, a are real numbers such that x+y = x^3 +y^3 = x^5 +y^5 = a. Find all possible values of a.

1) One solution is x = x^3 = x^5 and y = y^3 = y^5

x can be 0, 1 or -1 and y can be 0, 1 or -1.

so a can be -2, -1, 0 , 1 or 2

2) Are there any other solutions? Well,

(x+y) - (x^3 + y^3) = 0

so x^3 - x = y^3 - y

so x(x+1)(x-1) = y(y+1)(y-1) <= equation a

also, (x^3+y^3) - (x^5 + y^5) = 0

so x^5 - x^3 = y^5 - y^3

so (x^3)(x+1)(x-1) = (y^3)(y+1)(y-1) <= equation b

If x <> 0,1,-1 then equation a can be divided into equation b,

so x^2 = y^2

This yields another solution, x = -y, which gives a = 0

It also suggests x = y as a solution, but substituting into

(x+y) - (x^3 + y^3) = 0 gives x = 0 or 1 or -1, which is not a new solution.

SO, FINAL ANSWER, a can only be -2, -1, 0 , 1 or 2