Putnam 2001, Problem B-2

Find all pairs of real numbers $(x, y)$ satisfying the system of equations
$$\frac{1}{x} + \frac{1}{2y} = (x^2 + 3y^2)(3x^2 + y^2)$$
$$\frac{1}{x} - \frac{1}{2y} = 2(y^4 - x^4).$$

Solution:
By adding and subtracting the two given equations, we obtain the equivalent pair of equations
$$2/x = x^4 + 10x^2y^2 + 5y^4$$
$$1/y = 5x^4 + 10x^2y^2 + y^4.$$

Multiplying the former by $x$ and the latter by $y$, then adding and subtracting the two resulting equations, we obtain another pair of equations equivalent to the given ones,
$$3 = (x+y)^5, \quad 1 = (x-y)^5.$$

It follows that $x = (3^{1/5} + 1)/2$ and $y = (3^{1/5} - 1)/2$ is the unique solution satisfying the given equations.