Let X be the first half of Graeme's number, and Y the second half, so

G = X - Y

From this, X³ - Y³ = 20, and XY = 2 (proof left to reader)

Now, cubing the equation G = X - Y gives us:

G³ = X³ - 3X²Y + 3XY² - Y³

G³ = (X³ - Y³) - 3XY(X - Y)

G³ = 20 - 3(2)G

G³ + 6G - 20 = 0

This third degree equation has one real solution: 2