Thread: Puzzle with numbers

There are two real numbers and the sum of their reciprocals is equal to -1 and the sum of their cubes is equal to 4. Can you find the numbers?

Could you publish the solution to this problem?

-- James

Found the solution to this problem here,

http://www.qbyte.org/puzzles/p009s.html

Solution:

1/x + 1/y = -1 (1)
x3 + y3 = 4 (2)

(1) implies x + y = -xy
(2) implies (x + y)3 - 3xy(x + y) = 4
Hence -(xy)3 + 3(xy)2 - 4 = 0

By inspection, xy = -1 is a solution of this cubic equation.
Factorizing, we have (xy + 1)(xy - 2)2 = 0.
Hence xy = -1, x + y = 1, or xy = 2, x + y = -2.

If xy = -1 and x + y = 1, then x, y are roots of the quadratic equation u2 - u - 1 = 0.
(Consider the sum and product of the roots of (u - A)(u - B) = u2 - (A + B)u + AB = 0.)
Hence u = (1 ± root 5)/2.

If xy = 2 and x + y = -2, then x, y are roots of u2 + 2u + 2 = 0.
This has complex roots: u = -1 ± i.

Therefore the real solutions are x = (1 ± root 5)/2, y = (1 minus or plus root 5)/2.

-- James.

A very interesting puzzle which gave me idea to solve an other puzzle. Thanks for posting this.