The September, 2007 Problem of the Month

Find a pair of unequal numbers, x and y so that each one is the square of the other.

Hint:
My idea of numbers is very inclusive.

Solution:

If x = y² and y = x², then y = (y²)² = y⁴. So y⁴ – y = 0, or y(y³ – 1) = 0, so finally we factor as y(y – 1)( y² + y + 1) = 0. The solutions are y = 0, y = 1, and

y = [-1 +/– sqrt(-3)]/2. If we let y = [-1 + i*sqrt(3)]/2, then x = [-1 – i*sqrt(3)]/2, and we can readily verify y = x² and x = y².

( [-1 + i*sqrt(3)]/2)² = [1 – 2*i*sqrt(3) + i²*3]/4 = [1 – 2*i*sqrt(3) +(-1)*3]/4

= [-2 – 2*i*sqrt(3)]/4 = [-1 – i*sqrt(3)]/2

Similarly,

( [-1 – i*sqrt(3)]/2)² = [1 + 2*i*sqrt(3) + i²*3]/4 = [1 + 2*i*sqrt(3) +(-1)*3]/4

= [-2 + 2*i*sqrt(3)]/4 = [-1 + i*sqrt(3)]/2

Submit comments in PSA 216 to Problem of the Month care of Richard Reynolds, or via email at rich@math.asu.edu

See http://math.asu.edu/~rich/puzzles/main.html for a web page view of this problem and past problems with solutions.