To solve this problem, think of two numbers, x and y, such that

x^{2} = y and

y^{2} = x

This is a pair of equations, which you can solve for x and y. Two easy solutions are

x=0,y=0

and

x=1,y=1,

but we're looking for a solution in which x is not equal to y, so these won't do.

So let's assume neither x nor y is 0 and neither x nor y is 1. Using the substitution method, substitute y^{2} in place of x in the first equation. So we start with:

x^{2} = y, then make the substitution, giving us

y^{4} = y, then divide both sides by y (which isn't zero), giving us

y^{3} = 1

y isn't equal to 1, but there are two other cube roots of 1, and they're both complex numbers.

Let b be one of the cube roots of 1.

b^{6} = (b^{3})^{2} = 1^{2} = 1 (That was pretty obvious)

b^{6} can also be written (b^{2})^{3}, so b^{2} is another cube root of 1.

(You know b^{2} isn't equal to b, because the only solutions of b=b^{2} are 0, and 1, and we already said b isn't 0 or 1.)

(b^{2})^{2} = b^{4} = b(b^{3}) = b

So the square of b is b^{2}, and the square of b^{2} is b. The two numbers, each of which is the square of the other, are the two non-real cube roots of 1. FYI, these numbers are:

-1/2 + (sqrt(3)/2)i and

-1/2 - (sqrt(3)/2)i

If you know how to multiply complex numbers, you can verify that each one of these is the square of the other!