Question 577452
Suppose the roots are p and q. Therefore we want


*[tex \LARGE f(x) = (x-p)(x-q) = (x-p^2)(x-q^2)].


Hence either p = p^2 and q = q^2, or p = q^2 and q = p^2. The first case implies that p and q must be either 0 or 1, so we have


*[tex \LARGE f(x) = x^2] or *[tex \LARGE f(x) = x(x-1)] or *[tex \LARGE f(x) = (x-1)^2]


The second case implies p = p^4, or p^3 = 1. The roots for p are 1, -1/2 + i*sqrt(3) and -1/2 - i*sqrt(3)/2 (same for q).


We have taken care of the p = 1 case, so suppose that p is -1/2 + i*sqrt(3)/2. Then


*[tex \LARGE q = p^2 = (-\frac{1}{2} + \frac{i \sqrt{3}}{2})^2 = -\frac{1}{2} - \frac{i \sqrt{3}}{2}] (easy way to square it is note that p = e^(2i*pi/3))


Therefore the polynomial f(x) is


*[tex \LARGE f(x) = (x - (-\frac{1}{2} + \frac{i \sqrt{3}}{2}))(x - (-\frac{1}{2} - \frac{i \sqrt{3}}{2})) = x^2 - x - 1]


So there are four quadratic equations that satisfy (ignoring arbitrary constants we can multiply with).