Determine a quadratic function f(x) that satisfies this equation:
f(x)*f(x-1)= f(x^2)
Let r be a root of f(x). Then r+1 is a root of f(x-1).
These also need to be roots of f(x^2), but then their negatives will also be roots. Then the set of four roots are -r-1, -r, r, and r+1.
From this the second root of f(x) is -r-1 and the second root of f(x-1) is -r.
Then f(x)=(x-r)*(x+r+1)=x^2+x-(r^2+r)
and f(x-1)=(x+r)*(x-(r+1))=x^2-x-(r^2+r)
and f(x^2)=(x-r)*(x+r+1)*(x+r)*(x-(r+1))=x^4+(-2r^2-2r-1)x^2+r^2*(r+1)^2
But then we can evaluate f(x^2) by plugging x^2 into f(x) getting f(x^2) = x^4+x^2-(r^2+r)
The two expressions for f(x^2) must be equal, then we must have matching coefficients: -2r^2-2r-1 = 1 and r^2*(r+1)^2 = -(r^2+r)
Both equations for r have common solutions where r^2+r=-1, so then substituting this into f(x) gives a final answer of f(x)=x^2+x+1.
Solution
