20 The quadratic function $f(x)$ that satisfies the condition $f\left(x^{2}\right)=f(f(x))=(f(x))^{2}$ has
A. 0
B. 1
C. 2
D. infinitely many
20 B. Let $f(x)=a x^{2}+b x+c(a \neq 0)$, then $f\left(x^{2}\right)=a x^{1}+b x^{2}+c$, and $(f(x))^{2}=$ $a^{2} x^{4}+2 a b x^{3}+\left(2 a c+b^{2}\right) x^{2}+2 b x x+c^{2}$. Since $f\left(x^{2}\right)=(f(x))^{2}$, comparing the coefficients of the same powers on both sides of the equation, we should have $a=a^{2}, 2 a b=0, 2 a c+b^{2}=b, 2 b c=0, c^{2}=c$. Since $a \neq 0$, it is easy to get $a=1, b=c=0$. Therefore, the quadratic function that satisfies the condition $f\left(x^{2}\right)=(f(x))^{2}$ is only $f(x)=$ $x^{2}$. For this function, $f(f(x))=x^{4}=f\left(x^{2}\right)=(f(x))^{2}$. That is, there is only 1 quadratic function that satisfies the given condition.