So f (f (x1)) = f (x1) = x1; f(f(x2))=f(x2)=x2，

Therefore, x 1, x2 satisfies the equation f(f(x))=x, that is, f(f(x))=x has at least two real roots.

(2) because that two real root of the equation f(x)=x are x 1, x2 and x2-x1>; 2,

So x2+(b- 1) x+c = 0 satisfies the root x 1, and x2 satisfies (x2-x1) 2 = (x1+x2) 2-4x1x2 >; 4,

So there is (b- 1) 2 > 4(c+ 1), and whether c+ 1 is a positive number or a non-positive real number, there is b 2 >；; 4c，

That is, f (x) = x 2+bx+c has two different zeros.

Therefore, f (x1) = f (x3); F(x4)=f(x2) (as shown)

Furthermore, except that x 1 and x2 satisfy the equation f(f(x))=x, x3 and x4 satisfy the equation f(f(x))=x,

? That is, only f (f (x3)) = x1; f(f(x4))=x2，

So there is: x4