elementary superfunctions

[Kouznetsov]

Challenge:
Is there an elementary superfunction of a polynomial that has no real fixed point?

I thnk so. How about transformation of line 5, \( b=2 \)?
Let \( H_0(z)=z^2 \) ; \( P(z)=z+1 \); \( Q(z)=z-1 \);
Let \( H_1(z)=P(H_0(Q(z)))=1+(z-1)^2 \);
Equation \( z=1+z^2-2z+1 \);
(BAD)is equivalent of equation \( z^2-z+2=0 \)
(BAD)Gives the fixed points \( z=1/2 \pm \sqrt{1/4-2}=1/2\pm i \sqrt{7/4} \)
Sorry I lost the signum! should be
is equivalent of equation \( z^2-3z+2=0 \)
Gives the fixed points \( z=1/2 \pm \sqrt{9/4-2}=1/2\pm \sqrt{1/4} \)
I try again:
\( P(z)=z+a \); \( Q(z)=z-a \);
Let \( H_1(z)=P(H_0(Q(z)))=a+(z-a)^2 \);
Equation \( z=a+z^2-2az+a^2 \);
is equivalent of equation \( z^2-(2a+1)z+a^2+a=0 \)
Gives the fixed points \( z=(2a+1)/2 \pm 1/2\sqrt{(4a^2+4a+1-4(a+a^2)} \)
\( z=(2a+1)/2 \pm 1/2\sqrt{1} \)
Henryk, it is not so easy... Sorry... but.. Wait... May I use complex \( a \) or the basefunction is supposed to be real? Do you mean "real polynomial"?