The reverse is true, this is not!

G0ldenlem0n wrote: The reverse is true, this is not! I changed to official question, try now

toohardtochoose wrote: P(x) is polynomial such that, polynomial P(P(x)) is strictly monotone in all real number line. Prove that polynomial P(x) is also strictly monotone in all real number line. Let $Q(x)=P(P(x))$ Since stricly monotonous, real roots of $Q'(x)$, if there are, must all have even multiplicity. Let us assume that $P(x)$ is not stricly monotonous. So $P'(x)$ has at least one real root $r$ with an odd multiplicity. $Q'(x)=P'(P(x))P'(x)$ is such that $Q'(r)=0$ and so $r$ must have an even multiplicity. Since multiplicity is odd in $P'(x)$, we need $P(r)$ be a root of $P'(x)$ too. Let $P(x)-P(r)=(x-r)^nH(x)$ with $n\ge 1$ and $H(r)\ne 0$ Then $P'(x)=n(x-r)^{n-1}H(x)+(x-r)^nH'(x)$ and so $n\ge 2$ and is even (in order multiplicity of $r$ in $P'(x)$ be odd). 1) case 1 : $P(r)=r$ $P'(P(x))=n(P(x)-r)^{n-1}H(P(x))+(P(x)-r)^nH'(P(x))$ $=n(x-r)^{n^2-n}H(x)^{n-1}H(P(x))+(x-r)^{n^2}H(x)^nH'(P(x))$ So multiplicity of $r$ in $P'(P(x))$ is $n^2-n$ And multiplicity of $r$ in $Q'(x)=P'(P(x))P'(x)$ is $n^2-1$, which is odd. Impossible since $Q(x)$ is stricly monotonous 2) Case 2 : $P(r)\ne r$ We can also write $P'(x)=(x-r)^{n-1}(x-P(r))^mK(x)$ for some $K(x)$ such that $K(r)\ne 0$ and $K(P(r))\ne 0$ Then $P'(P(x))=(P(x)-r)^{n-1}(P(x)-P(r))^mK(P(x))$ $=(P(x)-r)^{n-1}(x-r)^{mn}H(x)^mK(P(x))$ And so multiplicity of $r$ in $P'(P(x))$ is $mn$ And multiplicity of $r$ in $P'(x)$ $n-1$ So multiplicity of $r$ in $Q'(x)$ is $mn+n-1$, which is odd (since $n$ is even). Impossible since $Q(x)$ is stricly monotonous And so no such $r$ and $P(x)$ is strictly monotonous. Q.E.D.