MHB Proving that $P(P(x))=x$ Has No Real Solution

[anemone]

Let $P(X)$ be a polynomial with real coefficients for which the equation $P(X)=x$ has no real solution.

Prove that the equation $P(P(x))=x$ has no real solution either.

 

[Nono713]

Spoiler

Consider the polynomial $f(x) = P(x) - x$. If it has no real roots, then clearly either $f(x) > 0$ for all $x \in \mathbb{R}$, or $f(x) < 0$ for all $x \in \mathbb{R}$, since if it assumes both positive and negative values then it must assume the value zero at some point, being continuous. As a consequence, if $P(x) = x$ has no real solutions, then either $P(x) > x$ for all $x \in \mathbb{R}$ or $P(x) < x$ for all $x \in \mathbb{R}$.

Assume that $P(x) > x$ for all $x \in \mathbb{R}$, then for any $x \in \mathbb{R}$ it follows that $P(P(x)) > P(x) > x$ since $P(x) \in \mathbb{R}$, hence $P(P(x)) \ne x$ for all $x \in \mathbb{R}$ so $P(P(x)) = x$ has no real solutions either. The case for $P(x) < x$ is proved similarly.

 

[kaliprasad]

Let $P(X)$ be a polynomial with real coefficients for which the equation $P(X)=x$ has no real solution.

Prove that the equation $P(P(x))=x$ has no real solution either.

Spoiler

There is no real X for which

$P(X) = x$

for $P(P(x)) = x$ , P(x) cannot be real so complex

$P(x)$ with real coefficient cannot be complex with real $x$

so $P(P(x)) =x $ does not have real solution

 

[anemone]

Thanks to Bacterius and kaliprasad for the solution and participating in this challenge of mine.

Here is another solution:

Spoiler

Let $Q(x)=P(x)-x$. Then $Q(x)$ is a polynomial that never vanishes. We argue that it must always have the same sign.

Supose if possible that $Q(a)<0<Q(b)$ for some reals $a$ and $b$. Since $Q(x)$ being a polynomial is continuous, the Intermediate Value Theorem applies and there must be a number $c$ between $a$ and $b$ for which $Q(c)=0$, yielding a contradiction.

Thus, either $Q(x)>0$ for all $x$ or else $Q(x)<0$ for all $x$. Then

$P(P(x))-x=P(P(x))-P(x)+P(x)-x=Q(P(x))+Q(x)$ for all real $x$.

Since $Q$ never changes sign, both $Q(x)$ and $Q(P(x))$ have the same sign (either positive or negative) so their sum cannot vanish. Hence $P(P(x))\ne x$ for any real $x$.

 

[Nono713]

Spoiler

There is no real X for which

$P(X) = x$

for $P(P(x)) = x$ , P(x) cannot be real so complex

$P(x)$ with real coefficient cannot be complex with real $x$

so $P(P(x)) =x $ does not have real solution

I don't follow this argument. Anemone, did you mean $P(x) = x$ in the problem statement? Is $X$ just a typo for $x$ or are $X$ and $x$ distinct?

 

[anemone]

I don't follow this argument. Anemone, did you mean $P(x) = x$ in the problem statement? Is $X$ just a typo for $x$ or are $X$ and $x$ distinct?

Ops...yes, I meant $P(x) = x$. Sorry for the typo mistake...

 

[kaliprasad]

I don't follow this argument. Anemone, did you mean $P(x) = x$ in the problem statement? Is $X$ just a typo for $x$ or are $X$ and $x$ distinct?

I took x and X different. if this was a typo then my argument does not hold.