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$p(x)$ and $q(x)$ are polynomials which satisfy the identity $p(q(x)) = q(p(x))$ for all real $x$ . If the equation $p(x) = q(x)$ has no real solutions, prove that the equation $p(p(x)) = q(q(x))$ has no real solutions.

What I Tried :- I just assumed $p(x) = a_nx^n + \dots + a_1x + a_0$ and $q(x) = b_nx^n + \dots + b_1x + b_0$.

It was also given that $p(x) = q(x)$ has no real solutions, so in the context of real numbers, I can claim $p(x) \neq q(x)$ for all $x$. But I am not sure how that is going to help, and I have not used this information yet.

After this, we have $p(b_nx^n + \dots + b_1x + b_0) = q(a_nx^n + \dots + a_1x + a_0)$.

But now, I am stuck. Expanding more is going to make it complicated.

I am also thinking of showing this by contradiction somehow, that I am assuming first there exists a real root to $p(p(x)) = q(q(x))$, but I am not sure how to do it.

Can anyone help me? Thank You.

Anonymous
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1 Answers1

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$p-q$ is a continuous function. If it has no real zeros then it is always positive or always negative. Suppose $p-q>0$. Then $p(p(x)) >q(p(x))=p(q(x))> q(q(x))$ so $p(p(x)) \neq q(q(x))$. The other case is similar.

Kavi Rama Murthy
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  • Can you just explain why "If a real polynomial has no real zeros then it is always positive/always negative?" Otherwise this looks like a great straightforward proof. – Anonymous Oct 18 '21 at 09:51
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    [+1] A remark : I would like to see examples of 2 polynomials such that $p(q(x))=q(p(x))$ and such that equation $p(x)=q(x)$ has no real solution. Just to be sure that we are not reasoning on the void set... – Jean Marie Oct 18 '21 at 09:51
  • Intermediate Value Property shows that if a continuous function attains a positive value as well as a negative value it must attain the value $0$ at some point. @Anonymous – Kavi Rama Murthy Oct 18 '21 at 09:52
  • @KaviRamaMurthy oh I got it it just follows from IVT. – Anonymous Oct 18 '21 at 09:54
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    @JeanMarie $p(x)=x, q(x)=x+1$. – Kavi Rama Murthy Oct 18 '21 at 09:56
  • @Kavi Rama Murthy Thanks ! – Jean Marie Oct 18 '21 at 09:59
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    @JeanMarie Although this is not the case, there's nothing wrong about reasoning about the empty set. A lot of theory has been developed about odd perfect numbers to try to prove its non-existence. – jjagmath Oct 18 '21 at 11:06
  • @jjagmath You are right. Nevertheless, it is a question that could be asked as a question $#0$... before embarking into a general proof... – Jean Marie Oct 18 '21 at 11:53