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Define $$f(x)=x^4 + 2 x^3 + 3 x^2 + 3x + 1.$$

Show that if $f(f(x)) = x$ then $f(x) = x$.


We can write $f(x)=x^4+2x^3+3x^2+x+1$ and $x=f(x)^4+2f(x)^3+3f(x)^2+f(x)+1$, and subtracting the two equations yields $$-(f(x)^4+2f(x)^3+3f(x)^2+2f(x)+1)=x^4+2x^3+3x^2+2x+1 .$$ Factoring yields $-(f(x)^2+f(x)+1)^2=(x^2+x+1)^2$, so either $$x^2+x+1=f(x)^2+f(x)+1 \qquad \textrm{or} \qquad x^2+x+1=-(f(x)^2+f(x)+1) .$$ How do we now show that if $f(f(x)) = x$ for some (real) $x$ then in fact $f(x) = x$?

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    $\begingroup$ $f(x)$ does not have any global inverse. $\endgroup$ May 14 at 4:50
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    $\begingroup$ Are you trying to solve $f(x)=x$, or $f(f(x))=x$? Your title says the former, while the question body says the latter. Which is it? $\endgroup$ May 14 at 5:04
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    $\begingroup$ @ProblemDestroyer Your $\,f\,$ is not bijective, thus not invertible, so there is no $\,f^{-1}\,$ to speak of. This was pointed out in an early comment. $\endgroup$
    – dxiv
    May 14 at 5:30
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    $\begingroup$ The degree-$16$ polynomial $g(x) := f(f(x)) - x$ is positive everywhere and in particular has no (real) roots, so the claim that $f(f(x)) = x$ implies $f(x) = x$ for real $x$ is vacuously true. $\endgroup$ May 14 at 5:35
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    $\begingroup$ @ProblemDestroyer I've (significantly) adjusted the wording of the problem statement based on your comments. Please ensure that I've preserved the intended meaning and revert/adjust accordingly. N.b. I did not change the computational error in the first display equation of your attempt, since I'm not certain what you intended for that line. $\endgroup$ May 15 at 18:23

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We can generalize the phenomenon observed for this particular function $f$: For any monic polynomial $p$ of degree $d > 0$---in our case, $p(x) = x^2 + x + 1$, $d = 2$---define the polynomial $$f(x) := p(x)^2 + x ,$$ which has degree $2 d$. By construction the solutions of $f(f(x)) = x$ are the roots of the polynomial $$f(f(x)) - x = p(p(x)^2 + x)^2 + p(x)^2,$$ which has degree $4 d^2$ If $a$ is a root of $p$ of order $k$, so that $p(x) = (x - a)^k r(x)$ for some polynomial $r$, substitution shows that it's also a root of $f(f(x)) - x$ of order (at least) $2 k$, hence $f(f(x)) - x$ is divisible by $p(x)^2 = f(x) - x$. Moreover, $f(f(x)) - x$ admits a polynomial factorization \begin{align*} f(f(x)) - x &= p(x)^2 \cdot \frac{f(f(x)) - x}{p(x)^2} \\ &= (f(x) - x) \cdot \frac{p(f(x))^2 + p(x)^2}{p(x)^2}, \end{align*} that is, $$\boxed{f(f(x)) - x = (f(x) - x) (q(x)^2 + 1)} , \qquad q(x) := \frac{p(f(x))}{p(x)} .$$ (Here $\deg q = 2 d^2 - d$.) (For $d = 1$, it seems $q = p + 1$ and for $d = 2$, $q = p^3 + p' p + 1$.)

In particular, for real $x$ the factor $q(x)^2 + 1$ is always positive, so if $x$ is a real solution of $f(f(x)) = x$, i.e., a real root of $f(f(x)) - x$, then it must be a root of $f(x) - x$, i.e., a solution of $f(x) = x$.

Example ($d = 2$) Write $$p(x) = x^2 + b x + c .$$ Then, $$f(x) = x^4 + 2 b x^3 + (b^2 + 2 c) x^2 + (2 b c + 1) x + c^2$$ and \begin{multline}q(x) = x^6 + 3 b x^5 + 3(b^2 + c) x^4 + (b^3 + 6 b c + 2) x^3 \\+ 3(b^2 c + c^2 + b) x^2 + (3 b c^2 + b^2 + 2 c) x + (c^3 + b c + 1) .\end{multline}

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  • $\begingroup$ Thanks a lot Sir thus clear pretty mych everything :) $\endgroup$ May 14 at 16:02
  • $\begingroup$ You're welcome, I'm glad you found it helpful. $\endgroup$ May 14 at 16:41
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The question is not so clear. Anyway, there are no real solution to the equation $f(x)=x$. Namely from $$ x^4 + 2x^3 + 3x^2 + 3x + 1 = x $$ you get $$ x^2 \left( {x^2 + 2x + 1} \right) + \left( {3x^2 + 2x + 1} \right) = 0 $$ and $$ x^2 \left( {x^2 + 2x + 1} \right) \ge 0 $$ for every $x \in \mathbb R$ while $$ \left( {3x^2 + 2x + 1} \right) > 0 $$ for every $x \in \mathbb R$.

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  • $\begingroup$ Sir this method i know , i mean i was assuming if we have a real x satisfying the solution to the given equation ( suppose which may be not exists ) then solving ff(x) = x should lead us to conclusion that it should be f(x) = x only . Hence i was trying to show that , not that there is no such x $\endgroup$ May 14 at 5:05

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