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PROBLEM
Let $P(x)$ be an odd degree polynomial in $x$ with real coefficients. Show using Induction that the equation $P(P(x))=0$ has at least as many distinct real roots as the equation $P(x)=0$.

MY APPROACH

  • Base Case: for $deg= 1$ we have $P_1 (x) = (x-a)$ which has root as $x= a$ , now $P_1 (P_1 (x)) = 0$ has root at $x=2a$. Hence base case is True.
  • Induction Hypothesis : We assume its true for $deg= odd= k >1$ polynomial that is if $P_k (x) = (x-a_1)(x-a_2)(x-a_3)...(x-a_t)(Q(x))$ ,where $Q(x)$ is $deg=(k-t)$ polynomial not factorizable in real numbers and $a_1 \neq a_2.. \neq a_t$ such that $P_k (P_k(x))$ has at least $t$ real distinct roots as the original polynomial .
    To Show : for $deg = k+2 (odd)$ its being true .
    WLOG lets say the next two roots be either both complex or both real and distinct we are excluding the case of equal roots as we have not lead that to be the case in induction hypothesis.
    But from this I am not able to conclude that $P_{k+2} (P_{k+2}(x))$ will have at least $t+2$ real distinct roots.

Any help would be appreciated!

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  • $\begingroup$ Do you want solutions ONLY in induction? $\endgroup$ Commented May 15, 2022 at 8:53
  • $\begingroup$ Yeah because i have proven otherwise $\endgroup$ Commented May 15, 2022 at 8:54
  • $\begingroup$ Do you mean that you have proved in way other than induction already? $\endgroup$ Commented May 15, 2022 at 8:54
  • $\begingroup$ Yeah Sir , i was interested in how induction helps in solving problems leaving number theory $\endgroup$ Commented May 15, 2022 at 8:56
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    $\begingroup$ Please don't assume all users are male ("Sir")—it further worsens the unwelcoming environment that women have in mathematics. $\endgroup$ Commented May 27, 2022 at 16:55

1 Answer 1

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Ah, you want an induction solution. Sorry this is not it.

Say $a_1,a_2,...,a_k$ are all real different solution to $P(X)=0$. Since $P$ is real odd degree polynomial, for each $i$ exists real $b_i$ such that $P(b_i)=a_i$. So $b_1,b_2,...,b_k$ are all solutions to $P(P(x))=0$. Now if $b_i = b_j$ for some $i\ne j$ we get $$a_i = P(b_i)= P(b_j) = a_j$$ a contradiction. So $P(P(x))=0$ has also at least $k$ different real solutions.

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  • $\begingroup$ Thanks for this method Sir , i was able to solve using this from first try . But i was thinking of seeing how we can do by induction as i saw many stack post using induction on polynomial problems :) $\endgroup$ Commented May 15, 2022 at 16:46

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