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Find all $p\in\mathbb R[x]$ s.t. $$p'\left(x^2\right)\cdot p'''(x)=3\cdot p''\left(x^2\right)\cdot x^2.$$

Attempt:

$$p(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$$ $$p'(x)=4a_4x^3+3a_3x^2+2a_2x+a_1$$ $$p''(x)=12a_4x^2+6a_3x+2a_2$$ $$p'''(x)=24a_4x+6a_3$$ $$p\left(x^2\right)=a_4x^8+a_3x^6+a_2x^4+a_1x^2+a_0$$ $$p'\left(x^2\right)=8a_4x^7+6a_3x^5+4a_2x^3+2a_1x$$ $$p''\left(x^2\right)=56a_4x^6+30a_3x^4+12a_2x^2+2a_1$$ The initiall expression becomes: $$\underbrace{\left(8a_4x^7+6a_3x^5+4a_2x^3+2a_1x\right)}_{/:(2x)}\underbrace{\left(24a_4x+6a_3\right)}_{/:3}=3\underbrace{\left(56a_4x^6+30a_3x^4+12a_2x^2+2a_1\right)}_{/:2}x^2$$ $$\left(4a_4x^6+3a_3x^4+2a_2x^2+a_1\right)\left(8a_4x+2a_3\right)=28a_4x^7+15a_3x^5+6a_2x^3+a_1x$$ $LHS:$ $$32a_4^2x^7+8a_4a_3x^6+24a_3a_4x^5+6a_3^2x^4+16a_2a_4x^3+4a_2a_3x^2+8a_1a_4x+2a_1a_3$$ $$32a_4^2=28a_4\iff 8a_4a_4=7a_4$$$$8a_4a_3=0$$$$24a_3a_4=15a_3\iff8a_4a_3=5a_3$$$$6a_3^2=0\implies a_3=0$$$$16a_2a_4=6a_2\iff8a_4a_2=3a_2$$$$4a_2a_3=0$$$$8a_1a_4=a_1$$ For $a_1,a_2,a_4\ne 0$ we get a contradiction:$$a_4=\frac{7}{8}=\frac{3}{8}=\frac{1}{8}\Rightarrow\Leftarrow$$

Are constants and null-polynomial the solutions?

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    $\begingroup$ Won't \begin{eqnarray*} p'(x^2)=4a_4x^6+3a_3x^4+2a_2x^2+a_1 \end{eqnarray*} ? $\endgroup$ Feb 14, 2020 at 0:04
  • $\begingroup$ @DonaldSplutterwit, so, I just plug $x^2$ into derivation itself? $\endgroup$ Feb 14, 2020 at 0:24
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    $\begingroup$ I think so. Let us know if it works out now. $\endgroup$ Feb 14, 2020 at 0:27

1 Answer 1

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If $\deg(p)=n$ then we have $$2(n-1)+n-3 = 2(n-2)+3$$

so $n=3$ and thus $p(x)=ax^3+bx^2+cx+d$ with $a\ne 0$. Since $p'(0)p'''(0)=0$ and $p'''(0)=6a\ne 0$ so $c=p'(0)=0$. Now we have, for all $x$ $$(3ax^4+2bx^2)6a=3(6ax^2+2b)x^2$$ so $a=1$ and $b=0$. So $p(x) = x^3+d$ where $d$ is arbitrary.

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