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Let $p(x) = (x + 1)^m - x^m - 1$ and $q(x) = (x^2+x+1)^2$. Coefficients if these polynomials are in $\mathbb{R}$

So, the question is: "For which $m$ p(x) can be divided by $q(x)$ without remainder?".

I tried to use this idea: let's simulate GCD's algorithm for $p(x)$ and $q(x)$ and see for which $m$ their gcd is dividable to $q(x)$, but didn't got any intresting results.

So, maybe exists a better solution. I'll be very grateful.

Someone
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    Do you know complex numbers? – Aphelli Nov 01 '20 at 15:49
  • Yes, of course. – Someone Nov 01 '20 at 15:51
  • Show that $q$ divides $p$ iff $p(j)=p'(j)=0$. – Aphelli Nov 01 '20 at 15:52
  • Are you saying that $p$ is divisible by $q$ if $p$ has a multiple root $j$? Can you explain why is that so? – Someone Nov 01 '20 at 15:56
  • @Mindlack Your hint makes no sense at all... – xxxxxxxxx Nov 01 '20 at 16:38
  • Sorry, I didn’t remember that $j$ wasn’t a standard notation for “the” primitive third root of unity (with positive imaginary part). You can check that a polynomial with real coefficients $r$ is divisible by $x^2+x+1$ iff $r(j)=0$, because $j^2+j+1=0$ and $x^2+x+1$ is irreducible. Moreover, as $x^2+x+1$ is irreducible, $q$ divides $p$ iff $x^2+x+1$ divides $p$ and $p’$, so iff $p(j)=p’(j)=0$. What doesn’t make sense about this? – Aphelli Nov 01 '20 at 16:49
  • Same method(s) in the linked dupe work, using $, P-\frac{1}{m}(x!+!1)P' = x^{\large m-1}-1\ \ $ – Bill Dubuque Nov 01 '20 at 19:33
  • @Mindlack The fact that it was given without any context, and so there was no indication that you meant for $j$ to be a primitive third root of unity. – xxxxxxxxx Nov 03 '20 at 05:46

1 Answers1

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Hint:

As the roots of $q(x)$ are the complex cubic roots of unity $\;\omega=\mathrm e^{\tfrac{2i\pi}3}$ and $\omega^2=\bar\omega=\mathrm e^{\tfrac{4i\pi}3}$, and $p(x)$ has to be divisible by the square of $q(x)$, you have to check that $\omega$ and $\omega^2$ are double roots of $p(x)$, i.e. that they're roots of $p(x)$ and $p'(x)$.

Some more details:

As $p'(x)=m(x+1)^{m-1}-mx^{m-1}$, we have the conditions

  • $(\omega+1)^m-\omega^m-1=(-\omega^2)^m-\omega^m-1=0$,
  • $(-\omega^2)^{m-1}-\omega^{m-1}=0$.
Bernard
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  • Didn't get how we can check that they're roots of $p(x)$. And where is dependence from $m$ in your solution. I would be very grateful if you could write a more detailed solution to this problem. – Someone Nov 01 '20 at 16:03
  • @Someone $\omega = -\frac12+\frac{i\sqrt3}2$ so $\omega+1=\frac12+\frac{i\sqrt3}2$ which you can also express in polar form. – saulspatz Nov 01 '20 at 16:06
  • @Someone: if they're roots of $q(x)$ and $q(x)$ divides $p(x)$, they're automatically roots of $p(x)$. Note I didn't post a full solution, only a hint. – Bernard Nov 01 '20 at 16:19