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Let $f$, $g$ $\in \mathbb{C}[X]$ be two polynomials, $f=x^m+px^n+q$, $g=x^m+qx^n+p$ with $p\ne q$. When do we have a common root?

Let $a\in\mathbb{C}$ be the common root. So $f(a)=g(a)=0$. $$f(a)-g(a)=0 \implies (a^n-1)(p-q)=0$$ For $a^n=1$ we get $f(a)=g(a)=a^m+p+q$, so the condition we were looking for would be $p+q=0$.

I'd like to know if my reasoning was right in this case, as I'm not 100% convinced about it. Thank you!

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  • $\begingroup$ Is the leading term in both cases $x^m$? $\endgroup$ – Dr. Sonnhard Graubner May 9 '18 at 17:08
  • $\begingroup$ Aren't we rather looking for $a^n=1\land a^m+p+q=0$? This certainly requirs $|p+q|=1$ $\endgroup$ – Hagen von Eitzen May 9 '18 at 17:10
  • $\begingroup$ Any nth root of unity works for $a$ $\endgroup$ – marwalix May 9 '18 at 17:10
  • $\begingroup$ The leading term is $x^m$ $\endgroup$ – user528021 May 9 '18 at 17:11
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We need to solve

$$\begin{cases} a^m +pa^n+q=0\\ a^m +qa^n+p=0\end{cases}$$

This gives $a^n=1$ (because $p\neq q$ by assumption) so $a$ has to be an $n^{th}$ root of unity and therefore

$$p+q=-\zeta_k^m$$

Where $\zeta_k=e^{2i\pi\over k}$

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  • $\begingroup$ $a$ is a $n^{th}$ root of unity but that still leaves us with $a^m+p+q=0$ $\endgroup$ – user528021 May 9 '18 at 17:35
  • $\begingroup$ The question was when do these polynomials have a common root and the answer is : “whenever $p+q=-\zeta^m$ with $\zeta$ one of the n $n^{th}$ roots of unity” $\endgroup$ – marwalix May 9 '18 at 17:38
  • $\begingroup$ I see. Thank you! $\endgroup$ – user528021 May 9 '18 at 17:40

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