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For any positive integer $ m $, the polynomials are defined as follows $(m>1)$:

$$f_m(x) = x^{2m+1} - (2m+1)x^{m+1} + (2m+1)x^m - 1$$

My goal is to prove that for any two distinct integers $ n $ and $ m $, the polynomials $ f_n(x) $ and $ f_m(x) $ have a gcd given by:

$$ \gcd(f_n(x), f_m(x)) = (x-1)^3 $$

In other words, I want to determine if they do not have any common complex roots other than 1.

What I've tried:

I have been using the Euclidean algorithm to compute the gcd for some small cases with computer. However, I haven't been able to detect any clear patterns in their coefficients when using Bézout's identity to express the gcd. So I am seeking for a theoretical proof.

Thanks for your time!

Reiss
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    How to ask a good question. In particular, please add what you have tried. – Martin Brandenburg Oct 11 '24 at 15:27
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    By the duplicate, $f_m=g_m(x-1)^3$, where $g_m$ is irreducible of degree $2(m-1)$. This yields your claim. – Dietrich Burde Oct 11 '24 at 15:58
  • Hmmm the linked duplicate hasn't received an answer covering all cases, right? – Martin Brandenburg Oct 11 '24 at 16:14
  • Yes, you are right. The duplicate question is still open in some cases. One could try to find a complete solution and add it there. I think it is possible to find something in the literature, but I didn't search for it yet. – Dietrich Burde Oct 11 '24 at 18:57
  • @DietrichBurde Could you kindly assist in finding a complete solution in the literature? Any guidance or references on how to fully resolve this would be appreciated. Thank you in advance for your time and help! – Reiss Oct 16 '24 at 09:12
  • Did you search already this site? Since derivatives of arithmetic progressions are irreducible, see here, you could do it one more time to divide by $(x-1)^3$, see the comment by The SimplyFire. – Dietrich Burde Oct 16 '24 at 09:53
  • @DietrichBurde Thank you! However, I’m still struggling to see how this relates to my case, for example $f_3(x) = (x^6 + x^5 + x^4 - 6x^3 + x^2 + x + 1)(x-1) = (x^5 + 2x^4 + 3x^3 - 3x^2 - 2x - 1)(x-1)^2 = (x^4 + 3x^3 + 6x^2 + 3x + 1)(x-1)^3$ , which intuitively feels different from the derivatives of arithmetic progressions. Could you clarify the connection between these two types of polynomials? – Reiss Oct 16 '24 at 11:53

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