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The Factor Square Property (FSP) is the divisibility of the polynomial $f(x^2)$ by $f(x)$.

  1. Is $x^2+x+1$ the only FSP irreducible polynomial of degree $2$ ?

  2. Are there other linear polynomial besides $x$ and $x-1$ with FSP?

  3. Do we have other FSP irreducible polynomials of degree $3$ or $4$? Any of these have integer coefficients??

  4. Are there any other observations you can make about polynomials with FSP?

So this question has been posted before in: Link 1 Link 2

But the solutions use cyclotomic polynomials. Is there an easier solution? The Question is from Ross MathCamp, so, I suppose they will give some observation based question which doesn't require any kind of Complex Numbers I suppose.

Mathejunior
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1 Answers1

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Hint: Assuming that $f(x)$ is a monic and non-constant polynomial, $f(x)\mid f(x^2)$ implies that all the complex roots of $f(x)$ are roots of $f(x^2)$ too, hence that the set of roots of $f(x)$ is closed with respect to squaring. By considering the $n$-th cyclotomic polynomial with $n$ being odd, we always have that $\Phi_n(x)$ is a divisor of $\Phi_n(x^2)$, hence there are polynomials with the FSP with arbitrarily large degree (the degree of $\Phi_n(x)$ is $\varphi(n)$).

Jack D'Aurizio
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  • I don't know about cyclotomic polynomials. I shall learn it but is there any other way? Geeky Ross doesn't give problems which require​ complex numbers and stuffs. They often give problems which are observation based. – Mathejunior Nov 01 '17 at 18:53
  • @Mathbg: I guess there are many "naive" approaches to this problem, but I would raise a point. What is the purpose in avoiding the fundamental theorem of Algebra when studying Algebra? – Jack D'Aurizio Nov 01 '17 at 19:10
  • Sir, I learnt about cyclotomic polynomials and I don't find any more naive approach than to use this. But since the admission test for Ross Camp generally requires naive approaches, I am just inquisitive about hearing such a solution. – Mathejunior Nov 02 '17 at 04:35
  • @JackD'Aurizio I follow your answer, except why does n have to be odd when considering the nth the cyclotomic polynomial? – Nico A Jan 09 '18 at 01:23
  • @TreFox: otherwise $\Phi_n(x)\mid \Phi_n(x^2)$ is not granted. For instance $\Phi_4(x)=x^2+1$ is not a divisor of $x^4+1$. – Jack D'Aurizio Jan 09 '18 at 14:47
  • @JackD'Aurizio That makes sense, but how can I prove it for all even-numbered cyclotomic polynomials? – Nico A Jan 09 '18 at 15:08
  • You cannot. $\Phi_{2m}(x)$ is not a divisor of $\Phi_{2m}(x^2)$ (it is enough to consider where the roots of these polynomials lie). – Jack D'Aurizio Jan 09 '18 at 15:17
  • @JackD'Aurizio Sorry! I think you misunderstood me. How can I prove that $\Phi_{2n}(x)$ is not a divisor of $\Phi_{2n}(x^2)$ for all $n$? – Nico A Jan 10 '18 at 00:46
  • @TreFox: by comparing the roots. – Jack D'Aurizio Jan 10 '18 at 01:12