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.


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 $\varphi(n)$).

  • $\begingroup$ 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. $\endgroup$ – Mathejunior Nov 1 '17 at 18:53
  • $\begingroup$ @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? $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 19:10
  • $\begingroup$ 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. $\endgroup$ – Mathejunior Nov 2 '17 at 4:35
  • $\begingroup$ @JackD'Aurizio I follow your answer, except why does n have to be odd when considering the nth the cyclotomic polynomial? $\endgroup$ – Nico A Jan 9 '18 at 1:23
  • $\begingroup$ @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 $x^4+1$. $\endgroup$ – Jack D'Aurizio Jan 9 '18 at 14:47

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