We consider the two polynomials with real coefficients: $${\displaystyle P_{n}(X)=X^{4n}+X^{3n}+X^{2n}+X^{n}+1 \quad \mbox{ with } n\in\mathbb{N}^{*}}$$ and $${\displaystyle Q(x)=X^{4}+X^{3}+X^{2}+X+1}$$ then:
Choose the correct option. more than one may be correct
$a)$ If $n$ is multiple of $5$ then $Q$ divides $P_n$
- $b)$ If $n\equiv 1\ [5]$ then $Q$ divides $P_n$
- $c)$ If $Q$ divides $P_n$ then $n$ is even
- $d)$ None of the above statements is correct
My proof:
note that: $P_n(x)=Q(x^n)$
- $c)$ is false since $n=1$ (odd) and we've $Q=P_{1} \iff Q\mid P_{1}$ and $P_{1}\mid Q$
- i think $b)$ is true :
suppose that $n\equiv 1\ [5]$ then $\exists k\in\mathbb{Z}\ \mbox{ s.t.}\quad n=5k+1$
$$P_n(X)=X^{4(5k+1)}+X^{3(5k+1)}+X^{2(5k+1)}+X^{5k+1}+1$$
and since $${\displaystyle Q(x)=X^{4}+X^{3}+X^{2}+X+1=\dfrac{X^{5}-1}{X-1}}$$ then we can factorized in $\mathbb{C}\setminus\{1\}$:
note that : $$\displaystyle \mathcal{U}_5\setminus\{1\}=\left\{ e^{\frac{2i\pi}{5}},e^{\frac{4i\pi}{5}},e^{\frac{6i\pi}{5}},e^{\frac{8i\pi}{5}} \right\}$$
then:
$${\displaystyle Q(x)=(X-\alpha)(X-\alpha^{2})(X-\alpha^{3})(X-\alpha^{4})\quad \mbox{ with }\alpha=e^{\frac{2i\pi}{5}} } $$
note that :
if $P_n(\alpha)=P_n(\alpha^{2})=P_n(\alpha^{3})=P_n(\alpha^{4})=0$ then $(X-\alpha)(X-\alpha^{2})(X-\alpha^{3})(X-\alpha^{4})\mid P_n$
Thus, it suffices to show that $$P_n(\alpha)=P_n(\alpha^{2})=P_n(\alpha^{3})=P_n(\alpha^{4})=0$$
- $P_n(\alpha)=e^{{\frac{8i\pi}{5}}(5k+1)}+e^{{\frac{6i\pi}{5}}(5k+1)}+e^{{\frac{4i\pi}{5}}(5k+1)}+e^{{\frac{2i\pi}{5}}(5k+1)}+1=Q(\alpha)=0 $
the same for others.
- $a)$ false
$d)$ false
Is my proof correct ?