Finding all $n \in N$ that $P(x) = x^n - 1$ is divisible by $Q(x) = x^4 -x^3 + x^2 - x + 1$

Question

I started by finding roots of $Q(x)$: $$Q(x) = x^4 - x^3 + x^2 - x + 1 = 0 \; \; // * (x + 1) \land x \neq - 1$$ $$(x + 1)(x^4 - x^3 + x^2 - x + 1) = x^5 + 1 = 0 \; \land x \neq -1$$ $$x^5 + 1 = 0 \Leftrightarrow x^5 = -1 \Leftrightarrow x = \sqrt[5]{-1}$$ Applied de Moivre formula: $$x_k = 1 \left(\cos\left( \frac{(2k+1)\pi }{5}\right) + i\sin\left( \frac{(2k+1)\pi }{5}\right) \right), \; k = 0, 1, 2, 3, 4$$ $$x_0 = \cos\left( \frac{\pi }{5}\right) + i\sin \left( \frac{\pi }{5}\right) = e^{i\frac{\pi}{5}}$$ $$x_1 = \cos\left( \frac{3\pi }{5}\right) + i\sin \left( \frac{3\pi }{5}\right) = e^{i\frac{3\pi}{5}}$$ $$x_2 = \cos\left( \frac{5\pi }{5}\right) + i\sin \left( \frac{5\pi }{5}\right) = -1 \; \text{, extraneous since } x \neq - 1$$ $$x_3 = \cos\left( \frac{7\pi }{5}\right) + i\sin \left( \frac{7\pi }{5}\right) = e^{i\frac{7\pi}{5}}=e^{-i\frac{3\pi}{5}} =\overline{x_1}$$ $$x_4 = \cos\left( \frac{9\pi }{5}\right) + i\sin \left( \frac{9\pi }{5}\right) = e^{i\frac{9\pi}{5}}= e^{-i\frac{\pi}{5}} = \overline{x_0}$$ No that I have the roots of $Q(x)$, I substitute $x_0$ and $x_1$ in turn for $x$ in $P(x)$ polynomial: $$e^{i\frac{n\pi}{5}} - 1 = 0$$ $$e^{i\frac{3n\pi}{5}} - 1 = 0$$ What I'm not certain of is how do I proceed. Do I just write out equations for several values of $n$ and conclude that $\; n = 10k,\; k = 1, 2, 3...$?

Answer 1

HINT.- $x^4-x^3+x^2-x+1=\dfrac{x^5+1}{x+1}$ and $x^{10}-1=(x^5+1)(x^5-1)$.

Because of $x^n-1$ is not divisible by $x^4-x^3+x^2-x+1$ for $1\le n\le9$, you can deduce that than $x^{10n}-1$ is as required. Thus the answer is $n$ multiple of $10$.

Answer 2

Let $R_n(x)$ be the remainder of the division of $P(x)$ by $Q(x)$.

We have $R_0(x)=R_{10}(x)=0$ and $R_n(x)\ne0$ for $0 < n < 10$. See WA.

Write $x^{n+10}-1=(x^{n+10}-x^n)+(x^n-1)=x^n(x^{10}-1)+(x^n-1)$.

Since $R_{10}(x)=0$, we have $R_{n+10}(x)=R_n(x)$ for all $n$.

Since $R_n(x)\ne0$ for $0 < n < 10$, we have $R_n(x)\ne0$ when $n$ is not a multiple of $10$.

Bottom line: $R_{n}(x)=0$ iff $n$ is a multiple of $10$.