$P(x) = 1+x+x^2+x^3+x^4+x^5$ Then find remainder when $P(x^{12})$ is divided by $P(x)$

Question

$P(x) = 1+x+x^2+x^3+x^4+x^5$

Then find remainder when $P(x^{12})$ is divided by $P(x)$.

$P(x)$ is a geometric progression, so I summed it up, but how do I find remainder when $P(x^{12})$ is divided by $P(x)$?

Answer 1

Let $Q(x)$ and $R(x)$ be the quotient and remainder when $P(x^{12})$ is divided by $P(x)$.

Then, $P(x^{12}) = P(x)Q(x)+R(x)$ and $\deg R \le 4$.

The roots of $P(x)$ are the $6$th roots of unity except for $1$, i.e. $\omega_k = e^{i k\pi/3}$ for $k = 1,2,3,4,5$.

Now, plug each of these roots in for $x$ and simplify:

$$P(\omega_k^{12}) = P(\omega_k)Q(\omega_k)+R(\omega_k)$$

$$P(1) = 0 \cdot Q(\omega_k)+R(\omega_k)$$

$$6 = R(\omega_k)$$

Hence, the remainder is $6$ at each of the $5$ values $\omega_k = e^{i k\pi/3}$ for $k = 1,2,3,4,5$. Since $R(x)$ is a polynomial with degree $\le 4$, and $R(x) = 6$ for $5$ distinct values of $x$, we have that $R(x)$ is constant, i.e. $R(x) = 6$ for all $x$.

Answer 2

Note that $x^6\equiv 1 \mod P(x)$ since $x^6-1=(x-1)P(x)$. Then, $x^{12}\equiv 1\mod P(x)$ as well. Then $P(x^{12})\equiv P(1)=6\mod P(x)$

Answer 3

$$x^6\equiv1\pmod{x^6-1}.$$

Therefore $$x^{12n}\equiv1\pmod{x^6-1}.$$

Therefore $$P(x^{12})=1+x^{12}+x^{24}+x^{36}+x^{48}+x^{60}\equiv 1+1+1+1+1+1=6\pmod{x^6-1}$$

Therefore ${x^6-1}$ divides $P(x^{12})-6.$

Therefore $P(x)=\dfrac{x^6-1}{x-1}$ divides $P(x^{12})-6$.

Answer 4

Without modular arithmetic you may use

• $(1): (x-1)P(x) = x^6 - 1$ and hence
• $(2): \left(x^6\right)^n - 1 = (x^6-1)\sum_{i=0}^{n-1}\left(x^6\right)^i \stackrel{(1)}{=}(x-1)P(x)Q_n(x) $ with $Q_n(x) = \sum_{i=0}^{n-1}\left(x^6\right)^i$

It follows

\begin{eqnarray*} \sum_{k=0}^5x^{12k} & = & 1 + \sum_{k=1}^5\left(\left( x^6\right)^{2k} -1\right) +5 \\ & = & \sum_{k=1}^5\left(\left( x^6\right)^{2k} -1\right) +6 \\ & \stackrel{(2)}{=} & (x-1)P(x)\sum_{k=1}^5Q_{2k}(x) +6 \\ \end{eqnarray*} So, the remainder is $\boxed{6}$.