+26387 What is the remainder when \((x+1)^{2010}\) is divided by \(x^2+x+1\)?
I assume
Look at the low powers of x.
Eventually they will cycle.
\(\begin{array}{|r|c|c|} \hline n & & \text{remainder} \\ \hline 0 & \dfrac{(x+1)^0}{x^2+x+1} & \color{red}1 \\ \hline 1 & \dfrac{(x+1)^1}{x^2+x+1} & x+1 \\ \hline 2 & \dfrac{(x+1)^2}{x^2+x+1} & x \\ \hline 3 & \dfrac{(x+1)^3}{x^2+x+1} & -1 \\ \hline 4 & \dfrac{(x+1)^4}{x^2+x+1} & -x-1 \\ \hline 5 & \dfrac{(x+1)^5}{x^2+x+1} & -x \\ \hline 6 & \dfrac{(x+1)^6}{x^2+x+1} & \color{red}1 \\ \hline 7 & \dfrac{(x+1)^7}{x^2+x+1} & x+1 \\ \hline 8 & \dfrac{(x+1)^8}{x^2+x+1} & x \\ \hline \ldots &\ldots & \ldots \\ \hline 2010 & \dfrac{(x+1)^{2010}}{x^2+x+1} & \color{red}1 \\ \hline \end{array}\)
\((x+1)^n / (x^2+x+1) \) is cyclic of order 6
So we get \((x+1)^n \equiv (x+1)^{n\pmod{6}}\pmod{x^2+x+1}\)
\((x+1)^{2010} \equiv (x+1)^{6·335+0} \equiv(x+1)^{0} \equiv 1 \pmod{x^2+x+1}\)
The remainder when \((x+1)^{2010}\) is divided by \(x^2+x+1\) is 1
