x²-x+1で割った余り

\begin{align*} x^{1000} & =x\left(x^{3}\right)^{333}\\ & =x\left(\left(x+1\right)\left(x^{2}-x+1\right)-1\right)^{333}\\ & =x\sum_{k=0}^{333}C\left(333,k\right)\left(-1\right)^{333-k}\left(\left(x+1\right)\left(x^{2}-x+1\right)\right)^{k}\\ & =x\left(C\left(333,0\right)\left(-1\right)^{333}\left(\left(x+1\right)\left(x^{2}-x+1\right)\right)^{0}+\sum_{k=1}^{333}C\left(333,k\right)\left(-1\right)^{333-k}\left(\left(x+1\right)\left(x^{2}-x+1\right)\right)^{k}\right)\\ & =x\left(-1+\sum_{k=1}^{333}C\left(333,k\right)\left(-1\right)^{333-k}\left(\left(x+1\right)\left(x^{2}-x+1\right)\right)^{k}\right)\\ & =-x+\sum_{k=1}^{333}xC\left(333,k\right)\left(-1\right)^{333-k}\left(\left(x+1\right)\left(x^{2}-x+1\right)\right)^{k} \end{align*} となるのでこれより、余りは\(-x\)となる。