Determine the remainder of the Euclidean division of the polynomial $x+x^{3}+x^{9}+x^{27}+x^{81}+$ $x^{243}$ by $x-1$ and by $x^{2}-1$.
Let $P(x)=x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}$. The Euclidean division theorem gives the existence of two polynomials $Q(x)$ and $R(x)$ such that $P(x)=(x-$ 1) $Q(x)+R(x)$. Since $x-1$ is of degree $1$, $R(x)$ is a constant polynomial and it is sufficient to evaluate it at one value to know it completely. By taking $x=1$ in the previous equality, we get $R(x)=P(1)=6$. We can also find this result by noting that $P(x)=(x-1)+\left(x^{3}-1\right)+\left(x^{9}-1\right)+\left(x^{27}-1\right)+\left(x^{81}-1\right)+\left(x^{243}-1\right)+6$ and using that $x^{n}-1$ is divisible by $x-1$ for all integers $n \geq 1$.

We reason similarly for the second part of the exercise: we write $P(x)=$ $\left(x^{2}-1\right) Q(x)+R(x)$, with $R(x)$ of degree less than or equal to 1. By evaluating at 1 and -1 in the previous equality, we obtain $R(x)=6 x$.