I3.1 若 $\mathrm{f}(x)=x^{4}+x^{3}+x^{2}+x+1$, 求 $\mathrm{f}\left(x^{5}\right)$ 除以 $\mathrm{f}(x)$ 的餘值 $a$ 。
If $\mathrm{f}(x)=x^{4}+x^{3}+x^{2}+x+1$, determine the remainder $a$ of $\mathrm{f}\left(x^{5}\right)$ divided by $\mathrm{f}(x)$.
Clearly $f(1)=5$.
By division algorithm, $\mathrm{f}(x)=(x-1) Q(x)+5$, where $Q(x)$ is a polynomial
$$
\begin{aligned}
\mathrm{f}\left(x^{5}\right) & =\left(x^{5}-1\right) Q\left(x^{5}\right)+5 \\
& =(x-1)\left(x^{4}+x^{3}+x^{2}+x+1\right) Q\left(x^{5}\right)+5 \\
& =\mathrm{f}(x)(x-1) Q\left(x^{10}\right)+5
\end{aligned}
$$

The remainder is $a=5$.