Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$

Let $x^{20}+x^{15}+x^{10}+x^5+1=(x^4+x^3+x^2+x+1)Q(x)+R(x)$.
$x^4+x^3+x^2+x+1=0$ has $4$ complex roots $a_1,a_2,a_3,a_4$. And these are also roots of $x^5=1$, so when $x=a1,a2,a3,a4$, the above equation becomes
$5=R(a_i)$ ($i=1,2,3,4$)
This is true when $R(x)=5$ for every $x$, and it is easy to show
that a polynomial of degree $3$ like this is unique. So the remainder is just $5$ (constant).
