Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumption is correct then $x^2+1=(x-i)(x+i)$
Now for $x=\pm i$ using synthetic division it leaves remainder 0
Am I at the right direction, please tell

HINT:
Note that $$\begin{align}x^{20}+x^{15}+x^{10}+x^5&=x^5((x^5)^3+(x^5)^2+x^5+1)\\&=x^5(x^5+1)((x^5)^2+1)\\&=x^5(x^5+1)(x^{10}+1)\end{align}$$ and $$(-i)^{10}=i^{10}=i^2=-1$$ so $x-i$ and $x+i$ are factors of the polynomial $x^{10}+1$.
