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: Show that $x^2+1$ divides $x^{10}+1$, and that $x^{10}+1$ divides $f(x)$.
$$
x^{20} + x^{15} + x^{10} + x^5 = x^5(x^5+1)(x^{10}+1)=x^5  \left((1 + x)(1 - x + x^2 - x^3 + x^4)\right) \left((1 + x^2) (1 - x^2 + x^4 - x^6 +
    x^8)\right)
$$