<s>[INST] 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
 [/INST] There is no need for synthetic division. Just evaluate $x^{20} + x^{15} + x^{10} + x^5 $ at $x=\pm i$:
$$
(\pm i)^{20} + (\pm i)^{15} + (\pm i)^{10} + (\pm i)^5
=
1 + \mp i - 1 + \pm i
= 0
$$
Therefore, $x\pm i$ divides $x^{20} + x^{15} + x^{10} + x^5 $ and so does their product, because their are coprime.
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