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
 A: Hint: Show that $x^2+1$ divides $x^{10}+1$, and that $x^{10}+1$ divides $f(x)$.
A: 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.
A: By Euclid lemma you can express 
$ f(x)=g(x)q(x) + r(x) $
where $ g(x)= x^2 + 1$ and degree of $ r(x) $ is less than that of $g(x) $
hence $ r(x) $ can be assumed as $ax+b$ rest I leave to you
And your direction is absolutely right just prove it mathematically
A: Yes, your method works: You checked that $i$ is a root by using synthetic division; since it is, you know that $x-i$ is a factor. Next, you checked whether $-i$ is also a root by using synthetic division; since it is, you also know that $x-(-i)=x+i$ is a factor. Putting this information together, you know that the expression has a factor of $(x-i)(x+i) = x^2 + 1$ as desired.
A: 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 $x^{10}+1$.
A: Now, even if we don't have any insight we can  always resort to long division:
$(x^2 + 1)| \begin{cases}x^{18} & & - x^{16} &&+ x^{14} &... & & \\
x^{20} & & & && x^{15} &&&&& x^{10}&&&&& x^5\\
x^{20} & & x^{18}\\
--&--&--&--&--&--\\
&& -x^{18}\\
&& -x^{18} && - x^{16}&\\
--&--&--&--&--&--\\
&&&&x^{16}&+x^{15}\\
...\\
\end{cases}$
We'll call that plan B.
(Which will give us $(x^2 + 1)(x^{18}-x^{16} + x^{14} + x^{13} - x^{12} - x^{11} + x^{10} + x^{9} -x^7 + x^5)$ but it's long and tedious and not very insightful.  And wouldn't be practical for proving something like $x^2 + 1$ divides $x^{100} + x^{75} + x^{50} + x^{25}$ which ... I think it does.)
....
But if we try to be insightful:
The way I see it there are two things to realize.
1) $a^3 + a^2 + a + 1 =$
$(a^3 + a^2) + (a+1)=$
$a^2(a+1) + (a+1) = $
$(a^2 + 1)(a+1)$.
To make this more general (although really hard to read:)
$a^{nk - 1} + a^{nk - 2} + .... + a + 1= $
$(a^{nk-1} +  .... + a^{(n-1)k}) + (a^{(n-1)k -1} + ..... + a^{(n-2)k}) + ..... + (a^{k-1} + ... + 1) =$
$a^{(n-1)k}(a^{k-1} + .... + 1) + a^{(n-2)k}(a^{k-1} + .... + 1) + ... + a^k(a^{k-1} + ..... + 1) + (a^{k-1} + .... + 1) = $
$(a^{(n-1)k} + a^{(n-2)k} + .... 1)(a^{k-1} + a^{k-2} + ... + 1)$.
... Yeesh ... 
but that means 
$x^{20} + x^{15} + x^{10} + x^5 = $
$x^5(x^{15} + x^{10} + x^5 + 1) = $
$x^5(x^{10} + 1)(x^5 + 1)$.
... and factoring the $(x^{10} + 1)$ looks like it has potential.
2)  $(a -1)(a^{k-1} + a^{k-2} + .... + 1) = a^k -1$
is well known.  But we can "flip signs" depending on parity:
$(a+1)(a^{k-1} - a^{k-2} +  a^{k-3} - ... + a - 1) = a^k - 1$ if $k$ is even
$(a+1) (a^{k-1} - a^{k-2} +  a^{k-3} - ... -a + 1)= a^k + 1$ if $k$ is odd.
So 
$(x^{10} + 1) = ((x^2)^5 + 1) = (x^2 + 1)(x^8 - x^6 + x^4 - x^2 + 1)$.
....
And that's it
$x^{20} + x^{15} + x^{10} + x^5 = $
$x^5(x^{10} + 1)(x^5 + 1)=$
$x^5(x^2 + 1)(x^8 - x^6 + x^4 -x^2 + 1)(x^5 + 1)$
Which has $x^2 + 1$ as a factor.
...
A: $$
x^{20} + x^{15} + x^{10} + x^5 = x^5(x^5+1)(x^{10}+1)=x^5  \left((1 + x)(1 - x + x^2 - x^3 + x^4)\right) \left((1 + x^2) (1 - x^2 + x^4 - x^6 +
    x^8)\right)
$$
