<s>[INST] Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.

Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
 [/INST] There are a few ways to do this problem : one of which is a more general method which turns out to be very elaborate in this case. Let $p(x) = x^3 - x$.
The first is to note that $x^3 \equiv x \mod p(x)$, therefore $x^{2n+1} \equiv x \mod p(x)$ for all $n \geq 0$(show this by induction). Thus, the given expression simplifies to $$
q(x) = x^{81} + x^{49}  + x^{25} + x^9+x \equiv x+x+x+x+x \equiv 5x \mod p(x)
$$ 
which is the answer.

For the more general method, we have the Chinese remainder theorem for polynomials. We find the remainder when $q(x)$ is divided by $x, x+1$ and $x-1$, which are the relatively prime factors of $x^3 - x$. The CRT guarantees a unique polynomial modulo $x^3-x$ that leaves those remainders modulo $x,x+1$ and $x-1$ respectively, which then will be the remainder.
For example, it is clear that $q(x) \equiv 0 \pmod {x}$, and that $q(1) = 5$ and $q(-1) = -5$.
Thus, $q(x) \equiv -5 \pmod{x+1}$ and $q(x) \equiv 5 \pmod{x-1}$. 
While the CRT provides us with an algorithm for computing the remainder in this case, we may also proceed by brute force : the remainder must be $ax^2+bx+c$, a degree two polynomial since $x^3-x$ is of degree $3$.
By substituting $x = 0$, $x= 1$ and $x=-1$ we get $c = 0 ; a+b = 5$ and $a-b = -5$. Thus, from here we get $a=c=0$ and $b=5$, thus the remainder is $5x$.
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