What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ in terms of $x$?I tried factoring $x$ from both polynomials but I don't know what to do next since there'd be a $1$ in the second polynomial.
Any help would be appreciated!
 A: Let your $87$-th degree polynomial be $P(x)$. You're to find $Q(x)=ax^2+bx+c$ where
$$
P(x)=(x^3-x)D(x)+Q(x).
$$
Then:
$$
5=P(1)=0\cdot D(1)+Q(1)\implies Q(1)=5;\\
0=P(0)=0\cdot D(0)+Q(0)\implies Q(0)=0;\\
-5=P(-1)=0\cdot D(-1)+Q(-1)\implies Q(-1)=-5.
$$
From $Q(0)=0$, it is easy to see $c=0$. Using this and the other 2 conditions, we have:
$$
a+b=5,\quad a-b=-5\implies a=0,\quad b=5.
$$
In sum $Q(x)=5x$.
A: $xf(x^2)\,\bmod\, x(x^2\!-\!1)\, =\, x\,(\overbrace{f(\color{#c00}{x^2})\,\bmod\, x^2\!-\!1}^{\color{#c00}{\Large x^2\ \equiv\,\  1}})\, =\, xf(\color{#c00}{ 1})\, =\, 5x$
A: Modulo $x^3-x$ we have $x^3=x$ so $x^{2n+1}=x$ for any integer $n\ge 0$ (a trivial induction exercise). Hence your sum leaves remainder $5x$.
A: $$ P(x)=x^{7} + x^{27} + x^{47} +x^{67} + x^{87}=x(x-1)(x+1)q(x) + r(x)$$
Note that we have $r(1)= P(1)$, $r(0)= P(0)$ and $r(-1)=P(-1)$.
We can find $$r(x)=x^2+ax+b =5x $$ using the above information.  
A: Doing long division:
$$P(x)=\frac{x^{7} + x^{27} + x^{47} +x^{67} + x^{87}}{x^3-x}=\frac{x^{86}+x^{66}+x^{46}+x^{26}+x^6}{x^2-1}=\\
\frac{\sum_{i=0}^{9}(x^{86-2i}-x^{84-2i})+2\sum_{i=0}^{9}(x^{66-2i}-x^{64-2i})+3\sum_{i=0}^{9}(x^{46-2i}-x^{44-2i})+4\sum_{i=0}^{9}(x^{26-2i}-x^{24-2i})+5\sum_{i=0}^{2}(x^{6-2i}-x^{4-2i})+5}{x^2-1}=\\
Q(x)+\frac{5}{x^2-1}=Q(x)+\frac{5x}{x^3-x}.$$
A: $x^{2n}\equiv1 \pmod{x^2-1}$ so $(x^2-1) | (x^{2n}-1)$ so $(x^3-x) | (x^{2n+1}-x)$ 
so $x^{2n+1}\equiv x \pmod {x^3-x}$ 
so $x^7+x^{27}+x^{47}+x^{67}+x^{87}\equiv 5x \pmod {x^3-x}$
