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 $(x+1)$?
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$\begingroup$Hint
$$x^{81} + x^{49} + x^{25} + x^{9} + x=(x^{3}-x)Q(x)+R(x)$$where $$R(x)=ax^2+bx+c$$and $$R(0)=0\\R(1)=5\\R(-1)=-5$$
$\endgroup$ $\begingroup$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 $5x$.
$\endgroup$ $\begingroup$As the degree of the divisor is $3$, so the degree of the remainder is at most $2$. Let the remainder $R\left(x\right)$ be $ax^2+bx+c$. As $x^3-x=x\left(x-1\right)\left(x+1\right)$, we'll find that: $$\begin{cases} R\left(-1\right)=a-b+c=\left(-1\right)^{81}+\left(-1\right)^{49}+\left(-1\right)^{25}+\left(-1\right)^9+\left(-1\right)=-5 \\ R\left(0\right)=c=0^{81}+0^{49}+0^{25}+0^9+0=0 \\ R\left(1\right)=a+b+c=1^{81}+1^{49}+1^{25}+1^9+1=5 \end{cases}$$ After some calculation, you'll get $a=0,b=5,c=0$, so the remainder is $\boxed{5x}$.
$\endgroup$ $\begingroup$You may use
- $(x^2-1)|(x^{2k} -1)$ by applying $(x^2)^k - 1 = (x^2-1)(x^{2(k-1)} + \cdots + x^2 + 1)$
\begin{eqnarray*} x^{81} + x^{49} + x^{25} + x^{9} + x & = & x\left((x^2)^{40} -1 + (x^2)^{24} -1 + (x^2)^{12} -1 + (x^2)^{4} -1 + 4+1\right) \\ & = & x((x^2-1)p(x) + 5)\\ & = & x(x^2-1)p(x) + 5x \end{eqnarray*}So, the remainder is $5x$.
$\endgroup$ $\begingroup$The remainder is a quadratic polynomial $g(x)$ that is divisible by $x$ (since $f(x)$ is divisble by $x$ and has remainders $f(1)$ and $f(-1)$ from division on $(x-1)$ and $(x+1)$ respectively.
One may argue that:$$ g(x)=f(1)\frac{x+1}{2}+f(-1)\frac{x-1}{-2} + \gamma(x^2-1) $$
From the fact $g(0) = 0$, one can find the coefficient $\gamma=\frac{f(1)-f(-1)}2$.
$\endgroup$ $\begingroup$Hint:
Let $$x^{81}+x^{49}+x^{25}+x^9+x=p(x)(x^3-x)+ax(x-1)+bx(x+1)+c(x-1)(x+1)$$
where $a,b,c$ are arbitrary constants
Put $x=0,-1,1$ to find $a,b,c$
$\endgroup$ $\begingroup$$$x^{81} + x^{49} + x^{25} + x^{9} + x=\left(x^{81}-x\right)+\left(x^{49}-x\right)+\left(x^{25}-x\right)+\left(x^9-x\right)+5x.$$Can you end it now?
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