Let $f(x)$ be the polynomial of degree $2n+1$, $(x-1)^n\mid(f(x)+1)$ and $(x+1)^n\mid((f(x)-1))$, Find $f(x)$. Someone asked a question in my group,But i have no idea about it.
Let $f(x)$ be the polynomial of degree $2n+1$,  $(x-1)^n\mid(f(x)+1)$ and  $(x+1)^n\mid((f(x)-1))$, Find $f(x)$.
I noticed $f(1)=-1,f(-1)=1$.....Who can help me? Thanks!
Thanks for "marty cohen" and "Jack's wasted life".They proved the solution is not unique. But the question was printed on a famous book with no answer.I think it could not be a typo. So we exactly wanted might be a general formula of $f(x)$ for any $n$?
 A: $(x-1)^{n-1}$ and $(x+1)^{n-1}$ are both factors of $f'(x)$.  So if $f(x)$ has degree $2n-1$, then it equals $$f(x)=A\int (x^2-1)^{n-1}dx+B$$
By symmetry $x\iff -x$, the constant term of $f(x)$ is zero.  Then $A$ is the constant for which $f(x)+1$ is a multiple of $x-1$ - and hence a multiple of $(x-1)^n$.
A: $f(x)+1
=(x-1)^n g(x)
$
and
$f(x)-1
=(x+1)^n h(x)
$
where
$g$ and $h$
are of degree $n+1$.
Then
$(x-1)^n g(x)-1
=(x+1)^n h(x)+1
$
or
$(x-1)^n g(x)-(x+1)^n h(x)
=2
$.
But,
for any real $c$,
$2
=(x-1)^n g(x)-(x+1)^n h(x)
=(x-1)^n (g(x)+c(x+1)^n)-(x+1)^n (h(x)+c(x-1)^n)
$
or
$(x-1)^n (g(x)+c(x+1)^n)-1
=(x+1)^n (h(x)+c(x-1)^n)+1
$.
Therefore,
if
$(x-1)^n|f(x)+1$
and
$(x+1)^n|f(x)-1$,
then
$(x-1)^n|(f(x)+c(x+1)^n(x-1)^n)+1$
and
$(x+1)^n|(f(x)+c(x+1)^n(x-1)^n)-1$
for any real $c$.
This is obvious now.
So,
if there is one solution,
then there are
an infinite number of solutions.
A: First we have:
$$u(x)(x-1)^n - 1 = v(x)(x+1)^n + 1$$
Because of $((x-1)^n,(x+1)^n)=1$,there exists $u(x),v(x)$, such that:
$$u(x)(x-1)^n - v(x)(x+1)^n = 2$$
if we find such $u(x),v(x)$,then $(u(x)+t(x)*(x+1)^n)(x-1)^n - 1$ could be a valid $f(x)$.
Notice that there exists a $u(x)$, $deg(u(x)) < n$,then because:
$$u(x) = (1-v(x)(x+1)^n)(x-1)^{-n}$$
So:
$$u^{(i)}(-1) = (\prod_{j=0}^{i-1}(-n-j)) (-2)^{-n-i}$$
Use Taylor's expansion:
$$u(x) = \sum_{i=0}^{n-1}\frac{u^{(i)}(-1)}{i!}(x+1)^i$$
