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I have to find $P(0)$ from the polynomial with minimum degree given that $$(x-1)^3|(P(x)+1)$$ $$(x+1)^3|(P(x)-1)$$ Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively: $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$

Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$ and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.

quasi
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    Are you sure you copied the problem correctly? First of all, it is unclear what you mean by minimal polynomial, this only works in the context when an algebra is given (field extensions, matrices, etc.) Secondly, if this is indeed the problem, then it is trivial. The polynomial $(x-1)^3$ has only four divisors (up to nonzero constant multiple). The same holds for $(x+1)^3$. So you can simply check the cases by hand. It is really just four cases to check, as the degree of $P(x)-1$ and $P(x)+1$ are the same. – A. Pongrácz Aug 14 '18 at 19:38
  • Yes, It is correct, only I don't know if I translated it right, $f|g$ means that $f$ is divisible by $g$. Minimal polynomial= the lowest order polynomial possible that satisifies the conditions. –  Aug 14 '18 at 19:48
  • I don't understand, what you mean by $(x-1)^3$ has four divisors? –  Aug 14 '18 at 19:49
  • I guess (?) you are in the ring of polynomials over some field. Maybe the reals? You should specify these details when you pose a problem. But if my guess is correct, then all number theoretic notions work perfectly in that ring. Up to a unit mulltiple (units=nonzero constant polynomials) the divisors of $(x-1)^3$ are exactly $1, x-1, (x-1)^2, (x-1)^3$. – A. Pongrácz Aug 14 '18 at 20:18
  • Yes, of course over reals. I forgot to add that.. –  Aug 14 '18 at 20:38
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    @Sonkun: I'll post an answer soon, but note my edit, which was based on your comment. The correct notation for the phrase "$f$ is divisible by $g$" is $g|f$. You had it reversed. – quasi Aug 14 '18 at 23:31

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Suppose $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that \begin{align*} (x+1)^3&{\,\mid\,}(P(x)-1)\\[4pt] (x-1)^3&{\,\mid\,}(P(x)+1)\\[4pt] \end{align*} Equivalently, $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that \begin{align*} P(x)&=A(x)(x+1)^3+1\\[4pt] P(x)&=B(x)(x-1)^3-1\\[4pt] \end{align*} for some polynomials $A,B\in\mathbb{Q}[x]$.

Necessarily, $A,B$ have the same degree.

Let $u=(x+1)^3$ and let $v=(x-1)^3$.

If $A,B\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $Au-Bv=-2$, then we have $Au+1=Bv-1$, hence $P=Au+1$ is a polynomial in $\mathbb{Q}[x]$ of least degree satisfying the specified requirements.

The equation $Au-Bv=-2$ is equivalent to $-\bigl({\large{\frac{A}{2}}}\bigr)u+\bigl({\large{\frac{B}{2}}}\bigr)v=1$, hence we have $A=-2a$ and $B=2b$, where $a,b\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $au+bv=1$.

Since $u,v$ are relatively prime in $\mathbb{Q}[x]$, there are unique polynomials $a,b\in\mathbb{Q}[x]$ with

  • $\deg(a) < \deg(v) = 3$$\\[4pt]$
  • $\deg(b) < \deg(u) = 3$

such that $au+bv=1$, so these are the polynomials $a,b$ we want.

The polynomials $a,b$ can be found via the Extended Euclidean Algorithm.

Here are the steps . . . \begin{align*} (x+1)^3=(1)(x-1)^3+(6x^2+2)&\implies u-v=6x^2+2\\[4pt] 6(x-1)^3=(x-3)(6x^2+2)+16x&\implies (-x+3)u+(x+3)v=16x\\[4pt] 8(6x^2+2)=(3x)(16x)+16&\implies (3x^2-9x+8)u+(-3x^2-9x-8)v=16\\[4pt] \end{align*} from which we get \begin{align*} a&={\small{\frac{1}{16}}}(3x^2-9x+8)\\[4pt] b&={\small{\frac{1}{16}}}(-3x^2-9x-8)\\[4pt] \end{align*} which yields \begin{align*} A&=-2a=-{\small{\frac{1}{8}}}(3x^2-9x+8)\\[4pt] B&=2b={\small{\frac{1}{8}}}(-3x^2-9x-8)\\[4pt] \end{align*} hence \begin{align*} P&=Au+1\\[4pt] &=\left(-{\small{\frac{1}{8}}}(3x^2-9x+8)\right)(x+1)^3+1\\[4pt] &=-{\small{\frac{1}{8}}}(3x^5-10x^3+15x)\\[4pt] \end{align*}

quasi
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