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Let $a\neq0$ and $p(x)$ be a polynomial of degree greater than $2$. If $p(x)$ leaves remainders $a$ and $-a$ when divided respectively by $x+a$ and $x-a$. Find the remainder when $p(x)$ is divided by $x^2-a^2$

$$ p(x)=q(x).(x+a)+r_1=q(x).(x+a)+a\quad\big[r_1=p(-a)=a\big]\\ p(x)=s(x).(x-a)+r_2=s(x).(x-a)-a\quad\big[r_2=p(a)=-a\big]\\ p(x)=t(x).(x^2-a^2)+r=t(x).(x^2-a^2)+Ax+B\\ p(a)=aA+B=-a\\ p(-a)=-aA+B=a\\ B=0,\;A=-1\implies r=Ax+B=-x $$ I was wondering Is there another way to solve this problem ?

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Hint $\ \overbrace{ x\!-\!a,x\!+\!a\,\mid\, p\!+\!x\!}^{\large p(a)+a\ =\ 0\ =\ p(-a)-a\!\!\!\!\!}\!\iff\! (x\!-\!a)(x\!+\!a)\mid p\!+\!x\,$ over a field where $\,\color{#c00}{2a\neq 0}$

since then we have the Bezout GCD equation $\ x+a - (x-a)\, =\, \color{#c00}{2a}$

which implies that $\,x\!+\!a,\, x\!-\!a\,$ are coprime, so their lcm = product.


Alternatively, $ $ apply: $\ fg\bmod fh = f(g\bmod h) = $ mod Distributive Law (a form of CRT)

$x\pm a\mid p\!+\!x\,\Rightarrow\, p\!+\!x\bmod x^2\!-\!a^2 = (x\!-\!a)\left[\dfrac{p\!+\!x}{x\!-\!a}\bmod x\!+\!a\right] = (x\!-\!a)[0] = 0$

There are two of many ways to solve CCRT = Constant-case of CRT [Chinese Remainder Theorem]

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Write $$p(x) = k(x)(x^2-a^2)+bx+c$$

for some $b$ and $c$. Since $$p(-a)=a \Longrightarrow a = -ab+c$$ and since $$p(a)=-a \Longrightarrow -a = ab+c$$

Solving thhis system we get $\boxed{r(x) = -x}$.

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One possible approach would be to use partial fractions:

$$\frac{1}{x^2-a^2} = \frac{1}{2a} \left( \frac{1}{x-a} - \frac{1}{x+a}\right) .$$

On the other hand, by the given information,

$$\frac{p(x)}{x-a} = q_1(x) - \frac{a}{x-a}$$

and

$$\frac{p(x)}{x+a} = q_2(x) + \frac{a}{x+a}$$

for some quotient polynomials $q_1, q_2$.

Therefore,

$$ \frac{p(x)}{x^2-a^2} = \frac{1}{2a} \left( \frac{p(x)}{x-a} - \frac{p(x)}{x+a} \right) = \frac{1}{2a} \left( q_1(x) - q_2(x) - \frac{a}{x-a} - \frac{a}{x+a} \right) = \\ q(x) - \frac{x}{x^2 - a^2}$$ where $q(x) = \frac{1}{2a} (q_1(x) - q_2(x))$ is a polynomial. It follows that the remainder of dividing $p(x)$ by $x^2 - a^2$ is $-x$.

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.While your attempt is also right, there is this one, more in the spirit of the Chinese remainder theorem, which is a generalization of this problem.

A specific case of CRT which covers the above case is the following :

Let $f(x)$ and $g(x)$ (polynomials over a field, say) be relatively prime. Let $s(x)$ and $t(x)$ be polynomials satisfying $s(x)f(x) + t(x)g(x) = 1$ for all $x$. Then, for any pair of polynomials $p(x),q(x)$, the solution to the pair of polynomial equations: $$ P(x) \equiv p(x) \mod f(x) \\ P(x) \equiv q(x) \mod g(x) $$ is given by: $$ P(x) \equiv p(x)t(x)g(x) + q(x)s(x)f(x) \mod (fg(x)) $$ where $fg$ is the product of $f$ and $g$.

Indeed, note that $a \neq 0$, so $\frac 1a$ is well defined. Next, we have: $$\frac 1{2a}(x+a) +\frac{-1}{2a}(x-a) = 1 \tag{1} $$

and now, we essentially want to solve for the system : $$ p(x) \equiv a \mod (x+a) \\ p(x) \equiv -a \mod (x-a) $$

by the usual solution for such a pair of congruences given by CRT, from the equation $(1)$ we have : $$ p(x) \equiv \left(a \times \frac {-1}{2a}(x-a)\right) + \left(-a \times \frac 1{2a}(x+a)\right) \mod (x^2-a^2) $$

which gives $p(x) \equiv -x \mod (x^2 - a^2)$.

What you did was use the structure of the remainder as $Ax+B$ and then find $A$ and $B$ via substitution. In case the polynomials $f,g$ are not linear, this won't be feasible, so you will need CRT for that.

NOTE : $s(x),t(x)$ can be found via the Euclidean algorithm for polynomials, more on that here.

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The other answers give good elementary methods. Here's an approach from abstract algebra.

Assume that the allowable coefficients of $p$ are a field $F$ (e.g., $\mathbb{R}$ or $\mathbb{C}$). Define $S = \{ p \in F[x]: p(\pm a) = \mp a\}$. This set includes every polynomial of degree $3$ or higher that leaves the required remainders. Define $I = \{ p \in F[x]: p(-a) = p(a) = 0\}$. Note that $S = I - x$ (i.e., adding $x$ to any element of $S$ gives an element of $I$, and vice versa), and that $I$ is an ideal. $F[x]$ is a PID, so $I$ must contain all multiples of some primitive element. Linear primitive elements are easy to rule out, and clearly $x^2 - a^2 \in I$, so $I = \langle x^2 - a^2\rangle$. Thus, every element of $S$ is $-x$ plus a multiple of $x^2 - a^2$, and you're done.

This approach gives an obvious generalization:

Every polynomial $p(x)$ that satisfies $p(x_1) = y_1, p(x_2) = y_2, \ldots, p(x_n) = y_n$ for some fixed quantities $x_i$ and $y_i$ has the form $p(x) = L(x) + \prod_{i=1}^n (x - x_i)$, where $L(x)$ is the Lagrange interpolation polynomial.

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  • $\begingroup$ Downvoter: please explain? $\endgroup$ – Connor Harris Apr 12 at 19:03

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