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Question:

Find the remainder when $x^n - a^n$ is divided by $x^2-a^2$, where $n$ is odd.

I am not sure if my process is correct:

As the divisor is a quadratic, the remainder should be linear or constant.

  1. $P(-a) = -a^n-a^n = -Aa+B = -2a^n$
  2. $P(a) = a^n - a^n = Aa+B = 0$

adding 1 and 2:

$2B =-2a^n$
$B = -a^n$

subtracting 1 and 2:

$-2Aa=-2a^n$
$Aa = a^n$
$A = a^{n-1}$

So, the remainder = $a^{n-1}x-a^n$

I am unsure how to approach it in any way or if my current way is correct.

Audric Jardine
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    You approach is correct. – Yathi Nov 30 '23 at 06:08
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    Another method: mod $x^2-a^2$ we have $x^2\equiv a^2$ hence $x^{2k+1}=(x^2)^kx\equiv (a^2)^kx\equiv a^{n-1}x$. – coiso Nov 30 '23 at 06:09
  • I'm confused as what to what my remainder will look like. Is that sufficient to say my remainder will be $a^{n-1}x−a^n$? Or can I somehow go further? – Audric Jardine Nov 30 '23 at 06:15
  • Also, the question has a part b: If $n$ is even. For this, is the remainder $0$? I did the same process, ending up with $-Aa+B = 0$ and $Aa+B=0$. – Audric Jardine Nov 30 '23 at 06:18
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    Yes. What you have done is sufficient to say that the remainder is $a^{n-1}x-a^n$. When $n$ is even, following your method, we see that the remainder is 0. Alternatively, when $n$ is even, say $n=2k$, then $x^n-a^n= (x^2)^k-(a^2)^k= (x^2-a^2)h(x)$, for some polynomial $h(x)$, implying that $x^2-a^2$ divides $x^n-a^n$. Hence the remainder would be 0. – Yathi Nov 30 '23 at 08:11

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