The residue of $x^{15}-1$ divided by $x^2-1$

Question

I want to use the long division, but I don't know how to start

Hint: $;x^2-1;$ has two rather easy roots, one, and only one, of which is also a root of $;x^{15}-1;$ . — Timbuc, Oct 22 '14 at 03:17
$$x((x^2)^7-1^7)+x-1$$
Now use http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b — lab bhattacharjee, Oct 22 '14 at 03:17
We want the remainder when $P(x)=x^{14}+\cdots+1$ is divided by $x+1$. Let $P(x)=(x+1)Q(x)+r$. Set $x=-1$. — André Nicolas, Oct 22 '14 at 03:22

score 1 · Accepted Answer · answered Oct 22 '14 at 03:21

1

Hints:

Observe that $x^2-1=(x-1)(x+1)$;

$x^3-1=(x-1)(x^2+x+1)$,

$\dots$

Generally, $x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x^2+x+1)$

answered Oct 22 '14 at 03:21

Paul

1 Answers1