# What is the remainder when x^1999 is divided by (x^2-1) ?

Apr 9, 2017

$x$

#### Explanation:

Note that:

$\left({x}^{2} - 1\right) \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right)$

$= {x}^{2} \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right) - \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right)$

$= \left({x}^{1999} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{1997}}}} + \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{x}^{1995}}}} + \ldots + \textcolor{c y a n}{\cancel{\textcolor{b l a c k}{{x}^{5}}}} + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{{x}^{3}}}}\right) - \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{1997}}}} + \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{x}^{1995}}}} + \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{x}^{1993}}}} + \ldots + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{{x}^{3}}}} + x\right)$

$= {x}^{1999} - x$

So:

${x}^{1999} = \left({x}^{2} - 1\right) \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right) + x$

That is:

${x}^{1999} / \left({x}^{2} - 1\right) = {x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x$

with remainder $x$

The polynomial $x$ is of degree less than the degree of ${x}^{2} - 1$, so is the remainder we want.

Apr 9, 2017

$x$

#### Explanation:

Calling

${p}_{n} \left(x\right) = {x}^{n} + {c}_{n - 1} {x}^{n - 1} + \cdots + {c}_{0}$

as a generic $n$ degree polynomial we have.

${x}^{n} = {p}_{n - 2} \left({x}^{2} - 1\right) + a x + b$ with $n \ge 2$

so we have

${\left(- 1\right)}^{n} = a \left(- 1\right) + b$

and

$1 = a + b$

solving we have

$a = \frac{1 - {\left(- 1\right)}^{n}}{2}$ and $b = \frac{1 + {\left(- 1\right)}^{n}}{2}$

in the present case we have $n = 1999$ so

$a = 1 , b = 0$ and the remainder is $x$