Question

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

Answer 1

`x`

Explanation:

Note that:

`(x^2-1)(x^1997+x^1995+x^1993+...+x^3+x)`

`=x^2(x^1997+x^1995+x^1993+...+x^3+x)-(x^1997+x^1995+x^1993+...+x^3+x)`

`=(x^1999+color(red)(cancel(color(black)(x^1997)))+color(purple)(cancel(color(black)(x^1995)))+...+color(cyan)(cancel(color(black)(x^5)))+color(green)(cancel(color(black)(x^3))))-(color(red)(cancel(color(black)(x^1997)))+color(purple)(cancel(color(black)(x^1995)))+color(blue)(cancel(color(black)(x^1993)))+...+color(green)(cancel(color(black)(x^3)))+x)`

`=x^1999-x`

So:

`x^1999 = (x^2-1)(x^1997+x^1995+x^1993+...+x^3+x)+x`

That is:

`x^1999/(x^2-1) = x^1997+x^1995+x^1993+...+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.

Answer 2

`x`

Explanation:

Calling

`p_n(x)=x^n+c_(n-1)x^(n-1)+ cdots+c_0`

as a generic `n` degree polynomial we have.

`x^n = p_(n-2)(x^2-1) + ax+b` with `n ge 2`

so we have

`(-1)^n = a(-1)+b`

and

`1 = a+b`

solving we have

`a = (1-(-1)^n)/2` and `b=(1+(-1)^n)/2`

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

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