An algebra problem by Rama Devi

By division Algorithm ,

Dividend = (Divisor * Quotient) + Remainder.

From the problem , we know that Dividend is $x^{1999}$ , Divisor is $x^2 - 1$

We know that the degree of the remainder is always one less than the degree of the Divisor .Since the degree of the Divisor is 2, the degree of the remainder is 1 , which is in the form $ax+b$ .

Dividend = (Divisor * Quotient) + Remainder.

$x^{1999}$ = ( $x^2 - 1$ ) * $Q$ + $ax+b$

$x^{1999}$ = $(x+1)(x-1)$ * $Q$ + $ax+b$

Now put $x = 1$

$1^{1999}$ = $0$ + $a+b$

$a + b$ = $1$ .............................................................. $\boxed{1}$

Now put $x=-1$

${-1}^{1999}$ = $0$ + $b-a$ .

$b - a$ = $-1$ ............................................................ $\boxed{2}$

From $\boxed{1}$ and $\boxed{2}$ , We get the values of $a$ and $b$ as $a=1$ and $b=0$ .

Therefore the remainder is $ax+b$ , which is $1(x)+0$ , which is $x$

Therefore the required answer is $\boxed{x}$

I think a shorter solution is- $x^{2}=1$ therefore, $x^{1999}=x^{1998}.x=x$

Consider the following:

\(\begin{array} {} \color{blue}{\dfrac{x^3}{x^2-1}} = \dfrac {x^3-x+x}{x^2-1} = x + \dfrac{x}{x^2-1} & \Rightarrow \text{remainder} = x \\ \color{red}{\dfrac{x^5}{x^2-1}} = \dfrac {x^5-x^3+x^3}{x^2-1} = x^3 + \color{blue}{\dfrac{x^3}{x^2-1}} & \Rightarrow \text{remainder} = x \\ \dfrac{x^7}{x^2-1} = \dfrac {x^7-x^5+x^5}{x^2-1} = x^5 + \color{red}{\dfrac{x^5}{x^2-1}} & \Rightarrow \text{remainder} = x \\ \Rightarrow \dfrac{x^{1999}}{x^2-1} & \Rightarrow \text{remainder} = \boxed{x} \end{array} \)

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

$= x(x^{1998} - 1) + x$

$= x(x^{2} - 1)Q(x) + x$