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

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Solution

(x)= x^1999
factors of x^2-1=(x+1)(x-1)
we know,
dividend=divisorquotient+remainder
since p(x) is a polynomial of degree 1999 so it will leave a linear remainder in form of ax+b where a and b are constants.
x^1999=divisor(x+1)(x-1)+ax+b
put x=1
so, 1=a+b ..................(i)
put x=-1
so, -1=-a+b .................(ii)
solving equations
we get,
a=1 and b=0
so remainder=ax+b=1*x+0=x
so when x^1999 is divided by x^2-1 then it leaves remainder as x.

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