Question

Let f(x) = n^2 - 1, where n is an integer and n >= 1. Is f(x) always divisible

Answer 1

Answer:

Yes.

Step-by-step explanation:

(Assume x is not 1.)

$\frac{x^n-1}{x-1}$ is always divisible for integers greater than 1.

Let's use synthetic division:

1  |   1x^n     + 0x^(n-1)    +0x^(n-2) + ....+0x^3+0x^2+0x -1

  |                  1                 1                      1         1        1     1

---------------------------------------------------------------------------------

      1              1                 1                      1         1         1      0

We see the remainder is 0 which means that $(x-1)$ divides $(x^n-1)$.

The quotient is $x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^2+x+1$.