Hello, free_to_fly!

I think I have a proof for #3 . . .

3. If n - 1 is divisible by three, show that n³ - 1 is divisible by nine.

We are told that n - 1 is divisible by 3.

. . Hence: .n - 1 .= .3a, for some integera.

Multiply both sides by (n² + n + 1): .(n - 1)(n² + n + 1) .= .3a(n² + n + 1)

. . and we have: .n³ - 1 .= .3a(n² + n + 1) .[1]

The quadratic factor is: .n² + n + 1 .= .(n - 1)(n + 2) + 3

. . Since n - 1 .= .3a, it becomes: .3a(n + 2) + 3 .= .3(an + 2a + 1)

. . So the quadratic factor is a multiple of 3: .3b, for some integerb.

Substitute into [1] and we have: .n³ - 1 .= .3a(3b) .= .9ab, a multiple of 9.

Therefore: .n³ - 1 is divisible by 9.