epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Collection of Proofs 121
THEOREM 10.
by 9,
The sum of the cubes of three consecutive integers is divisible
Alleged Proof For the base step observe that 0^ + 1^ + 2^ = 9 is indeed
a multiple of 9. Suppose that n^ + (n + 1)^ + (n + 2)^ = 9k for some integer
number fe. Then:
(n+l)^+(n + 2)^ + (n + 3)^ = n^+(n+l)^ + (n + 2)^+(n + 3)^-n^
= 9/c + 9n^ + 27n + 27
= 9(/c + n^ + 3n + 3),
which is a multiple of 9.
•