Here is the additional proof I am including

**(*)**:

__Claim__: a² is a multiple of 3 → a is a multiple of 3

__Proof__: Assume a is not a multiple of 3, for some integer a. So a = 3k is not true for any integer k. So a = 3k + 1 or a = 3k + 2 for some integer k.

Case I: Assume a = 3k + 1.

a² = 9k² + 6k + 1

a² = 3(some integer) + 1

a² does not equal 3k for any integer k

Therefore a² is not a multiple of 3

Case II: Assume a = 3k + 2.

a² = 9k² + 12k + 4

a² = 9k² + 12k + 3 + 1

a² = 3(some integer) + 1

------a² = 9k² + 12k + 4------

Therefore a² is not a multiple of 3.