abstract algebra?

Question

For an arbitrary integer , verify that
3|a((2a^2) +7)
Hint: An integer a has exactly one of three possible forms.

Answer 1

For integer a verify that

3 divides a(2a² + 7)
______________

Case 1: Dividing a by 3 gives a remainder of zero.
Then there exists an integer b such that
a = 3b

a(2a² + 7) = 3b(2a² + 7)

The expression is multiplied by 3 and therefore

3 | a(2a² + 7)

Case 2: Dividing a by 3 gives a remainder of 1.
Let
a = 3b + 1

a(2a² + 7) = (3b + 1)[2(3b + 1)² + 7]
= (3b + 1)[2(9b² + 6b + 1) + 7]
= (3b + 1)[18b² + 12b + 2 + 7]
= (3b + 1)[18b² + 12b + 9]
= (3b + 1) * 3[6b² + 4b + 3]
= 3(3b + 1)[6b² + 4b + 3]

Again the expression is multiplied by 3 and therefore

3 | a(2a² + 7)

Case 3: Dividing a by 3 gives a remainder of 2.
Let
a = 3b + 2

a(2a² + 7) = (3b + 2)[2(3b + 2)² + 7]
= (3b + 2)[2(9b² + 12b + 4) + 7]
= (3b + 2)[18b² + 24b + 8 + 7]
= (3b + 2)[18b² + 24b + 15]
= (3b + 1) * 3[6b² + 8b + 5]
= 3(3b + 1)[6b² + 8b + 5]

Again the expression is multiplied by 3 and therefore

3 | a(2a² + 7)

There are the only three cases so

3 | a(2a² + 7)

qed