Question

show that square of any positive integer is of the form 3q or 3q + 1 for some integer q

Answer 1

Answer:

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a2 = (3q)2 or (3q + 1)2 or (3q + 2)2

a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4

= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1

= 3k1 or 3k2 + 1 or 3k3 + 1

Where k1, k2, and k3 are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

Answer 2

Answer:

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a2 = (3q)2 or (3q + 1)2 or (3q + 2)2

a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4

= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1

= 3k1 or 3k2 + 1 or 3k3 + 1

Where k1, k2, and k3 are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

All the (?)2 is the square.