Question

show that any positive integer is of the form 3q, 3q+1,3q+ 2 for some integer q

Answer 1

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

Let 'a' be any positive integer and p = 3 
such that

a = 3q + r .....where 0 ≤ r ≥ 3.

∴ r = 0,1,2

Now (i) if r = 0
∴a = 3q + 0
∴a = 3q

(ii) if r = 1
∴a = 3q + 1

(iii) if r = 2
∴a = 3q + 2

∴ Any positive integer is of the form 3q, 3q+1,3q+ 2 for some integer q.