show that square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q

Let a be any positive integer and b = 6.

Then, by Euclids algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k ₁ + 1, where k ₁ is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k ₂ + 1, where k ₂ is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k ₃ + 1, where k ₃ is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.