Use division algorithm to show that the square of any positive integer is of the form /3p or 3p+ 1. ​

Sure, let's use the division algorithm to show that the square of any positive integer is of the form 3p or 3p+1.

The division algorithm states that for any integer 'a' and positive integer 'b', there exist unique integers 'q' and 'r' such that a = bq + r, where 0 ≤ r < b.

Let's consider a positive integer 'n'. We can express 'n' in the form of 3q, 3q+1, or 3q+2 (where q is a quotient). This is because when 'n' is divided by 3, the remainder 'r' can only be 0, 1, or 2.

1. If 'n' is of the form 3q, then n^2 = (3q)^2 = 9q^2 = 3(3q^2). So, n^2 is of the form 3p (where p = 3q^2).

2. If 'n' is of the form 3q+1, then n^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1. So, n^2 is of the form 3p+1 (where p = 3q^2 + 2q).

3. If 'n' is of the form 3q+2, then n^2 = (3q+2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1. So, n^2 is of the form 3p+1 (where p = 3q^2 + 4q + 1).

So, we can conclude that the square of any positive integer is of the form 3p or 3p+1.