(Q13). Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1 ?

Let a be any positive integer

Let b = 3

By dividing algorithm rule for a and b : a = 3q + r, for q > 0,and 0 ≤ r <

The possible remainders are 0, 1,

So, a = 3q , 3q+1, 3q+2, 3q+3, where q is the

Let us consider a = 3q+0,3q+1

a = 3q+0

a = 3q

a² = (3q)²

a² = q²

a² = (3q²)

Here 3(3q²) is in the form of 3p

p = q²

Let us consider a = 3q+0,3q+1

a = 3q+1

a² = (3q+1)²

a² = (3q)²+()² + 2(3q)()

a² = (3q²)+ + q

a² = (3q²+2q) + 1

Here a is in the form of p+1

p = q² + 2q