Use eculids division lemma to show …

Let n be an arbitrary positive integer.
On dividing n by 3, let q be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
n = 3q + r, where {tex}0 \leq r < 3{/tex}.
The possibilities of remainder = 0,1 or 2
n² = (3q + r)² [∵ (a + b) ² = a² + 2ab + b²]
{tex}\therefore{/tex} n² = 9q² + r² + 6qr ... ....(i), where {tex}0 \leq r < 3{/tex}.
Case I When r = 0.
Putting r = 0 in (i), we get
n² = 9q²
= 3(3q²)
n² =  3m, where m = 3q² is an integer.
Case II When r = 1.
Putting r = 1 in (i), we get
n² = (9q² + 1 + 6 q)
= 3(3q² + 2q) + 1
n²=  3 m + 1, where m = (3q² + 2q) is an integer.
Case lll When r = 2.
Putting r = 2 in (i), we get
n² = (9q² + 4 + 12q)
= 3(3q² + 4q + 1) + 1
n²= 3m + 1, where m = (3q² + 4q + 1) is an integer.
From all the above cases it is clear that the square of any positive integer is of the form 3m or (3m + 1) for some integer m.