Let a be a positive integer.

Then, by Euclid ’s division lemma, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that

a = 5m + r, where 0 ≤ r < 5⇒ a2= (5m + r2) = 25mr + r2+ 10mr

[∵(a+b)2= a2+ 2ab + b2]

⇒ a2= 5(5m2+ 2mr) + r2...(i)where, 0 ≤ r < 5

Case I

When r = 0, then putting r = 0 in Eq.(i), we geta2= 5(5m2) = 5qwhere, q = 5m2is an integer.

Case II

When r = 1, then putting r = 1 is Eq.(i), we geta2= 5(5m2+ 2m) + 1⇒ q = 5q + 1where,q = (5m2+ 2m) is an integer.

Case III

When r = 2,then putting r = 2 in eq.(i),we geta2= 5(5m2+ 4m) + 4 = 5q + 4where, q = (5m2+ 4m) is an integer.

Case IV

When r = 3, then putting r = 3 in Eq.(i), we geta2= 5(5m2+ 6m) + 9 = 5 (5m2+ 6m) + 5 + 4= 5(5m2+ 6m + 1) + 4 = 5q + 4where, q = (5m2+ 8m + 3) is an integer.

Case V

When r = 4, then putting r = 4 in Eq.(i), we get

a2= 5(5m2+ 8m) + 16 = 5 (5m2+ 8m) + 15 + 1⇒ a2= 5(5m2+ 8m + 3) + 1 = 5q + 1

where, q = (5m2+ 8m + 3) is an integer.

Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.