Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a2) will be
Case – I: a2 – (3q)2 = 9q2 =3(3q2) = 3p (p = 3q2)
Case-II: a2 = (3q + l)2 = 9q2 + 6q+ 1 = 3[3q2 + 2q] + 1 = 3p+l (Where p = 3q2+ 2q) or
Case – III: a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1
= 3[3q2 + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q2 + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.