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Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.

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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.

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