What is the reason behind taking b =3 in Question:
Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. [4 MARKS]
let ' a' be any positive integer and b = 3.
we know, a = bq + r , 0 < r< b.
now, a = 3q + r , 0 < r < 3.
the possibilities of remainder = 0,1 or 2
Case I - a = 3q
a2 = 9q2
= 3 x ( 3q2)
= 3m (where m = 3q2)
Case II - a = 3q +1
a2 = ( 3q +1 )2
= 9q2 + 6q +1
= 3 (3q2 +2q ) + 1
= 3m +1 (where m = 3q2 + 2q )
Case III - a = 3q + 2
a2 = (3q +2 )2
= 9q2 + 12q + 4
= 9q2 +12q + 3 + 1
= 3 ( 3q2 + 4q + 1 ) + 1
= 3m + 1 where m = 3q2 + 4q + 1)
From all the above cases it is clear that square of any positive integer ( as in this case a2 ) is either of the form 3m or 3m +1.
The reason for taking b=3 is because of the question 3m and 3m+1 that is of the form a=bq+r
HOPE IT MAKES CLEAR.
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