use euclid's division Lemma show that the square of any positive integer is of the form 3m or 3m+1 for some integer m
Answers
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.
To show that the square of any positive integer is of the form m or 3m + 1 for some integer m.
By Euclid's division lemma,
a = bq + r, 0 ≤ r ∠ b.
Let b be equal to 3.
Hence, a = 3q + r, 0 ≤ r ∠ b.
That is, r can be equal to 0, 1 or 2.
When r = 0,
a = 3q + 0 = 3q
a^2 = (3q)^2
a^2 = 9q^2
= 3(3q^2)
= 3m, where m = 3q^2.
When r = 1,
a = 3q + 1
a^2 = (3q + 1)^2
a^2 = 9q^2 + 6q + 1 + 1
a^2 = 3(3q^2 + 2q) + 1
= 3m + 1, where m = 3q^2 + 2q.
When r = 2,
a = 3q + 2
a^2 = (3q + 2)^2
a^2 = 9q^2 + 12q + 4
a^2 = 9q^2 + 12q + 3 + 1
a^2 = 3(3q^2 + 4q + 1) + 1
= 3m + 1, where m = 3q^2 + 4q + 1.
∴ The square of any positive integer is either of the form 3m or 3m + 1.