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Let a=3b+r ( a, b are positive integers)
So r = 0,1,2,
Case 1:If r = 0
Then a = 3b
On squaring we get
a2 = (3b)2
a2 = 9b2
a2 = 3 × 3b2
a2 = 3m ( m=b² is a positive intger)
******************************************************
Case 2: r = 1
a = 3b + 1
On Squaring we get
a2 = (3b + 1)2
a2 = (3b)2 + 2 × (3b) × 1 + 1
a2 = 9b2 + 6b + 1
a2 = 3(3b2 + 2b) + 1
a2 = 3m + 1
(Here m = 3b2 + 2b is an integer)
******************************************
Case 3: r=2
Then a = 3b + 2
On quaring we get
a2 = (3b + 2)2
a2 = 9b2 + (2 × 3b × 2) +4
a2 = 9b2 + 12b + 3 + 1
a2 = 3(3b2 + 4b + 1) + 1
Here m = 3b2 + 4b + 1 is a positive integer
Hence we can conclude that square of any positive integer is either in the
form of 3m or 3m+1( m being a positive integer)
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