Question

Using euclids division lemma show that square of any positive integer is either of form 3m or 3m + 1

Answer 1

let us take, 'x'= 3q , 3q+1, 3q+2
when, x=3q
        x2 =  (3q) 2
         x2 = 9q2 
        x2  = 3(3q2)
we see that 3q2= m
so we have done the first equation 3m

when , x=3q+1
           x2= (3q+1)2
                                 [since, (a+b)2 = a2+2ab+b2]
           x2= 9q+6q+1
           x2= 3(3q+2q)+1
in this we see that 3q+2q= m
    therefore, this satisfy the equation m+1
hence, we have done this question

Answer 2

HI

Let a be any positive integer.

Therefore, a can be 3q or 3q+1 or 3q+2

where b = 3 and r = 0,1,2

(3q)² = 9q²

= 3(3q)²

= 3m ➡️ m = 3q²

(3q+1)² = (3q)² + 2(3q)(1) + (1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1 ➡️m = 3q² + 2q

(3q+2)² = (3q)² + 2(3q)(2) + (2)²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1 ➡️m = 3q² + 4q + 1

Therefore, the square of any positive integer is of the form 3m or 3m+1.

Hope it proved to be beneficial....