Question

use Euclids division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for the some integer m .

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

Answer 2

let a be any positive int.

and b = 3

we know D = dq + r

a = bq + r ......(1)

possible r (remainder) = 0 1 2

by putting the value of r ....

a = 3q + 0

= 3q .....(2)

a = 3q + 1 ..(3)

a = 3q + 2......(4)

by squaring eqn 2 3 and 4

by eqn 2 ...

a = (3q)^2

= 9 q ^2

= 3 ( 3q^2) { let () = m }

= 3m

by eqn 3 ...

a = (3q + 1) ^2 ... by identity ( a+ b ) ^2

= 9q ^ 2 + 6q + 1

=3(3 q^2 + 2q) + 1 { let () =m }

= 3m + 1

by eqn 4 ...

a = ( 3q +2 ) ^2

a = 9q ^2 + 12 q + 4

a = 9q ^2 + 12 q + 3 + 1

a = 3(3q ^2 + 4q + 3) +1 { let ()= m}

a = 3m+ 1

as by eqn 2 3 and 4 we got 3q and 3q+ 1 so therefore square of any positive integers in the firm of 3q or 3q + 1

HOPE ITS HELPFUL PLS MARK AS BRAINLIEST ....