Question

use EDL to show that the square any positive integers is either of the form 3m, Or 3m+ 1 for some integer m

Answer 1

$\textsf{Euclid's division lemma states that, }\\ \\ \\ \Large \textsf{a = bq + r}\\ \\ \\ \large \textsf{where 0 \leq r < b.}$

In this question, we have to take b = 3.

Thus the equation becomes

a = 3q + r

This means that 'a' divided by 3 leaves the value of 'r' as the remainder.

So 'r' has the values of 0, 1 and 2.

(1) Let r = 0.

$a=3q+0\\ \\ a=3q\\ \\ \\ a^2=(3q)^2\\ \\ a^2=9q^2\\ \\ a^2=3 \times 3q^2\\ \\ a^2 = 3m$

Thus proved that a^2 is of the form 3m for any integer a, where m = 3q^2, for any integer q.

(2) Let r = 1.

$a=3q+1\\ \\ \\ a^2=(3q+1)^2\\ \\ a^2=9q^2+6q+1\\ \\ a^2=3(3q^2+2q)+1\\ \\ a^2=3m+1$

Thus proved that a^2 is of the form 3m + 1 for any integer a, where m = 3q^2 + 2q, for any integer q.

(3) Let r = 2.

$a=3q+2\\ \\ \\ a^2=(3q+2)^2\\ \\ a^2=9q^2+12q+4\\ \\ a^2=9q^2+12q+3+1\\ \\ a^2=3(3q^2+4q+1)+1\\ \\ a^2=3m+1$

Thus proved that a^2 is of the form 3m + 1 for any integer a, where m = 3q^2+4q+1, for any integer q.

OR

$a=3q+2\\ \\ a=3q+3-1\\ \\ a=3(q+1)-1\\ \\ a=3k-1\ \ \ \ \ [k=q+1]\\ \\ \\ a^2=(3k-1)^2\\ \\ a^2=9k^2-6k+1\\ \\ a^2=3(3k^2-2k)+1\\ \\ a^2=3m+1$

Thus proved that a^2 is of the form 3m + 1 for any integer a, where m = 3k^2 - 2k, for any integer k.

Hence proved!!!

Answer 2

$\Huge\orange{\boxed{\boxed{\underline{ANSWER}}}}$

Let " a" be any positive integer and take b = 3.

Applying Euclid's division lemma :

a = 3q + r , 0 ≤ r < 3

r = 0 or 1 or 2

I'ST :

When, a = 3q

          a² = ( 3q )²

               =  9q²

               = 3 ( 3 q )²

               =   3m { where m = 3q² }

2'ND :

When,

a  =  3q + 1

a²  = ( 3q + 1 )²                ( using identity : ( a + b )² = a²+ 2ab + b²)

     = ( 3q )² + 2 ( 3q ) ( 1 ) + 1²

     = 9q² + 6q +1

   

     =  3 ( 3q² + 2q ) + 1

     =   3 m + 1        { where m = 3q² + 2q }

3'RD :

When a = 3q + 2

          a²=  ( 3q + 2 )²      ( using identity : ( a + b )² = a²+ 2ab + b²)

              =  ( 3q )² + 2 ( 3q ) ( 2 ) + ( 2 )²

              =  9q² + 12q + 4

              = 9q² + 12q + 3 + 1

             = 3 ( 3q² + 4q + 1 ) + 1

             =  3 m + 1                    { where m = 3q² + 4q + 1 }



Hence, a is of the form 3m or 3 m + 1