Question

use euclid's division Lemma to show that the square of any positive integer is either of the form 3m or 3m+1, or 3m+4​

Answer 1

Let a = Any positive integer

b = 3

r = 0, 1, 2

According to Euclid's Division Lemma: a = bq + r

Squaring both sides:

If, a = 3q

=> a² = (3q)²

=> a² = 9q²

=> a² = 3(3q²)

=> a² = 3m

If, a = 3q + 1

=> a² = (3q + 1)²

=> a² = 9q² + 6q + 1

=> a² = 3(3q² + 2q) + 1

=> a² = 3m + 1

If, a = 3q + 2

=> a² = (3q + 2)²

=> a² = 9q² + 12q + 4

=> a² = 3(3q² + 4q) + 4

=> a² = 3m + 4

So, square of any positive integer i.e a² is in the form 3m, 3m + 1, 3m + 4

Answer 2

Answer:

let a is any positive integer in which b=3.

if a is divisible by b then there exists positive remainders.

a= bq+r

= 3q+ r

now,

a=3q........(1)

or, a=3q+1......(2)

or, a=3q+2..........(3)

now,

in eqn. (1)

a=3q

squaring both sides,

(a )square =(3 q) square

(a )square = (9 q) square

so,a = 3m

now,in equation (2),

a=3q+1

squaring on both sides

(a )square=(3 q +1) square

(a) square=9 q square + 6q+1

(a) square=3q(3q+2)+1

a=3m+1

in equation (3)...

a=3q+2

(a)square=(3q+2)square

(a)square=9q square+12 q +4

a=3(3q square + 4 q)+4

a=3m+4

Hence,we can write square of any positive integer in the form of 3m ,3m+1or 3m+4

one. thing i tell u i write square in letter but u can put square.

hope u understand.