Question

use of euclid division lemma show square of any positive integer 3m or 3m+1​

Answer 1

HERE'S THE ANSWER,

Let x be any positive integer 3q, 3q+ 1,3q + 2,

CASE 1:-

= 3q

= (3q)^ [squaring]

= 9q^

= 3(3q^)

= 3m, where m = 3q^

CASE 2:-

= 3q+1

= (3q+1)^ [squaring]

= 9q^ +2 ×3q ×1 +1

= 9q^ +6q +1

= 3(3q^+2q) +1

= 3m+1, where m = 3q^+2q

CASE 3:-

= 3q+2

= (3q+2)^ [squaring]

= (3q)^ +2 ×3q ×2 +(2)^

= 9q^ +12q +4

= 3(3q^+4q+1) +1

= 3m+1,where m = 3q^+4q+1

therefore, these are the square of any positive integer of the form 3m or 3m + 1.

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Answer 2

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

=  9q² + 6q +1

= 3 (3q² +2q ) + 1

= 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

= 9q² + 12q + 4

= 9q² +12q + 3 + 1

= 3 (3q² + 4q + 1 ) + 1

= 3m + 1 ( where m = 3q² + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.

Hence, it is solved .