Question

4. Use Euclid's division lemma to show that the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m.
Hint : Let x be any positive integer then it is of the form 3q, 39 + 1 or 39+2. Now square
each of these and show that they can be rewritten in the form 3m or 3m + 1.1
interis of the form​

Answer 1

Answer:

Step-by-step explanation:

let x = 3q

then. x2=3q2

=9q2= 3(3q2)

in the form 3m where m is 3q2

now continue this until you get all 3m, 3m+1

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.