Question

use Euclid's division lemma to show that square of any positive integer is either of the form 3n or 3n+1 for some integer 'n'.​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

Euclid division lemma fromula is, a=bq+r

let consider that , that must be r<3, r={0,1,2}

consider that b=3

let r=0

now from given question,we need to do square of positive integer.

a=bq+r

a=(3n+0)²        {(a+b)²=a²+b²+2ab}

by using (a+b)² formula.

=3n²+0²+2×3×0

=9n+0+6×0

=9n+0+0

=3(3n).....consider 3n as n

so, it is in the form of 3n.

now,let consider that b-=3

r=1

a=(3q+1)²

a=3q²+1²+2×3q×1

a=9q+1+6q

a=3(3q+2q)+1.......consider that 3q+2q as 3n

so, it become 3n+1

now let, b=3

r=2

a=(3q+2)²

a= (3q)²+(2)²+2×3×2

a= 9q+4+12

a= 3(3q+1+4)+1........consider 3q+1+4 as n

now, it will be consider as n

now it resulted will be 3n+1

Hence it proved