Question

use euclid lemma to show that any square of any positive integer is of form 3p,3p+1​

Answer 1

Answer:

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Answer 2

Answer:

Let n be an positive integer

On dividing n by 3 ,let q be the quotient and r be the remainder.

Then,by Euclid division lemma ,we get

n=3q+r,where 0<r<3. (1)

n^2= 9q^2 + r^2+ 6qr

Case 1 . when r= 0

Putting r = 0 in 1 we get

n^2 = 9q^2 = 3(3q^2 ) = 3m, where m= 3q^2 is an integer.

Case 2.where r= 1

Putting r= 1 in (1) we get,

n^2=(9q^2+1+6q)

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

where m= (3q^2+ 2q) is an integer.

Case 3.Where r = 2

Putting r= 2 In (1)

n^2 =(9q^2+4+12q)

=3(3q^2+4q+1) +1

= 3m+1 where m= (3q^2+4q+1) is an integer.

Hence,the square of any positive is of the form 3m or 3m+1 for some integer.

Step-by-step explanation:

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