Question

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.​

Answer 1

Step-by-step explanation:

let n be an arbitrary integer

on dividing n by 3 let q be the quotient and r be the remainder

then by Euclid's lemma we have

n= 3q + r where 0 is greater tha or equal to r which is smaller than 3

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

case 1 when r = 0

n^2= 9q^2

= 3(3q^2) let 3q^2 be m

thus = 3m

case 2 when r = 1

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

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

let 3q^2 + 2q be m

thus

n^2 = 3m + 1

case 3 when r= 2

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

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

let 3q^2 +4q+1 be = m

thus

n^2= 3 m

hence proved

please mark it as the brainliest

Answer 2

❤️$\huge \star \red{ Solution\:..!! }$❤️

Let 'a' be any positive integer.

✨✨On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.✨✨

Such that ,

a = 3q + r , where r = 0 ,1 , 2

When, r = 0

∴ a = 3q

When, r = 1

∴ a = 3q + 1

When, r = 2

∴ a = 3q + 2

When , a = 3q

On squaring both the sides,

When, a = 3q + 1

On squaring both the sides ,

When, a = 3q + 2

On squaring both the sides,

✨✨Therefore , the square of any positive integer is either of the form 3m or 3m+1.✨✨