Question

Use Euclid lemma to show that the square of any positive integer is either of the form of 3mor 3m +1 for some integer m​

Answer 1

Answer:

Step-by-step explanation:

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,

a2 = (3q)2

a2 = 3(3q)

a2= 3m

When, a = 3q + 1

On squaring both the sides ,

a2= (3q+1)2

a2 = 9q2 + 6q + 1

a2= 3(3q2 + 2q) +1

a2 = 3m + 1

When, a = 3q + 2

On squaring both the sides,

a2= (3q+2)2

a2 = 9q2 + 4 + 12q

a2= 3(3q2 + 4q) +4

a2 = 3m + 4  

 which is even, hence, cannot be positive...

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

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

Hope this attachment helps you.