Use Euclid’s division lemma to show that the square of any positive

1.	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» Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b Taking b = 3, we get a = 3q + r; where 0 ≤ r < 3 When, r = 0 = ⇒ a = 3q When, r = 1 = ⇒ a = 3q + 1 When, r = 2 = ⇒ a = 3q + 2 Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m. ⇒ Squares of 3q = (3q)² = 9q² = 3(3q)² = 3 m where m is some integer. Square of 3q + 1 = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m +1, where m is some integer Square of 3q + 2 = (3q + 2)² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1 for some integer m. ∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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»

Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b

Taking b = 3, we get

a = 3q + r; where 0 ≤ r < 3

When, r = 0 = ⇒ a = 3q

When, r = 1 = ⇒ a = 3q + 1

When, r = 2 = ⇒ a = 3q + 2

Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m.

⇒ Squares of 3q = (3q)²

= 9q² = 3(3q)² = 3 m where m is some integer.

Square of 3q + 1 = (3q + 1)²

= 9q² + 6q + 1 = 3(3q² + 2q) + 1

= 3m +1, where m is some integer

Square of 3q + 2 = (3q + 2)²

= (3q + 2)²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1 for some integer m.

∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.