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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.

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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.

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