Prove that the square of any positive integer is of the form 3m or 3m

1.	Prove that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.
Answer» Let any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2. (From Euclid’s division lemma for b= 3) If n= 3q, Then, on squaring ⇒ n²= (3q)² = 9q² ⇒ n²= 3(3q²) ⇒ n²= 3m, where m is some integer [m = 3q²] If n= 3q+1, Then, on squaring ⇒ n²= (3q+1)² = 9q² + 6q + 1 ⇒ n²= 3(3q² +2q) + 1 ⇒ n²= 3m + 1, where m is some integer [m = 3q² +2q] If n= 3q+2, Then, on squaring ⇒ n²= (3q+2)² = 9q² + 12q + 4 ⇒ n²= 3(3q² + 4q + 1) + 1 ⇒ n²= 3m, where m is some integer [m = 3q² + 4q + 1] Thus, it is observed that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Prove that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Answer»

Let any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2. (From Euclid’s division lemma for b= 3)

If n= 3q,

Then, on squaring

⇒ n²= (3q)² = 9q²

⇒ n²= 3(3q²)

⇒ n²= 3m, where m is some integer [m = 3q²]

If n= 3q+1,

Then, on squaring

⇒ n²= (3q+1)² = 9q² + 6q + 1

⇒ n²= 3(3q² +2q) + 1

⇒ n²= 3m + 1, where m is some integer [m = 3q² +2q]

If n= 3q+2,

Then, on squaring

⇒ n²= (3q+2)² = 9q² + 12q + 4

⇒ n²= 3(3q² + 4q + 1) + 1

⇒ n²= 3m, where m is some integer [m = 3q² + 4q + 1]

Thus, it is observed that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.