Question

prove that the square of any positive integer is of the form 3m or,3m+1 but not of the form 3m+2​

Answer 1

SOLUTION :  

Since positive integer n is of the form of 3q , 3q + 1 and 3q + 2

Case : 1

If n = 3q,then

n² = (3q)²

[On squaring both sides]

n² = 9q²

n²= 3 (3q)²

n² = 3m , where m = 3q²

Case : 2

If n = 3q + 1,then

n² = (3q + 1)²

[On squaring both sides]

n² = (3q)² + 6q + 1²

[(a+b)² = a² + b² + 2ab]

n² = 9q² + 6q + 1

n² = 3q (3q + 2) + 1

n² = 3m +1 , where m = q(3q + 2)

Case : 3

If n = 3q + 2, then

n² = (3q + 2)²

[On squaring both sides]

n² = (3q)² + 12q + 4

[(a+b)² = a² + b² + 2ab]

n² = 9q² + 12q + 4

n² = 3 (3q² + 4q + 1) +1

n² = 3m + 1 , where q = 3q² + 4q + 1

Hence, n²  is of the form 3m, 3m + 1 but not of the form 3m +2.

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Answer 2

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Refer to the attachment