Question

show that the square of any positive integer is of the form 3m or 3m +1 for some integer m.​

Answer 1

$\huge\boxed{\fcolorbox{black}{pink}{Solution}}$

Let us consider a positive integer a

Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that

a = 3b + r……………………………(1)

where r = 0,1,2,3…..

Case 1: Consider r = 0

Therefore , a = 3b

On squaring both the side

${a}^{2} = {3b}^{2}$

${a}^{2} = {9b}^{2}$

${a}^{2} = 3 \times {3b}^{2}$

${a}^{2} = 3m$

$where \: m \: = \: 3 {b}^{2}$

Case 2: Let r = 1

Therefore , a = 3b + 1

Squaring on both the side we get

${a}^{2} = \: ({3b + 1})^{2}$

${a}^{2} = ({3b})^{2} + 1 + 2 \times 3b \times 1$

${a}^{2} = 9 {b}^{2} + 6b + 1$

${a}^{2} = 3( {3b}^{2} + 2b) + 1$

${a}^{2} = 3m + 1$

$Where \: m = 3 {b}^{2} + 2b$

Case 3: Let r = 2

Therefore , a = 3b + 2

Squaring on both the sides we get

${a}^{2} = ( {3b + 2})^{2}$

${a}^{2} = 9 {b}^{2} + 4 + (2 \times 3b \times 2)$

${a}^{2} = 9 {b}^{2} + 4 + 12b$

${a}^{2} = 9 {b}^{2} + 12b + 3 + 1$

${a}^{2} = 3( {3b}^{2} + 4b + 1) + 1$

${a}^{2} = 3m + 1$

$where \: m = 3 {b}^{2} + 4b + 1$

∴ square of any positive integer is of the form 3m or 3m+1.

Hence proved.

Answer 2

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