Square of any positive integer is of the form

Asked by admin @ 20/08/2021 in Math viewed by 263 People

use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m

Answered by admin @ 20/08/2021

Let 'a' be any positive integer.

On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.

Such that ,

a = 3q + r , where r = 0 ,1 , 2

When, r = 0

∴ a = 3q

When, r = 1

∴ a = 3q + 1

When, r = 2

∴ a = 3q + 2

When , a = 3q

On squaring both the sides,

${a}^{2} = 9 {q}^{2} \\ {a}^{2} = 3 \times (3 {q}^{2} ) \\ {a}^{2} = 3 \\ where \: m = 3 {q}^{2}$

When, a = 3q + 1

On squaring both the sides ,

${a}^{2} = (3q + 1)^{2} \\ {a}^{2} = 9 {q}^{2} + 2 \times 3q \times 1 + {1}^{2} \\ {a}^{2} = 9 {q}^{2} + 6q + 1 \\ {a}^{2} = 3(3 {q}^{2} + 2q) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 2q$

When, a = 3q + 2

On squaring both the sides,

${a}^{2} = (3q + 2)^{2} \\ {a}^{2} = 3 {q}^{2} + 2 \times 3q \times 2 + {2}^{2} \\ {a}^{2} = 9 {q}^{2} + 12q + 4 \\ {a}^{2} = (9 {q}^{2} + 12q + 3) + 1 \\ {a}^{2} = 3(3 {q}^{2} + 4q + 1) + 1 \\ {a}^{2} = 3m + 1 \\ where \: m \: = 3 {q}^{2} + 4q + 1$

Therefore , the square of any positive integer is either of the form 3m or 3m+1.

Square of any positive integer is of the form

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