Question

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.

Answer 1

Step-by-step explanation:

in place of m there is q that's it

Answer 2

Answer:

✒️Let 'a' be any positive integer.

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

Such that

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

When, r = 0

a = 3q + 0 ---------- (1)

When, r = 1

a = 3q + 1 ----------- (2)

When r = 2

a = 3q + 2 ------------ (3)

So ,

(1) when, a = 3q + 0

a = 3q

By squaring both side

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

By taking common

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

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

(2) When, a = 3q + 1

By squaring both side

${a}^{2} \: = \: {(3q \: + \: 1) }^{2} \\ by \: using \: identity \: {(a \: + \: b \: )}^{2} \: = \: \\ {a}^{2} \: + \: 2ab \: + \: {b}^{2} \\ {a}^{2} \: = \: {9q}^{2} \: + \: 2 \: \times \: 3q \: \times \: 1 \: + \: {1}^{2} \\ {a}^{2} \: = \: {9q}^{2} \: + \: 6q \: + \: 1$

By taking common

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

(3) When, a = 3q + 2

By squaring both side

${a}^{2} \: = \: {(3q \: + \: 2)}^{2} \\ by \: using \: identity \: {(a \: + \: b \: ) }^{2} \\ = \: {a}^{2} \: + \: 2ab \: + \: {b}^{2} \\ {a}^{2} \: = \: {3q}^{2} \: + \: 2 \: \times \: 3q \: \times \: {2}^{2} \\ {a}^{2} \: = \: {9q}^{2} \: + \: {12q}^{2} \: + \: 4 \\ {a}^{2} \: = \: ( {9q}^{2} \: + 12q \: + \: 3 \: ) \: + \: 1$

By taking common

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

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