Question

Question :
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➡️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.
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Answer 1

Answer:

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To prove

• The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Proof

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

• Equation (1) becomes

• a = 3b

On squaring both the side

• a^2 = (3b)^2

• a^2 = 9b^2

• a^2 = 3 × 3b^2

• a^2 = 3m

• Where m = 3b^2

Case 2: Let r = 1

• Equation (1) becomes

• a = 3b + 1

Squaring on both the side we get

• a^2 = (3b + 1)^2

• a^2 = (3b)^2 + 1 + (2 × (3b) × 1)

• a^2 = 9b^2 + 6b + 1

• a^2 = 3(3b^2 + 2b) + 1

• a^2 = 3m + 1

• Where m = 3b^2 + 2b

Case 3: Let r = 2

• Equation (1) becomes

• a = 3b + 2

Squaring on both the sides we get

• a^2 = (3b + 2)^2

• a^2 = 9b^2 + 4 + (2 × 3b × 2)

• a^2 = 9b^2 + 12b + 4

• a^2 = 9b^2 + 12b + 3 + 1

• a^2 = 3(3b^2 + 4b + 1) + 1

• a^2 = 3m + 1

• where m = 3b^2 + 4b + 1

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

Hence proved.

Done ✅

Answer 2

Answer:

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