Question

use Euclid's division lemma to show that the square of any positive integer is of the form 3p,3p+1.​

Answer 1

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$\large{\red{\underline \bold{\tt{To \:Prove :-}}}}$

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• Square of any positive integer is of the form 3p , 3p + 1

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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As per Euclid's Division Lemma

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➠ a = bq + r

Where ,

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0 ≤ r < b

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• Let "a" be a positive integer
• b = 3

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Hence a = 3q + r

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Where ,

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0 ≤ r < 3

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∴ r is an integer greater than or equal to 0 and less than 3

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Hence r can be either 0 , 1 or 2

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Case 1

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➠ Let r = 0

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➜ 3q + r

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➜ a = 3q

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⟮ Squaring both the side ⟯

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➜ a² = (3q)²

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➜ a² = 9q²

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➜ a² = 3 × 3q²

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➨ a² = 3m

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Where ,

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➠ p = 3q²

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Case 2

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➠ Let r = 1

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➜ a = 3q + r

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➜ a = 3q+ 1

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⟮ Squaring on both the side we get ⟯

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➜ a² = (3q + 1)²

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➜ a² = (3q)² + 1 + 2 × (3q) × 1

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➜ a² = 9q² + 6q+ 1

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➜ a² = 3(3q² + 2q) + 1

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➜ a² = 3p + 1

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Where ,

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➠ p = 3q² + 2q

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Case 3

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➠ Let r = 2

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➜ a = 3q + r

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➜ a = 3q + 2

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⟮ Squaring on both the sides we get ⟯

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➜ a² = (3q + 2)²

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➜ a² = 9q² + 4 + (2 × 3q × 2)

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➜ a² = 9q ² + 12q + 3 + 1

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➜ a² = 3(3q² + 4q + 1) + 1

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➜ a² = 3p + 1

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Where ,

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➠ p = 3q² + 4q + 1

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∴ Square of any positive integer is of the form 3p or 3p + 1

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Hence proved

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Answer 2

Step-by-step explanation:

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