Question

use euclid's division Lemma to show that square of any positive integer is either of the form 3m or 3m+1

Answer 1

Hey!!!

Good Afternoon

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Let a be any positive integer.

let b = 3(Divisor)

Thus by Euclid's Division Lemma,

=> a = 3q + r where
$0 \leqslant r < b$

Thus r = 0,1 and 2

Case 1, r = 0

=> a = 3q

Square both side

=> a² = 9q²

=> a² = 3m where m = 3q²
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Case 2, r = 1

=> a = (3q + 1)

Square both sides

=> a² = (3q + 1)²

=> a² = 9q² + 1 + 6q

=> a² = 3m + 1 where m = 3q² + 2q
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Case 3, r = 2

=> a = 3q + 2

=> a² = (3q + 2)²

=> a² = 9q² + 4 + 12q

=> a² = 9q² + 12q + 3 + 1

=> a² = 3m + 1 where m = 3q² + 4q + 1
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Thus from the above three cases

=> a = 3m or 3m + 1

HENCE SHOWN

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Hope this helps ✌️

Answer 2

Hey friend ..!!! here's your answer
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Let a be any positive integer . b= 3, By Euclid Alogorithm a = 3q + r

${(3m)}^{2} = {9m}^{2} \\ 3 \times {3m }^{2} \\ 3m \\ he nce \: proved \\ \\ {(3m + 1)}^{2} = {9m }^{2} + 1 + 6m \\ {9m }^{2} + 6m + 1 \\ {9m}^{2} + 3m + 3m + 1 \\ 3(3m^2 + m + m) + 1 \\ 3m + 1 \\ \\ hence \: proved$
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#hope its help you dear#
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