# Use euclids division lemma to show that square of any +ve integer is either of form 3m or 3m+1 for some integer m.. plzz guys tell me...

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by aayushghore

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by aayushghore

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X² is either 3 m or 3 m + 1 where X >= 1 and m >= 1

let z>=1

1) IF 3 | X X => 3 | X => X = 3 z X² = 3 * (3*z²)

THEN**X² is of form 3m**

2) IF 3 does not divide X X => 3 does not divide X => X = 3 z+1 or 3z+2

(i) X² = (3z+1)² = 3 (3z²+2z) + 1** of the form 3 m + 1**

(ii) X² = (3z+2)² = 3(3z²+4z) + 4 = 3(3z²+4z+1) + 1

** it is in form 3 m + 1**

Hence proved.

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any natural number, a square or otherwise is in form : 3 m , 3 m + 1, 3 m + 2

But if it is a square then it is in form : 3 m or 3m+1 only.

let z>=1

1) IF 3 | X X => 3 | X => X = 3 z X² = 3 * (3*z²)

THEN

2) IF 3 does not divide X X => 3 does not divide X => X = 3 z+1 or 3z+2

(i) X² = (3z+1)² = 3 (3z²+2z) + 1

(ii) X² = (3z+2)² = 3(3z²+4z) + 4 = 3(3z²+4z+1) + 1

Hence proved.

=====================================

any natural number, a square or otherwise is in form : 3 m , 3 m + 1, 3 m + 2

But if it is a square then it is in form : 3 m or 3m+1 only.