Question

use euclid's division Lemma to show that the square of any positive integer is either of the form 3M 3M + 1 for some integer m.​

Answer 1

Answer:

9m1 this questions answered

Answer 2

$\huge\sf\gray{To\;Prove}$

✭ The square of any positive integer is of the form 3m or 3m+1

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$\huge\sf\purple{Steps}$

Let ' a' be any positive integer and b = 3

We know, a = bq + r where 0 <  r< b

Now, a = 3q + r , 0<r < 3

So the remainder = 0,1 or 2

Case I

➝ a = 3q

➝ a² = 9q² .

➝ 3 x ( 3q²)

➝ 3m (where m = 3q²)

Case II

»» a = 3q +1

»» a² = ( 3q +1 )²

»» 9q² + 6q +1

»» 3 (3q² +2q ) + 1

»» 3m +1 (where m = 3q² + 2q )

Case III

➢ a = 3q + 2

➢ a² = (3q +2 )²

➢ 9q² + 12q + 4

➢ 9q² +12q + 3 + 1

➢ 3 (3q² + 4q + 1 ) + 1

➢ 3m + 1 ( where m = 3q² + 4q + 1)

$\large\sf Hence\:Proved!!$

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