Question

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

Answer 1

✰︎✪︎ AnswEr ✪︎✰︎

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

By Euclid's division Lemma , a = bq + r , Where

$0 \leqslant r \leqslant b$

So, r = 0 , 1 , 2

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➪︎ First case :-

If r = 0 , then a = 3q + 0 = 3q

a^2 = (3q)^2

= 9q^2

= 3( 3q^2 )

a^2 = 3m [ m is some integer ]

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➪︎ Second case :-

If r = 1 , then a = 3q + 1

a^2 = (3q + 1)^2

= 9q^2 + 1 + 6q

= 3 ( 3q^2 + 2q ) + 1

a^2 = 3m + 1 [ m is some integer ]

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➪︎ Third case :-

Ir r = 2 , then a = 3q + 2

a^2 = ( 3q + 2 )^2

= 9q^2 + 4 + 12q

= 3 ( 3q ^2 + 4q ) + 4

a^2 = 3m + 4 [ m is some integer ]

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Therefore ,

The square of any positive integer is of the form 3m , 3m + 1 ( or ) 3m + 4.....

Answer 2

plz refer to this attachment