Question

show that the square of any two positive integers is of the form 3m or 3m+1 for some integer m

Answer 1

(3q)2 = 9q2 = 3(3q2)
where m = 3q2
(3q +1)2 = (3q)2 + (1)2 + 2 × 3q × 1
9q2 + 1 + 6q
9q2 +6q + 1
3(3q2+2q)+1
where m = 3q2 + 2q
= 3m + 1

Answer 2

HOLA

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Let N be an arbitary integer

On dividing N by 3 we get quotient M and remainder R

So we have ( 3m ) ( 3m + 1 )

On squaring we have

$( \: 3m \: ) {}^{2} = \: 9m \: clearly \: postive \\ \\ ( \: 3m \: + \: 1 \: ) {}^{2} = \: 9m \: + \: 1 \: clearly \: \: postive$
Hence any positive square integer can be of the form ( 3m ) ( 3m + 1 )

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HOPE U UNDERSTAND ❤❤❤