Question

Show that any positive integer is of the form 3m or 3m+1 or 3m+2

Answer 1

$\huge{\underline{\mathfrak{\red {\ ANSWER;}}}}$

Let a be any odd positive integer.

We apply the division lemma with

a and b = 3 .

Since 0 ≤ r < 3 ,

the possible remainders are 0 , 1

and 2 .

That is , a can be 3q , or 3q + 1 , or

3q + 2 , where q is the quotient .

Now ,

a² = ( 3q )² = 9q²

Which can be written in the form

= 3 ( 3q² )

= 3m , [ since m = 3q² ]

It is divisible by 3 .

Again ,

a² = ( 3q + 1 )²

= 9q² + 6q + 1

= 3( 3q² + 2q ) + 1

= 3m + 1 [ since m = 3q² + 2q ,

3( 3q² + 2q ) is divisible by 3 ]

Lastly ,

a² = ( 3q + 2 )²

= 9q² + 12q + 4

= ( 9q² + 12q + 3 ) + 1

= 3( 3q² + 4q + 1 ) + 1

Which can be written in the form

3m + 1 , since

3( 3q² + 4q + 1 ) is divisible by 3.

Therefore ,

The square of any positive integer

is either of the form 3m or 3m + 1

for some integer m .

I hope this helps you.

: )