Show that the square of any …
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Yogita Ingle 6 years, 3 months ago
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 .
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