show that n square-1 is divisible by 8,if n is an odd positive integer.
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I hope someone will give a Solution by using Euclid's Division lemma !
A hint is :
-> Any odd square is of the form ( 4q + 1 )
And hence, when you simplify for n^2 - 1, you'll get your result !! However -_- I'm providing a more conceptual solution using the Density Property of Integers ^^"
Well, between any two consecutive multiples of four, the odd is on either extreme, which implies that either of ( n + 1 ) or ( n - 1 ) is divisible by 4 and the rest is even, i.e., divisible by 2 and hence, ( n^2 - 1 ) as a whole is divisible by 8
A hint is :
-> Any odd square is of the form ( 4q + 1 )
And hence, when you simplify for n^2 - 1, you'll get your result !! However -_- I'm providing a more conceptual solution using the Density Property of Integers ^^"
Well, between any two consecutive multiples of four, the odd is on either extreme, which implies that either of ( n + 1 ) or ( n - 1 ) is divisible by 4 and the rest is even, i.e., divisible by 2 and hence, ( n^2 - 1 ) as a whole is divisible by 8
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