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# maths

Fri 25 May, 2012 05:35 am
show that n^-1 is divisible by 8,if n is a odd positive integer.
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fresco

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Fri 25 May, 2012 06:12 am
@anu98,
You presumably mean (n^2)-1 =(n+1)(n-1).
If n is odd then n+1 and n-1 are consecutive even numbers
so the expression can be re-written 2a (2a +2) or 4(a sqd +a) where a is any positive integer.
But (a sqd +a)=a(a+1) and either a or a+1 must be even giving another factor 2.
Hence the expression is divisible by 4 x 2 = 8

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fresco

1
Fri 25 May, 2012 12:59 pm
@anu98,
A slightly shorter proof is that any two consecutive even numbers (n-1)(n+1) must involve the product of factors 2 and 4 since if 2 is a factor of the first number then 4 will be a factor of the second, or vice versa. Hence the product has a factor of 2 x 4 = 8
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