SOLUTION: I am struggling with this question, can someone help? thanks so much. Select any odd integer greater than 1, square it, and then subtract 1. Is the result evenly divisible by 8

Question 130501: I am struggling with this question, can someone help?
thanks so much.
Select any odd integer greater than 1, square it, and then subtract 1. Is the result evenly divisible by 8? Prove that this procedure always produces a number divisible by 8. (Suggestion: Any odd integer greater than 1 can be expressed as 2n + 1, where n is a natural number.)
Answer by amine(11) (Show Source): You can put this solution on YOUR website!
a odd number is x=2n+1
then x*x=(2n+1)(2n+1)=4n^2 + 4n + 1
then x^2-1=4n^2 + 4n
=4n(n+1)
the n will be a odd or a pair number
1)now if the n is a odd number then n=2m+1 will m>=0 then n+1=2m+2 ==>
n+1= 2(m+1)
with this
x^2-1=4n(n+1)
=4n(2(m+1)) = 8n(m+1) then divisible by 8
2) if the n is a pair number then n=2m
x^2-1=4n(n+1)=8m(n+1)
then divible by 8
cheers
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