Proving Divisibility of (n^2-1) for Odd Integers n

  • Thread starter Thread starter scott_bruenin
  • Start date Start date
  • Tags Tags
    Proof
scott_bruenin
Messages
1
Reaction score
0

Homework Statement


For each integer n, if n is odd then 8\left| (n^{2}-1)


Homework Equations


Def of an odd number 2q+1


The Attempt at a Solution



(2q+1)^{2} -1
4q^{2} +4q+1-1
4q^{2} +4q
Here is where I get stuck... should I factor out the 4 and say that q^{2} +q is an integer and therefore can be wrote as some integer r and therefore 8\left| 4r?
 
Physics news on Phys.org
Can you show q^2+q is divisible by 2 for any integer q? If so, then 4q^2+4q is divisible by 8.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top