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

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SUMMARY

The discussion centers on proving that for any odd integer n, the expression 8 divides (n² - 1). The proof begins with the definition of an odd integer as n = 2q + 1. By expanding (2q + 1)² - 1, the result simplifies to 4q² + 4q. The key point is to demonstrate that q² + q is always even, which confirms that 4q² + 4q is divisible by 8. Thus, the conclusion is that 8 divides (n² - 1) for all odd integers n.

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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?
 
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Can you show q^2+q is divisible by 2 for any integer q? If so, then 4q^2+4q is divisible by 8.
 

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