Direct Proof With Odd Integers

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Homework Statement


If m is an odd integer and n divides m, then n is an odd integer.


Homework Equations


Odd integers can be written in the form m=2k+1.
Since n divides m, there exists an integer p such that m=np


The Attempt at a Solution


We will assume that m is an odd integer and that n divides m. We will show that n is an odd integer. Since m is an odd integer, there exists an integer k such that m=2k+1. Since n divides m, there exists an integer p such that m=np.

I don't know where to go from here to arrive at n is equal to 2*an integer +1. Do I need to use cases?
 
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so you have m = np and m is odd. you want to show n is odd. so suppose it's even, then look what happens, you should get a contradiction.
 
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...
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