Integers as the sum of 3 integers.

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Homework Statement


Prove that every integer >17 can be written as the sum of 3 integers >1 that are pairwise relatively prime.

The Attempt at a Solution


I already proved the case for even integers. Now I am just working on the case for odd integers.
I know that it has to be the sum of 3 odd integers because it can't be the sum of 2 even and one odd. I started with 2x+1=x+(x+1) where x is a positive integer. now either x or x+1 is even and I guess I could break the even one up as a sum of 2 odd numbers and then go from their.
 
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The proof (as an explicit construction of those 3 integers) for odd integers can be done similar to the proof for even integers, just with more casework.
 
Did you finish the one about Prove that every integer >6 can be written as the sum of 2 integers >1 that are pairwise relatively prime? It is the same idea consider n/3-1,n/3,n/3+1 and adjust slightly in different cases.
 
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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