Show that for all integers n>2, n does not divide n^2+2

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Homework Statement


Show that for all integers n>2, n does not divide n^2+2.

2. The attempt at a solution
I believe this solution can be solved by induction, I just don't know how to phrase it recursively.

For all n>2, n^2+2 mod n ≠ 0

Base case n=3
3^2 + 2 =11
11 mod 3 = 2 ≠ 0.

Show that
(n-1)^2 +2 mod n-1 ≠ 0
 
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I don't think induction is the right way to do this. n divides n2, can you use that in this problem?
 
I can use that, but I'm not sure how it would be done. Intuitively it makes sense, but is there an axiom that I could use that proves this?
 
Suppose n does divide n2+ 2. Then there exist integer m such that mn= n2+ 2. Now divide both sides by n.
 
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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