Proving a subgroup is equivalent to Z

Gale
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Homework Statement



for n \in N, n \geq 1 Prove that (n^{3} +2n)Z + (n^{4}+3n^{2}+1)Z= Z

Homework Equations


I know subgroups of Z are of the form aZ for some a in Z and also that aZ+bZ= dZ, where d=gcd(a,b)

The Attempt at a Solution



So I was thinking if I could prove that the gcd of (n^3+2n) and (n^4+3n^2+1) was 1, then I could make the proof, but I'm struggling to figure out how to find a gcd of two polynomials... I also tried factoring to see if that led anywhere, but it didn't really...

Then I was thinking that if I could show that 1 was in the group, and since 1 generates Z, that would prove that the group was equivalent to Z... but then I wasn't actually sure that logic was sound.

Any help or some guidance in the right direction would be appreciated. Thanks!
 
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Your first idea is correct. Do you know the Euclidean algorithm and polynomial division? So that you can check your work, you should find that

-(n^3+2n)(n^3+2n) + (1+n^2)(n^4+3n^2+1) = 1
 
Thank you! For some reason I didn't think polynomial long division was the way to go... I'm always so unconfident when I do these sorts of proofs.
 
There is something to be said for confidence (or perhaps intuition?). Unfortunately, it seems like a lot of the time the only way to find that confidence/intuition is to fail/succeed at a tonne of questions.

Also, something you may notice is that (like this question) there are always many ways to attack a problem. We often cannot determine which way is correct until we have followed a path through to its conclusion and found that it is a dead end. Luckily, even dead-ends often provide insight and understanding into the structure of a problem: Even through failure, we are constantly learning.
 
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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