Division algorithm and unique Gaussian integers

Proggy99
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Homework Statement


Theorem
Let \alpha\neq0 and \beta be Gaussian integers. Then there are Gaussian integers \tau and \rho such that \beta=\tau\alpha+\rho and N\left(\rho\right)<N\left(\alpha\right)

Problem
Show that the Guassian integers \tau and \rho in the Theorem are unique if and only if \beta is a multiple of \alpha

Homework Equations





The Attempt at a Solution


Can someone please give me a jumping off point for this question because I am not sure how to proceed? I used the theorem in previous problems to actually solve for \tau and \rho given \beta and \alpha, but I am not sure how to go about this problem. Thanks
 
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Proggy99 said:

Homework Statement


Theorem
Let \alpha\neq0 and \beta be Gaussian integers. Then there are Gaussian integers \tau and \rho such that \beta=\tau\alpha+\rho and N\left(\rho\right)<N\left(\alpha\right)

Problem
Show that the Guassian integers \tau and \rho in the Theorem are unique if and only if \beta is a multiple of \alpha

Homework Equations





The Attempt at a Solution


Can someone please give me a jumping off point for this question because I am not sure how to proceed? I used the theorem in previous problems to actually solve for \tau and \rho given \beta and \alpha, but I am not sure how to go about this problem. Thanks

I did some poking around trying to find hints but can not seem to find any beyond proving uniqueness for the division algorith for real numbers. I can not seem to find a way to apply a similar method here. Help please.
 
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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