My Math Forum ring of integers problem

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February 11th, 2014, 12:26 AM   #1
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ring of integers problem

hi, i tried to solve the attachment problem. i tried to calculate the norm of u and its get complicate.

thank you for the help
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 February 19th, 2014, 04:04 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: ring of integers problem I don't see how the "norm" of u has anything to do with this. You are told that u is a unit which means it is invertible. What is the multiplicative inverse of u? What must be true of a and b for that to exist?
 February 22nd, 2014, 11:15 AM #3 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: ring of integers problem I'm not sure where YOU are going with this. As is easy to verify, the multiplicative inverse of u is: $\frac{1}{a^2 - b^2d}(a - b\sqrt{d})$ So clearly, we require that: $\frac{a}{a^2 - b^2d},\ \frac{b}{a^2-b^2d}$ both be integers. But arguing from the norm: $N(a+b\sqrt{d})= a^2 - b^2d$ since this norm is multiplicative, it is clear that N(u) is a unit in the ring of integers, that is to say N(u) = 1 or -1. This is a much stronger condition on the coefficients than them merely being integers, so I fail to see how your observation is pertinent. Arguing mod 4, it turns out that N(u) = 1 if exactly one of a or b is even, and d = 3 (mod 4). It also turns out if N(u) = -1, then both a and b are odd, and d = 2 (mod 4). Unfortunately, I haven't been able to make further progress on this.

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