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Given the set $A = \{a + ib\sqrt{3} \mid a, b \in \mathbb{Z}\}$ and $u, v \in A$ with $uv = 5 + 2i\sqrt{3}$, prove that one of $u$ and $v$ is $1$ or $-1$.

First, lets represent $u$ and $v$ as follows: $u = a + ib\sqrt{3}$ and $v = c + id\sqrt{3}$, where $a, b, c, d \in \mathbb{Z}$.

By doing some basic operations on what we are given, I got the following $abcd = 0$ and $(a^2 - 3b^2)(c^2 - 3d^2) = 13$. Because $a, b, c, d$ are integers, we only have four possibilities for the last multiplication: $1 \cdot 13, 13 \cdot 1, -1 \cdot (-13), -13 \cdot (-1)$. Now, we need to consider each of these cases and in every case we also need to deal with $abcd = 0$ by setting each of the numbers $a, b, c, d$ to $0$ separately.

After doing what I explained above, I first got $b = 0$ and $a = \pm 1$ and then $d = 0$ and $c = \pm 1$.

However, this solution seems to be a little bit too long, and I would like to have a more direct solution. So, if you have any ideas, please share them!

Thank you!

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  • $\begingroup$ If $z \in A$, what do you know about $\lvert z\rvert^2$? $\endgroup$ – Daniel Fischer Sep 4 '16 at 15:41
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Let $N(a+bi\sqrt 3) := a^2+3b^2$. Then you can check that $N(zz')=N(z)N(z')$ for any $z,z' \in A$. This is just because $N(z)=|z|^2$.

Since $uv=5 + 2i\sqrt{3}$, you get $N(uv)=N(u)N(v)=N(5 + 2i\sqrt{3})=5^2+4\cdot 3=37$. Since $N(u)$and $N(v)$ are positive integers, one of them must be $1$ because $37$ is prime.

Here is the end:

Let's say $N(u)=1=a^2+3b^2$. It follows that $b=0$ because it is an integer, so that $u=a=±1$.

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  • $\begingroup$ This is a very simple and nice solution! Thank you very much! $\endgroup$ – George R. Sep 4 '16 at 15:50
  • $\begingroup$ @GeorgeR. You're welcome! $\endgroup$ – Watson Sep 4 '16 at 15:50
  • $\begingroup$ I like this. But wouldn't it be more illuminating to a student if instead of implying N(u) was a function made up for the purpose of the exercise, you point out that N(u) is simply $|z|^2$ and a ready tool in ones arsinal? $\endgroup$ – fleablood Sep 4 '16 at 15:52
  • $\begingroup$ @fleablood: I agree with you, but I wrote "This is just because $N(z)=|z|^2$." $\endgroup$ – Watson Sep 4 '16 at 15:55
  • $\begingroup$ Oh, did I miss that? My "proof-reading" eyes are doing a very poor job this morning. $\endgroup$ – fleablood Sep 4 '16 at 16:01

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