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!

  • $\begingroup$ If $z \in A$, what do you know about $\lvert z\rvert^2$? $\endgroup$ – Daniel Fischer Sep 4 '16 at 15:41

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 $u=a=±1$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.