Irrational numbers $z_0,z_1$ such that $\sqrt{z_0^2+z_1^2}$ is an integer

Question

Question: If $z_0$ and $z_1$ are real irrational numbers I write $$q=z_0+z_1\sqrt{-1}$$ Surely $q$ is just a complex number. Under what condition will the number $|q|$ be an $\color{blue}{integer}$ ?

I have that $$|q| = \sqrt{z_0^2+z_1^2}$$ Consequently the only exceptions I can think of are the cases where $z_0^2+z_1^2=x^2$ for some positive integer $x$. For example $z_0=z_1=\sqrt{2}$; which is irrational. Then $$|q| = \sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=\sqrt{2+2}=\sqrt{4}=2$$ I denote the set of exceptional pairs $(z_0,z_1)$ by $\mathcal{E}$. I have shown that $(\sqrt{2},\sqrt{2}) \in \mathcal{E}.$ What else is in $\mathcal{E}?$

Answer 1

You’re precisely correct that elements of $\mathcal{E}$ satisfy $z_1^2+z_2^2=x^2$ for some $x\in\mathbb Z$. One way to generate such numbers is to generalize what you have done: take some integer $k$ and look at $(k\sqrt{2},k\sqrt{2})$. This gives us $2k^2+2k^2=4k^2=(2k)^2$.

However, there are other numbers in your set. Let $x$ be irrational and let $a$ be a integer. Then $\bigg(\sqrt{\frac{1}{2}a^2+x},\sqrt{\frac{1}{2}a^2-x}\bigg)$ works.

We can push further though. If $a$ is an integer and $p+q=k^2$ for some $k\in \mathbb Z$ the. We can replace the $\frac{1}{2}$ in the previous example with $p$ on one side and $q$ on the other.

This still doesn’t exhaust everything though, and I don’t think that there is a good characterization of these numbers beyond something obvious equivalent to your definition, such as “$(a,b)$ is in the set if $a$ and $b$ are the square roots of numbers $c$ and $d$ that sum to a perfect square.”

Answer 2

Following your intuition, you can get the following members for $\mathcal{E}$:

For $n\in\Bbb{N}$, write $n^2$ as $n^2=x+y$, where $x$ and $y$ are not perfect squares. Then $(\sqrt{x},\sqrt{y})\in\mathcal{E}$.

But obviously these doesn't exhaust $\mathcal{E}$.

Answer 3

Suppose $\sqrt{z_0^2+z_1^2}$ is a natural number $n$. Then $(z_0,z_1)$ corresponds to points the unit circle, namely $(\pm z_0/n,\pm z_1/n)$ and vice versa.

From here, the set of points $(z_0,z_1)$ in questions consists exactly of all points of the unit circle with both irrational coordinates, scaled by the square of an integer, that is $$\mathcal E=\{n^2q\mid q\in S^1\text{ where $q$ has irrational coordinates and $n$ is a positive natural number}\}.$$