Fix a natural number $n\ge2$. Can we find $n$ algebraic integers $a_1,\dots,a_n$ in the field of complex numbers such that $a_ia_j$ is a unit for all $i\ne j$?

$\begingroup$ Isn't $0$ the only nonunit, so that any distinct integers work? $\endgroup$– Peter TaylorCommented Jun 27 at 7:36

$\begingroup$ @PeterTaylor $2$ is not a unit in the ring of algebraic integers. Units are algebraic numbers whose monic minimal polynomial is integral with $\pm 1$ constant coefficient. $\endgroup$– YCorCommented Jun 27 at 7:40

$\begingroup$ I imagine that for each $n$ there exists a unit $u$ such that $a_i=u^i$ works. But I could check it only for $n=3$ (e.g., $u$ root of $X^3X+1$). $\endgroup$– YCorCommented Jun 27 at 7:57

$\begingroup$ @YCor, $x^2 + x  1$ works up to $n=3$; $x^4  x  1$ works up to $n=5$; $x^6  x^2  1$ works up to $n=7$. $\endgroup$– Peter TaylorCommented Jun 27 at 8:26

4$\begingroup$ Look up exceptional sequences and exceptional units for this question in a fixed number field. $\endgroup$– Chris WuthrichCommented Jun 27 at 9:01
2 Answers
Yes. Choose two large primes $p$ and $q$ with $q>p$ and set $z=\exp(2i\pi/(pq))$. Then $a_m=z^m$ for $0\le m<p$ will be such that $a_ia_j$ is a unit for $i\ne j$, with $0\le i,j<p$.

14$\begingroup$ For those, like me, who forgot why this is true: writing $\zeta_n$ for a primitive $n$th root of unity, $1\zeta_n$ is a unit in $\mathbb{Z}[\zeta_n]$ when $n$ has at least two distinct prime factors (Washington, Introduction to Cyclotomic Fields (1982) Springer GTM 83, proposition 2.8), so then by Galois action $1\zeta_n^i$ is a unit if $i$ is prime to $n$, and therefore $\zeta_n^i\zeta_n^j$ is a unit if $ij$ is prime to $n$. $\endgroup$– GroTsenCommented Jun 27 at 9:08
There is also a sequence of totally real examples (generalized golden ratios) which arise from the dynamics of the simplest quadratic rational $f(x)=x\frac{1}{x}$ whose inverse are $F_{\pm}(x)=\frac{x \pm \sqrt{x^2+4}}{2}$.
Let $\phi_0=1, \phi_{n+1}=F_+(\phi_n)=\frac{\phi_n+\sqrt{\phi_n^2+4}}{2}$, so $\phi_1$ is the golden ratio. $\phi_n$ are totally real algebraic units of degree $2^n$ and the defining irreducible polynomials are given by $p_0=x1, p_n(x)=x^{2^{n1}}p_{n1}(x\frac{1}{x}).$ One can show that
$Res(p_n,p_{nj}):=\prod_{p_n(\alpha)=0, p_{nj}(\beta)=0} (\alpha\beta)=\prod(\phi_n'\phi_{nj}'')=1.$
So all the $\phi_n'\phi_{nj}''$ are units, where $\phi_n' (\phi_{nj}'')$ are conjugates of $\phi_n(\phi_{nj})$.
The number fields $\mathbb{Q}(\phi_n) \subset \mathbb{Q}(\phi_{n+1})$ form a nested quadratic tower.

$\begingroup$ The prime $3$ (also $2$ and $17$) remains prime in $\mathbb{Q}(\phi_n)$ for all $n$, and $0,1,\phi_1, \ldots \phi_n$ form exceptional units. (in view of Myerson's comment above). $\endgroup$– CHUAKSCommented Jul 3 at 9:39

$\begingroup$ mathoverflow.net/users/112259/chuaks this sequence of totally real generalised golden ratios also more than answers my question and is very elegant. Is there any literature on this sequence that you can point me to. I would have ticked your answer as the answer had Henri Cohen not answered first, and it seems I cannot tick two answers. Strictly, Cohen's answer is a complete answer to what I asked but the advantage of your sequence is that as the desired length of sequence is increased so the sequence can be extended. This potentially has useful applications. $\endgroup$ Commented Jul 16 at 10:57

$\begingroup$ @Peter Kropholler This comes from my (unpublished) study on dynamics of the quadratic map $f(x)=x1/x$. I don't know where or whether it had been considered before. If $f(y)=y1/y=x$, then $y=x+1/y=x+\cfrac{1}{x+1/y}=x+\cfrac{1}{x+\cfrac{1}{x+...}}$, so the inverse $F_+(1)=\phi$ and $F_+(x)$ is the golden ratio map. Sine $f'(x)>1$, $f$ is expanding, so $\phi_{n+1}\phi_n$ is decreasing but $\phi_n \rightarrow \infty$. One can also generalized and look at inverse of $f(x)=x+a1/x, a>0$, then $\phi_n(a) \rightarrow 1/a$ and this has an interesting dynamical/cosmological interpretation. (ctd) $\endgroup$– CHUAKSCommented Jul 19 at 12:50