# Sets of algebraic integers whose differences are units

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 $$a_i-a_j$$ is a unit for all $$i\ne j$$?

• Isn't $0$ the only non-unit, so that any distinct integers work? Commented Jun 27 at 7:36
• @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.
– YCor
Commented Jun 27 at 7:40
• 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 $X^3-X+1$).
– YCor
Commented Jun 27 at 7:57
• @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 $n=7$. Commented Jun 27 at 8:26
• Look up exceptional sequences and exceptional units for this question in a fixed number field. Commented Jun 27 at 9:01

Yes. Choose two large primes $$p$$ and $$q$$ with $$q>p$$ and set $$z=\exp(2i\pi/(pq))$$. Then $$a_m=z^m$$ for $$0\le m will be such that $$a_i-a_j$$ is a unit for $$i\ne j$$, with $$0\le i,j.

• 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 $i-j$ is prime to $n$. Commented 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 $$p_0=x-1, p_n(x)=x^{2^{n-1}}p_{n-1}(x-\frac{1}{x}).$$ One can show that

$$Res(p_n,p_{n-j}):=\prod_{p_n(\alpha)=0, p_{n-j}(\beta)=0} (\alpha-\beta)=\prod(\phi_n'-\phi_{n-j}'')=1.$$

So all the $$\phi_n'-\phi_{n-j}''$$ are units, where $$\phi_n' (\phi_{n-j}'')$$ are conjugates of $$\phi_n(\phi_{n-j})$$.

The number fields $$\mathbb{Q}(\phi_n) \subset \mathbb{Q}(\phi_{n+1})$$ form a nested quadratic tower.

• 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). Commented Jul 3 at 9:39
• @Peter Kropholler This comes from my (unpublished) study on dynamics of the quadratic map $f(x)=x-1/x$. I don't know where or whether it had been considered before. If $f(y)=y-1/y=x$, 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 $f(x)=x+a-1/x, a>0$, then $\phi_n(a) \rightarrow 1/a$ and this has an interesting dynamical/cosmological interpretation. (ctd) Commented Jul 19 at 12:50