# For any $n$, does there exist a number field with at least $n$ solutions to the unit equation

Let $n$ be a positive integer.

Does there exist a number field $K$ such that the number of solutions of the unit equation $$a+b =1, \quad a,b\in O_{K}^\ast$$ is at least $n$? Can we write down such a number field explicitly?

I know that the number of solutions is always finite in a fixed number field.

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– user2035 Oct 9 '11 at 14:14
For arbitrary $n$, you can find $m$ odd with $\phi(m)\geq n$. Then, $a=\zeta_m+1$, $b=-\zeta_m$ gives $\phi(m)\geq n$ solutions for $K=\mathbf Q(\zeta_m)$. – user2035 Oct 9 '11 at 15:06
Got it! That's nicer than what I'd come up with. – David Speyer Oct 9 '11 at 15:17
That is pretty nice indeed. If you post it as an answer I could accept it. – Taicho Oct 9 '11 at 15:21
I do not understand this last point: if $n$ is even, $n+1$ is odd...? Also $m=3^k$ are odd numbers having very explicitly unbounded $\phi(m)$. – user2035 Oct 9 '11 at 16:06

A slightly more general form of the above mentioned lemma states: whenever $m$ has at least two distinct prime factors and $\zeta_m$ is a primitive $m$-th root of unity, $1-\zeta_m$ is a unit in $\mathbf Z[\zeta_m]$.

Choosing $a=1-\zeta_m$ and $b=\zeta_m$ for the various primitive roots of unity, we get $\varphi(m)$ solutions for $K=\mathbf Q(\zeta_m)$. So any such $m$ satisfying $\varphi(m)\geq n$ will do.

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Another answer: let $f(x)$ be any monic polynomial with integer coefficients satisfying $f(0)=\pm 1$ and $f(1)=\pm 1$. Then all zeros $u$ of $f(x)$ are units (in the splitting field of $f(x)$), and each $1-u$ is also a unit.

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This construction has a-fortiori's answer as a special case: when $f(x)$ is the $m$-th cyclotomic polynomial and $m$ has at least two distinct prime factors, $f(0) = 1$ and $f(1) = 1$. Of course there is then also the task of directly constructing such polynomials beyond cyclotomic examples... – KConrad Oct 9 '11 at 18:29
Keith, what do you mean by constructing such polynomials? It is trivial to construct polynomials satisfying $f(0)=\pm 1$ and $f(1)=\pm 1$. – Richard Stanley Oct 11 '11 at 1:19
It is easy to see that the polynomials f in K[X] such that f(0) = 1 and f(1) = 1 is precisely the set of polynomials 1+ x(x-1) g, where g is an element of K[X]. (It's similar for $\pm 1$.) – Shaye Oct 13 '11 at 15:51

Someone (Elkies?) pointed out recently here on MO that, if $u$ is a unit, then the roots $a,b$ of $x(1-x)=u$ are units satisfying $a+b=1$. Start with your favorite $u$ and iterate.

Bonus question: It's known that the number of solutions of the unit equation is bounded in terms of the rank of the group of units, hence the degree. What's the smallest degree of a number field where the unit equation has $n$ solutions?

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Elkies's comment is here: mathoverflow.net/questions/76206/… – user2035 Oct 9 '11 at 20:53