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. 
 A: 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?  
A: As in Felipe's answer, one should really ask for the number of solutions as a function of the rank of the unit group, not the degree of the number field. More generally the question can be posed for $S$-units, where the rank is $s-1$ with $s=\#S$. Then even working over $\mathbb Z_S$ is very interesting. For $\mathbb Z_S^*$ it is known that the number of solutions is bounded by $C^s$ for some absolute constant $C$. (This is due to Evertse [1].) In the reverse direction, which is maybe more relevant to your question, there is a paper of Erdos, Stewart, and Tijdeman [2] in which they prove that there exist arbitrarily large finite sets of primes $S$ such that $u+v=1$ has at least $C^{s^{1/2-\epsilon}}$ solutions in $\mathbb Z_S^*$. They also present some heuristic evidence suggesting that the correct upper and lower bounds should be  $C^{s^{2/3\pm\epsilon}}$. Extrapolating, this suggests that there should be number fields of arbitrarily high degree $n$ with at least $C^{n^{2/3-\epsilon}}$ solutions in $O_K^*$, but only finitely many with $C^{n^{2/3+\epsilon}}$ solutions.
Addendum Lucia has pointed out that Konyagin and Soundararajan [3] have improved [2] by showing that there are arbitrarily large sets of rational primes $S$ so that the unit equation has $C^{s^{2-\sqrt2-\epsilon}}$ solutions in $\mathbb Z_S^*$. Note that $2-\sqrt2\approx 0.586$ is roughly halfway between $\frac12$ and $\frac23$, since $\frac12(\frac12+\frac23)\approx0.583$.
[1] J.-H. Evertse, MR 735341 On equations in $S$-units and the Thue-Mahler equation, Invent. Math. 75 (1984), no. 3, 561--584.
[2] P. Erdös, C. L. Stewart, and R. Tijdeman, MR 937987 Some Diophantine equations with many solutions, Compositio Math. 66 (1988), no. 1, 37--56.
[3] S. Konyagin and K. Soundararajan, MR 2321000 Two $S$-unit equations with many solutions, J. Number Theory 124 (2007), no. 1, 193--199.
A: The unit equation is discussed in detail in Chapter 5 of Bombieri and Gubler's book Heights in diophantine geometry.  In particular, they note (see Example 5.2.6) that in the cyclotomic field ${\Bbb Q}(e^{2\pi i/p})$ there are exactly $(p-1)(p-2)$ non-real solutions to the unit equation $x+y=1$.  This observation is attributed to H.W. Lenstra.  
In Example 5.2.7 it is observed that there seem to be many more real solutions to the unit equation in cyclotomic fields.  For example the maximal real subfield of ${\Bbb Q}(e^{2\pi i/19})$ contains $28 398$ solutions to the unit equation.  
Another intriguing example is 5.2.8.  Let $\alpha$ be the real root $\alpha>1$ of the Lehmer polynomial 
$$ 
x^{10}+x^{9}-x^7-x^6-x^5-x^4-x^3+x+1. 
$$ 
Then the field ${\Bbb Q}(\alpha)$ (with unit group of rank $5$) contains $2532$ solutions to the unit equation. 
A: 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.
A: 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.
