I wonder if it is possible to show, without using the Schmidt subspace/Roth theorem/Baker's bounds on linear forms in logarithms or other very deep results, that, in a number field, not all integral elements are sums of two units/most of them are not.


The known results that state that there are few elements that are sums of two units in number fields (see e.g. Fuchs, Tichy, Ziegler, "On quantitative aspects of the unit sum number problem", Arch. Math. 93 (2009)) seem to use either Thue-Siegel-Roth theorem or Baker's bounds on linear forms in logarithms. Both I find very difficult to reach. Since I rely on a similar result in a statement I have made [1], [2], [3], the motivation behind the question is a wish to find a simpler argument (or to convince oneself that existence of (many) elements that are not sums of two units is a claim that is itself of strength that is seemingly out of reach to significantly easier means than Thue-Siegel-Roth/Baker theorems).


References:


[1] Zinevičius A., *On the congruent number problem over integers of real number fields*, Albanian J. Math. **8** (2014), 49-53.


[2] Zinevičius A., *On the congruent number problem over integers of cyclic extensions*, Math. Slovaca **66** (2016), 561-564,


[3] Zinevičius A., *Corrigendum to "On the congruent number problem over integers of cyclic extensions"*, Math. Slovaca **69** (2019), 1233-1233.



Simple argument regarding sums of two units in a number field?
$\newcommand\p{\mathfrak{p}}$
$\newcommand\OL{\mathcal{O}}$
$\newcommand\P{\mathfrak{P}}$


Here is a solution which is essentially an elaboration on Felipe's answer.
Instead of working with squares, consider working with $m$th powers instead. 


**Lemma:** If $(1 - v u^m)$ is exactly divisible by a prime $\p$ of $\OL\_K$, then, assuming $K(v^{1/m}) \ne K$, the prime $\p$ is not inert in $L = K(v^{1/m})$.


**Proof:** In $\OL\_L$, the ideal $\P = (\p,1 - v^{1/m} u)$ has norm $\p$.


By Cebotarev, the density of primes $\p$ which remain inert in $\OL\_L$ is non-zero. Hence by the analytic
 arguments Felipe alluded to, the set of principal ideals of the form $(1 - v u^m)$ with $v$ ranging over a the (finite) set of non-zero representatives in $\OL^{\times}\_K/\OL^{\times m}\_K$ has density zero. So it remains to deal with ideals of the form $(1 - u^m)$, where we now have flexibility in choosing $m$.


**Lemma:** Let $\ell$ be any prime. Suppose that $m = |(\OL\_K/\ell)^{\times}|$. Then the density of principal ideals of the form $(1 - u^m)$ is at most $1/\ell$.


**Proof:** Since $u^m \equiv 1 \mod \ell$, this is the same as saying that the density of principal ideals divisible by $\ell$ is at most $1/\ell$.


Taken together, it follows that the density of principal ideals of the form $(1 - u)$ has density at most $1/\ell$ for any prime $\ell$, and hence has density zero.


