Blog - Sum-product, unit distances, and number fields | Erdős Problems

By Thomas Bloom on 31 May 2026 In this blog post I will give my personal view on the recent counterexamples to the unit distance conjecture and sum-product conjecture over the reals (see [90] and [52] respectively). My goal is to sketch the constructions and try and give some intuition as to where they came from and why they work. My main target audience is the me-of-a-month-ago, who did not know much algebraic number theory, and who needs the relevant parts of the basic theory in this area explained, but wants to know exactly where the quantitative improvements come from. (I know a bit more algebraic number theory now, but still much less than I'd like!) My focus is on the combinatorial side, and I will stop with an appeal to the literature as soon as we need to do any serious number theory. (In particular I will not attempt to discuss even the statement, let alone the proof, of the Golod-Shafarevich theorem.)Any faults in this post are entirely my own, and if you are confused by any aspect of the proofs sketched here I encourage you to review the original sources. Comments and corrections are welcome in the comment section.The original OpenAI disproof of the unit distance conjecture can be read here, with a human-written companion paper here and an explicit (and improved) version of the argument by Sawin here. The disproof of the sum-product paper by me, Sawin, Schildkraut, and Zhelezov, is here.Warmup roundConsider the following natural statement in additive combinatorics: if $A+A=\{ a+b : a,b\in A\}$ and $A-A=\{a-b : a,b\in A\}$ then how well can we bound $\lvert A+A\rvert$ above by $\lvert A-A\rvert$? In other words, for what constant $c$ can we say $\lvert A+A\rvert \leq \lvert A-A\rvert^c$? This is trivially true with $c=2$, since $\lvert A-A\rvert\geq \lvert A\rvert$ and $\lvert A+A\rvert\leq \lvert A\rvert^2$. Also obviously, weird things might happen for some small sets.

But is it possible with $c$ arbitrarily close to $1$ - in other words, with $c=1+\epsilon$, provided $\lvert A\rvert$ is large enough in terms of $\epsilon$? Initial experimentation suggests that it might be - the sum set tends to be smaller than the difference set, and in all the natural examples one might consider, anything that forces $A+A$ to be large also forces $A-A$ to be large.So maybe one could conjecture that $\lvert A+A\rvert \leq \lvert A-A\rvert^{1+o(1)}$. This is false, however, thanks to the 'tensor power trick'. First note that I never actually said where the $A$ lives - maybe you were thinking of a set of integers, but in additive combinatorics all of these concepts make sense in any abelian group, so we can also consider finite $A\subset \mathbb{Z}^d$ for arbitrarily large $d$.How does this help? Well first find some example where $A+A$ is large compared to $A-A$, just by fluke/law of small numbers. For example, if $A=\{0, 2, 3, 4, 7, 11, 12, 14\}$ then $A+A$ has $26$ elements but $A-A$ has $25$ elements. So here $\lvert A+A\rvert \geq \lvert A-A\rvert^c$ where $c=\log_{25}(26)\approx 1.012$. So what, you might say - this is one particular finite $A$, and I knew that small sets might be weird, so I covered this by saying 'exponent at most $1+\epsilon$ only for sufficiently large sets $A$. The point is that we can blow up any fixed example like this to arbitrary large examples with the same behaviour: if we let $B=A^d\subset \mathbb{Z}^d$ then the sum and difference sets are also just the cartesian products, so $\lvert B+B\rvert=26^d$ and $\lvert B-B\rvert =25^d$. This means that there

exist arbitrarily large sets $B$ (in some abelian group) such that\[\lvert B+B\rvert \geq \lvert B-B\rvert^{1.012\cdots}.\]Thus the initial naive guess was wrong; the best exponent is $>1$ by at least $0.012$, and no restricting to 'sufficiently large sets' is going to save it.(If you're interested in this type of problem, and earlier examples of this kind of tensor power construction used in additive combinatorics, see e.g. this page or [GRH07] by Gyarmati, Ruzsa, and Hennecart.)The constructions in both the sum-product and unit distance counterexamples have a similar flavour: one finds a 'trivial' construction, only winning by some constant factor, and then 'blowing up' this constant win by taking $d$-dimensional versions, where $d\to \infty$. In the example above this was easy to do, since given any abelian group $G$ we can just form the direct product $G^d$ and everything scales as expected. In the sum-product and unit distance problems, however, we have to (a) construct sets in $\mathbb{R}$ and $\mathbb{R}^2$ respectively, not in some $\mathbb{R}^d$, and (b) make sure that not just addition but some additional operation (multiplication and distance respectively) also scales in some predictable way.The former is actually not that big of an issue, but the latter requires some serious work. Fortunately, all of this work was done a long time ago, by algebraic number theory.Algebraic number theory refresherI'll begin with an informal refresher of algebraic number theory, or at least that tiny fraction relevant to us. All of the below can be found in any graduate course on algebraic number theory. You may want to skip to the summary at the end of this section, and only read back if that is unfamiliar to you. One striking aspect is that, despite the central role played by primes, ideals, the class group, etc in algebraic number theory we don't need to discuss them at all, or even define them, to explain the constructions.

A number field $K$ is a field of characteristic $0$, and so contains $\mathbb{Q}$, which has finite dimension over $\mathbb{Q}$. This dimension is called the degree of $K$. The finite dimension means that every $x\in K$ is algebraic over $\mathbb{Q}$ - that is, it is the root of a polynomial with coefficients in $\mathbb{Z}$. If there is such a polynomial which is monic then $x$ is called an algebraic integer. The algebraic integers in $K$ form a ring in $K$, denoted $\mathcal{O}_K$. (Note that this contains the usual integers, since $\mathbb{Z}\subset \mathbb{Q}\subset K$ and every $n\in\mathbb{Z}$ is the root of $x-n$.) This is not obvious (e.g. it is not obvious, under our definition, that the sum of two integers is another integer) but it is one of the first results proved in any course on algebraic number theory.Since $\mathbb{C}$ contains the algebraic closure of $\mathbb{Q}$, we can view $K$ as a subset of $\mathbb{C}$ - but, importantly, there is not a unique way of doing so. For example, if we let $K=\mathbb{Q}(\sqrt{2})$ then $\{1,\sqrt{2}\}$ forms a basis over $\mathbb{Q}$, so every $x\in K$ can be written as $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$. This looks like a well-defined element of $\mathbb{C}$ (in fact of $\mathbb{R}$), except when we recall that $\sqrt{2}$ is not uniquely defined - there are two distinct roots of $x^2-2$ in $\mathbb{R}$: $1.41\cdots$ and $-1.41\cdots$. As soon as we fix which one of these we mean, we fix how $K$ is embedded into $\mathbb{R}$, but there is a choice here. In general, if $K$ has degree $d$, there are exactly $d$ embeddings (field homomorphisms) of $K$ into $\mathbb{C}$. If all of these are actually maps into $\mathbb{R}$ (like with $\mathbb{Q}(\sqrt{2})$) then $K$ is called totally real.

Any embedding which does not map into $\mathbb{R}$ is called a complex embedding - these naturally form pairs via the conjugate map, since if $x\mapsto \sigma(x)$ is an embedding then $x\mapsto \overline{\sigma(x)}$ is also an embedding (and if $\sigma(x)\not\in \mathbb{R}$ these must be different). Therefore we often write $r_1$ for the number of real embeddings and $r_2$ for the number of complex embeddings up to conjugation, so that $d=r_1+2r_2$.These embeddings give us a natural way to view $K$ geometrically as a high-degree geometric space. For convenience let's suppose that $K$ is totally real, so we only have $\mathbb{R}$ to worry about (but everything below works for arbitrary number fields, you just have to be careful with conjugate embeddings, and some $\mathbb{R}$s become $\mathbb{C}$). Then we have the natural map (sometimes called the Minkowski map)\[x\mapsto (\sigma_1(x),\ldots,\sigma_d(x)),\]where $\sigma_1,\ldots,\sigma_d$ are the $d$ embeddings of $K$ into $\mathbb{R}$. Importantly, just like $\mathbb{Z}$ forms a $1$-dimensional lattice inside $\mathbb{R}$, the ring of integers $\mathcal{O}_K$ forms a $d$-dimensional lattice inside $\mathbb{R}^d$ under this map.The covolume of a lattice $L$ is the volume of the parallelepiped formed by its basis vectors. It basically measures how tightly packed the lattice $L$ is - so $\mathbb{Z}^d$ has covolume $1$, for example. Since $\mathcal{O}_K$ is a lattice in $\mathbb{R}^d$, this is a natural parameter to associate with $K$, and we call it the discriminant $\Delta_K$ of $K$ (this is a lie - for good algebraic reasons it's actually defined to be the square of this covolume (and also perhaps with a factor of $2^{r_2}$) but this will be irrelevant for our purposes, so just think of $\Delta_K$ as 'the covolume

of the $\mathcal{O}_K$ lattice').We will view $\mathbb{R}^d$ with the norm $\|x\|=\|x\|_\infty$, and will call this the 'size', so that the 'size' of $x\in K$ is $\max_{1\leq i\leq d}\lvert \sigma_i(x)\rvert$. Importantly, points in $\mathcal{O}_K$ are $1$-separated in this norm; this is best proved via the norm map $N(x)=\prod \sigma_i(x)$, which takes algebraic integers to integers. If $x\neq y$ are two algebraic integers then $x-y$ is a non-zero algebraic integer, and so\[1\leq \lvert N(x-y)\rvert = \prod_i \lvert \sigma_i(x)-\sigma_i(y)\rvert,\]so there is at least one coordinate where $x$ and $y$ differ by at least $1$.This separation means, via standard geometry of numbers, we understand how to count the size of the lattice $\mathcal{O}_K$ intersected with various convex sets defined in terms of this norm - in particular, if\[B^+(X)= \{ x\in \mathcal{O}_K : \| x\| \leq X\},\]then $\lvert B^+(X)\rvert\approx X^d$, where the $\approx$ hides losses of $O(1)^d$ and the covolume of the lattice, which is $\Delta_K^{O(1)}$.For our applications, we will also need another lattice. The integer ring $\mathcal{O}_K$ is not, in general, a group under multiplication, since it does not contain multiplicative inverses. The unit group $\mathcal{O}_K^\times$ is the set of algebraic integers $x$ whose multiplicative inverse $x^{-1}$ (which obviously exists somewhere in $K$) is also an algebraic integer.One might first think that there is not much interesting to say here - in $\mathbb{Z}$, for example, there are only two units, $1$ and $-1$. In general, any root of unity is a unit - it is an algebraic integer, as a root of $x^n-1$, and its multiplicative inverse is another root of unity. But there are only finitely many roots of unity in any number field, so is this it?