Recent Progress on a Problem of Sárközy

I would like to thank Brandon Hanson, Giorgis Petridis, and Kevin Ford for helpful contributions to this post.

A well-known problem of Sárközy is the following.

{\ } Conjecture 1: Let {p} be a prime and {S \subset \mathbb{F}_p^{\times}} be the set of squares. Suppose {A+B = S}. Then either {|A| = 1} or {|B| = 1}. {\spadesuit} {\ }

As the squares are a multiplicative subgroup, it is natural to guess they cannot be written additively as a sumet. Conjecture 1 is in a similar spirit of a long list of conjectures concerning the independence of multiplication and addition, such as the twin prime conjecture, the abc-conjecture, and the sum-product conjecture.

Progress towards Conjecture 1, as of last week, was summarized in this mathoverflow post and this other mathoverflow post (and the papers referenced within). Sárközy proved that if {A+B = S}, then

\displaystyle \frac{\sqrt{p}}{3 \log p} \leq |A| , |B| \leq \sqrt{p} \log p,\ \ \ \ \ (1)

which we recall below. In fact, Shparlinski (in Theorem 7) improved (1) and then later Shkredov showed (in Corollary 2.6)

\displaystyle (1/6 - o(1)) \sqrt{p} \leq |A| , |B| \leq (3 + o(1)) \sqrt{p}.

Brandon Hanson and Giorgis Petridis, utilizing the polynomial method, recently made significant progress towards Conjecture 1.

{\ } Theorem 1 (Hanson-Petridis): Suppose {A+B = S}. Then

\displaystyle |A| |B| = |S|.

In particular every element of {S} has a unique representation of the form

\displaystyle a + b , \ \ \ a \in A \ \ , b \in B. \ \ \ \spadesuit

{\ }

Their techniques handle the case where {S} is replaced by any nontrivial multiplicative subgroup, but we focus on the squares for simplicity. In particular, if {(p-1)/2} is a prime, then Conjecture 1 is established. This implies Conjecture 1 is true for infinitely many primes, provided there are infinitely many Sophie Germain primes (yet another conjecture based on the independence of multiplication and addition). Making use of (1) we are able to prove this unconditionally. Here {\pi(x)} is the prime counting function.

{\ } Corollary 1: For all but {o(\pi(x))} primes less than {x}, Conjecture 1 holds. {\spadesuit} {\ }

Proof: Let {p} be a prime. Suppose {A+B = S} with {|A| , |B| > 1}. Then by Theorem 1 and (1), {(p-1)/2} has a divisor between {\frac{\sqrt{p}}{3 \log p} } and {\sqrt{p} \log p}. By Theorem 6 (followed by Theorem 1 (v)) in Kevin Ford’s work on the distribution of integers with divisors in an given interval, there are at most {o(x / \log x)} such primes. Corollary 1 then follows from the prime number theorem. {\spadesuit} {\ }

If {S = A+B}, we always have trivial bound

\displaystyle (p-1)/2 = |A+B| \leq |A||B|, \ \ \ \ \ (2)

which can be compared to the following bilinear estimate.

{\ } Theorem 2: Let {\chi} be the quadratic character modulo {p}. Then

\displaystyle |\sum_{a \in A , b \in B} \chi(a +b ) | \leq \sqrt{p |A| |B|}.

In particular if {S = A+B}, then

\displaystyle |A| |B| \leq p. \ \ \ \spadesuit

{\ }

The proof of Theorem 2, which we give, is standard Fourier manipulations (see chapter 4 of Tao and Vu for more details). As we will see below, Hanson and Petridis make no use of this perspective.

{\ } Proof: The second statement follows from the first as if {S= A + B}, then

\displaystyle \sum_{a \in A , b \in B} \chi(a+b) = |A| |B|.

By Fourier inversion, it is enough to show

\displaystyle \sum_{a \in A , b \in B , \xi \in \mathbb{F}_p} e_p(\xi (a +b)) \widehat{\chi}(\xi) \leq \sqrt{p |A| |B|}, \ \ \ e_p(\theta) : = e^{2 \pi i \theta / p} ,\ \ \ \ \ (3)


\displaystyle \widehat{f}(\xi) = \frac{1}{p} \sum_{x \in \mathbb{F}_p} f(x) e_p(-\xi x).

Note { \widehat{\chi}(0) = 0}, and the usual estimate for Gauss sums implies

\displaystyle |\widehat{\chi}(\xi)| \leq p^{-1/2} , \ \ \ \xi \neq 0.

Combining with (3), it is enough to show

\displaystyle \sum_{\xi \in \mathbb{F}_p^{\times}} |\widehat{1_A}(\xi)||\widehat{1_B}(\xi)| \leq p^{-1} \sqrt{|A| |B|} .

But this follows from Cauchy-Schwarz and Parseval. {\spadesuit} {\ }

Combining (2) and Theorem 2, we see that if {S = A+B}, then

\displaystyle (p-1)/2 \leq |A||B| \leq p.\ \ \ \ \ (4)

Thus Theorem 1 asserts that the lower bound in (4) is the only possibility. We now proceed towards a proof of Theorem 1. The starting point is a classical theorem of Fermat.

{\ } Theorem 3 (Fermat): Let {a \in \mathbb{F}_p}. Then {a \neq 0} if and only if {a} is a root of {x^{p-1} - 1}. {\spadesuit} {\ }

There are many proofs of this elementary fact, for instance it is a quick consequence of Lagrange’s theorem. As a consequence, we have the following.

{\ } Corollary 2: Let {a \in \mathbb{F}_p^{\times}}. Then {a} is a square if and only if {a} is a root to

\displaystyle x^{(p-1) / 2} - 1. \ \ \ \spadesuit \ \ \ \ \ (5)

{\ }

Proof: Every square is a root of (5) as if {a = x^2} for some nonzero {x} then by Theorem 3,

\displaystyle a^{(p-1)/2} = x^{p-1} = 1.

Thus the squares are a subset of the roots of (5). On the other hand there are {(p-1)/2} squares and at most {(p-1)/2} roots and so the set of squares is precisely the set of roots of (5). {\ }

In is worth noting that there is a significant gap in the degree of the polynomial in (5) as opposed to in Theorem 3, which we make use of later. We now give a polynomial characterization of the cardinality of a set.

{\ } Lemma 1: Let {0 \leq n \leq p} and {A \subset \mathbb{F}_p}. Then {|A| \geq n} if and only if for any {d_0 , \ldots , d_{n-1} \in \mathbb{F}_p}, the equations

\displaystyle \sum_{a\in A} c_a a^k = d_k , \ \ \ 0 \leq k \leq n-1,

have a solution in the {c_a}. {\spadesuit} {\ }

Proof: This is a classical fact about Vandermonde matrices. {\spadesuit} {\ }

We now proceed to the proof of Theorem 1, which adopts the Stepanov method of auxiliary polynomials.

{\ } Proof of Theorem 1: Let {A , B \subset \mathbb{F}_p} and suppose {A+B = S}. We choose {|A| < (p-1)/2}, which is possible in light of (4). By Lemma 1, we may choose {c_a\in \mathbb{F}_p} so that {g \equiv 0}, where

\displaystyle g(x) = - 1 + \sum_{a \in A} c_a (x+a)^{|A| -1} .\ \ \ \ \ (6)


\displaystyle F(x) = -1 + \sum_{a\in A} c_a (x+a)^D, \ \ \ D = (p-1) / 2 + |A| -1.\ \ \ \ \ (7)

Our choice of {c_a} will ensure that each {b \in B} is a root of {F} with high multiplicity and that {F \neq 0}. By Corollary 2, since {a+b \in S},

\displaystyle F(b) = -1 + \sum_{a\in A} c_a (b+a)^D =- 1 + \sum_{a\in A} c_a (b+a)^{|A| - 1} = g(b).

Thus each {b \in B} is a root of (7). Furthermore,

\displaystyle F'(x) = \sum_{a \in A} c_a D(x+a)^{D-1},

and so, again by Corollary 2,

\displaystyle F'(b) = Dg'(b) = 0.

We may apply this argument for higher derivatives to obtain

\displaystyle F^{(j)} (b) = 0 , \ \ \ 0 \leq j \leq |A| - 1.

Thus each {b \in B} is a root of {F} with multiplicity {|A|} and it follows that, if {F \neq 0},

\displaystyle |A| |B| \leq {\rm deg}(F).\ \ \ \ \ (8)

It turns out that our previous choice for {c_a} ensure that {{\rm deg}(F) = (p-1)/2} (and is thus nonzero). For instance, since {g \equiv 0}, considering the leading term in (6), we have

\displaystyle \sum_{a \in A} c_a = 0.

and so the leading term of {F}, which is the same, is also zero. The same argument works to show that the coefficient of {x^j} of {F} is zero for {(p-1)/ 2 + 1 \leq j \leq D}. Now the constant term of {g} in (6) is

\displaystyle -1 + \sum_{a \in A} c_a a^{|A| -1} = 0,

where is the coefficient of the {x^{(p-1)/2}} term in {F} is

\displaystyle {D \choose (p-1)/2} \sum_{a \in A} c_a a^{|A| - 1} = {D \choose (p-1) / 2} \neq 0.

Combining this with (8), we find

\displaystyle |A||B| \leq (p-1)/2.

The second assertion in the Theorem follows immediately (see also Proposition 2.3 in Tao and Vu). {\spadesuit} {\ }

Remark 1: Suppose {A+B = S} with {|A||B| = (p-1)/2}. The proof of Theorem 1 reveals that

\displaystyle F(x) = \alpha \prod_{b \in B} (x- b)^{|A|}, \ \ \ \alpha = {D \choose |A| - 1},

with {F} and {D} defined (7). Furthermore, we have the identity

\displaystyle \prod_{a \in A , b \in B} (x - (a+b)) = x^{(p-1)/2} - 1. \ \ \ \spadesuit

{\ }

A close variant of the extremal case left by Theorem 1 was studied by Lev and Sonn. Following Sárközy, we now prove (1) (with little concern for the precise constants). The key input is the Weil bounds, and so the square root barrier that appears in (1) is natural (as discussed in this previous blog post).

{\ } Proof of (1): Let {p} be a prime and suppose {A+B = S}. Without loss of generality, we may suppose {|B| \geq |A|}. By (2), it is enough to show

\displaystyle |B| \lesssim \sqrt{p} \log p.\ \ \ \ \ (9)

Let {\chi} be the Legendre symbol and consider, for {A' \subset A},

\displaystyle h(x) = 2^{-|A'|} \prod_{a \in A'} (\chi(x + a) + 1) .\ \ \ \ \ (10)

Thus {h \geq 0} and by our assumption {h(b) = 1} for any {b \in B} and so

\displaystyle |B| \leq \sum_{x \in \mathbb{F}_p} h(x) \ \ \ \ \ (11)

On the other hand, expanding the product in (10) and applying the Weil bounds of the form

\displaystyle \sum_{x \in \mathbb{F}_p} \chi(x+ a_1) \cdots \chi(x+ a_k) \leq (k-1) \sqrt{p},

for distinct {a_1 , \ldots , a_k}, we find

\displaystyle |B| \leq \sum_{x \in \mathbb{F}_p} h(x) \leq p 2^{-|A'|} + (|A'|- 1) \sqrt{p} , \ \ \ A' \subset A.\ \ \ \ \ (12)

This establishes (9) if {|A| \geq \log p / \log 4}, since then we may choose {\log p / \log 4 \leq |A'| \leq \log p / \log 4 + 1}. Otherwise, we apply (12) with {A' = A} and multiply both sides by {|A|}, and using (2), we find

\displaystyle (p-1) / 2 \leq p 2^{-|A|} + (|A| - 1) \sqrt{p}.

This implies, crudely, that for {|A| \geq 2},

\displaystyle p/4 \lesssim \sqrt{p} \log p,

which is absurd for {p} large (and letting the implied constants handle the case {p} small). {\spadesuit}

Spacing of Quadratic Fractions

The following is a joint blog post with Kyle Pratt, a fellow graduate student at UIUC.

Recall a theorem of Fermat, which asserts that a prime {p} is {\equiv 1 \pmod{4}} if and only if it is the sum of two squares. From Dirichlet’s theorem on primes in arithmetic progressions we conclude there are infinitely primes that can be written as the sum of two squares:

\displaystyle p = m^2+\ell^2 \ \ \ \ \ (1)

Fouvry and Iwaniec showed this can be strengthened in a strong way, by showing there are infinitely many primes of the form in (1) with {\ell} prime.

Kyle and I, mostly to understand Fouvry and Iwaniec’s proof better, considered (a couple of years ago) the analogous question with

\displaystyle \ell^2 - N m^2 , \ \ \ N > 0 , \ \ \ \ \ (2)

replacing {\ell^2 + m^2}. We were lead to a rather interesting spacing result of fractions of the form

\displaystyle \frac{v}{d}, \ \ \ \ \ v^2 - N \equiv 0 \, (\text{mod } d), \ \ \ \ D \leq d \leq 2D,

which we explain below in the fold, using some basic theory of quadratic forms as well as a Brook’s theorem from graph theory.

We (very) briefly recall the structure of the Fouvry-Iwaniec argument. They look to asymptotically estimate

\displaystyle \sum_{\ell^2 + m^2 \leq x} \Lambda(\ell) \Lambda(\ell^2 + m^2) ,\ \ \ \ \ (3)

where {\Lambda} is the Von Mongoldt function. They use Vaughan’s identity on {\Lambda(\ell^2 + m^2)} to decompose (3) into Type I and Type II sums for the sequence

\displaystyle a_n = \sum_{\ell^2 + m^2 =n} \Lambda(\ell).

Great complications from the type II sums ensue, but the fundamental idea is to utilize the quadratic form appearing in (3) to factor

\displaystyle a_{mn} = \sum_{a^2 + b^2 = m} \sum_{c^2 + d^2 = n} b_{m,n}

where {b_{m,n}} encodes a rather strange looking condition that is eventually managable. In essence, this follows from the Brahmagupta-Fibonacci identity of the form

\displaystyle (a^2+b^2)(c^2 + d^2) = (ad-bc)^2 + (ac+bd)^2.

Properties of Gaussian integers underly their analysis.

In this post we are more concerned with the Type I estimates. We look for a bound of the form

\displaystyle \sum_{d \leq x^{\theta}} \Bigg| \sum_{\substack{n \leq x \\ d | n}} a_n - \frac{1}{d} \sum_{n\leq x} a_n \Bigg| \ll x^{1 - \epsilon}. \ \ \ \ \ (4)

The largest {\theta} for which (4) holds is the so-called level of distribution. In the analogous Bombieri-Vinogradov theorem, one can take any {\theta<1/2} (thought only saving an arbitrary power of log), while it turns out here we can take the much stronger {\theta <1} here. This is of the same strength as the unproven Elliot-Halberstam conjecture, as we are allowed to take advantage of the structure of the quadratic form {\ell^2 + m^2}. In the end, the Bombieri-Vinogradov theorem relies on the spacing of distinct Farey fractions

\displaystyle \left|\frac{a}{q} - \frac{a'}{q'}\right| \geq \frac{1}{qq'}, \ \ \ (a,q) = (a',q') = 1.

In a similar manner, a spacing result for the fractions

\displaystyle \{\frac{v}{d} : D \leq d \leq 2D , \ v^2 + 1 \equiv 0 \ {\rm mod} \ d\},

for a fixed large {D}, can be used to provide a bound for (4).

Thus in our search for primes of the form in (2), we are lead to study the spacing of fractions of the form

\displaystyle \Big\{\frac{v}{d} : D \leq d \leq 2D , \ v^2 - N \equiv 0 \, (\text{mod }d)\Big\}, \ \ \ \ \ (5)

where {N} and {D} are fixed ({D} large). For simplicity we take {N} to be a discriminant of a quadratic number field, though this is not a significant assumption. The main spacing result looks as follows.

{ \ }

Proposition 1: The fractions in (5) can be partitioned into sets {F_1 , \ldots , F_t} such that {t = O_N(1)}, and for any distinct {x,y \in F_j} we have

\displaystyle \| x - y \| \gg \frac{1}{D}. \ \ \ \spadesuit

{ \ }

We remark this estimate is best possible up to logarithms. Proposition 1 combined with some reductions and the large sieve can be used to establish (4).

{ \ }

Proof of Proposition 1: Suppose {v^2 - N \equiv 0 \ (\text{mod }d)}. Then we have

\displaystyle d = ar^2 + 2brs + cs^2,\ \ \ \ \ (6)

for some {a,b,c,r,s\in \mathbb{Z}}. Note we can choose the tuples {(a,b,c)} to come from the {h(N)} inequivalent quadratic forms of discriminant {N}. One can check that after multiplication by units in the ring of integers of {\mathbb{Q}(\sqrt{N})}, we may assume {r \asymp_N s}. It follows that {r , s \asymp_N D^{1/2}}

We have, as in this paper of Hooley (page 106), that

\displaystyle \frac{v}{d} = \frac{\overline{r}}{s} - \frac{ar+bs}{sd}, \ \ \ r\overline{r} \equiv 1 \, (\text{mod }s).\ \ \ \ \ (7)


\displaystyle \Big|\frac{\overline{r}}{s} - \frac{\overline{r'}}{s'}\Big| =\Big| \frac{r s' - r' s}{ss'}\Big| \gg \frac{1}{D}, \ \ \ r s' - r' s \neq 0. \ \ \ \ \ (8)

Thus we want to show {\frac{ar+bs}{sd}} is much smaller than {1/D}. We have the trivial bound

\displaystyle \frac{ar+bs}{sd} \ll_N 1/D.\ \ \ \ \ (9)

This is not good enough to ensure a spacing of {1/D} as the error term can be bigger than the main term by a multiplicative constant. We cannot even ensure such a spacing for all the fractions in (5), but what we are able to do is partition the fractions into sets

\displaystyle F_1, \ldots , F_t , \ \ \ t = O_N(1),

such that we can improve (8) for the fractions that lie inside each set. Indeed for each {a,b,c, r,s} there is a unique {v} and {d} that satisfy (6) and (7). Thus there are at most {O_N(1)} choices of {v,d} for a fixed {r,s}.

{ \ }

Lemma 1: Suppose {(x,y) = 1} and {x,y \asymp_N D^{1/2}}. Then there are at most {O(k)} choices of coprime {x',y' \asymp D^{1/2}} so that {xy' - x'y = k}. {\spadesuit}

{ \ }

Proof of Lemma 1: Note that, by coprimality, {k=0} is impossible. We have {x' = x_0 + xt}, {y' = y_0 + yt}. By size considerations, there are {O(1)} choices for {t} and thus {x'} and {y'}. {\spadesuit}

{ \ }

We choose {K = K(N)} to be a sufficiently large constant. By Lemma 1, for a fixed fraction {v/d} from (5), there are at most {O_N(1)} choices {v',d'} so that

\displaystyle \Big|\frac{v}{d} - \frac{v'}{d'}\Big| \leq \frac{1}{D}.

We create a graph {G} with vertex set from (5) and {v/d \sim v'/d'} if {|r s' - r' s| \leq K}. Thus {G} has maximum degree {O_N(1)}. By Brooks’ theorem, we may find a coloring of {G} with {O_N(1)} color classes {F_1, \ldots , F_t}.

Inside a color class, we improve the lower bound in (8) to {K/D} while (9) remains the same and so {\|v/d - v'/d'\| \gg_N 1/D}, as desired. {\spadesuit}

Sidon-like Sets

What is the largest set avoiding a linear equation? This general question is a central one in additive combinatorics. Indeed the search of extremal structures gives it a combinatorial flavor, while the presence of an additive equation makes such a question ripe for additive techniques. Such questions include well-studied subjects such as sum-free sets, Sidon sets, and sets avoiding arithmetic progressions. In 1993, Ruzsa made an effort to put these examples into a general framework. In what follows, we focus on Sidon sets and their generalizations, as developed by Ruzsa in the same paper.

We fix {k \geq 2} and consider the linear equation

\displaystyle x_1 + \ldots + x_k = y_1 + \ldots + y_k \ \ \ \ \ (1)

Following, Ruzsa, we say a solution is trivial if the {x_i} are a permutation of the {y_i}. We set {r(N) = r_k(N)} to be the size of the largest set of {[N]} with only trivial solutions to (1). We set {R(N) = R_k(N)} to be the size of the largest set of {[N]} with no solutions to (1) in {A} with {x_1 , \ldots , x_k , y_1 , \ldots y_k} distinct. It follows from these definitions that

\displaystyle r(N) \leq R(N).

We will show below that {R_k(N) \ll_k r(N)}, while it is not known It is not known if {R_k(N) \ll r_k(N)}. Indeed one may feel that most of the solutions to (1) where the variables lie in a set {A} come when the variables are distinct, since otherwise we have imposed an additional constraint. This idea does work well for sets with additive structure, but has shortcomings for general sets. Furthermore, it was shown by Timmons that {r_3(N) \neq (1 + o(1)) R_3(N)}.

For a set {A \subset [N]}, we let {n(A)} be the number of nontrivial solutions to (1) in {A} and let {N(A)} be the number of solutions to (1) with distinct variables.

We briefly remark that in Wooley’s efficient congruencing proof of Vinogradov’s mean value theorem, he estimates quantities similar to {N(A)} (where {A} is the set of {k^{\rm th}} powers), as applications of Hensel’s lemma require the variables to be distinct (modulo a certain prime). Indeed there are some overlapping techniques (see Lemma 1 below).

To start, we first consider the case {k=2}. A set with {n(A) = 0} is known in the additive combinatorics literature as a Sidon set. Thus {r_{A-A}(x) \leq 1} for {x \neq 0}. There is a nice construction due to Ruzsa (modifying a previous construction of Erdős), which is just one of several known constructions.

Example 1 (Large Sidon Set): Let {p} be a prime. Let {g} be a primitive root in {\mathbb{F}_p}. One can check the set of residues in the cyclic group of order {p(p-1)}, defined via the Chinese remainder theorem, by

\displaystyle x = k \ {\rm mod} \ p-1 , \ x = g^k \ {\rm mod} \ p , \ \ \ 0 \leq k \leq p-1,

forms a Sidon set in the cyclic group modulo {p(p-1)} and hence in {\mathbb{Z}}. {\spadesuit}

On the other hand, we have the following upper bound.

Theorem 1: Let {A \subset [N]} be a Sidon set. Then

\displaystyle |A| \leq N^{1/2} + N^{1/4} + 1. \ \ \ \spadesuit

Proof: We follow an argument of Green, which will rely on some basic Fourier analysis (we follow notation from Chapter 4 of Tao and Vu’s book on additive combinatorics). Let {1 \leq u \leq N} and consider {A} as a subset of {Z : = \mathbb{Z} / (N+u) \mathbb{Z}}. Let {I = \{1 , \ldots , u\}}. We have by Parseval,

\displaystyle (N+u)^2\sum_{x \in Z} 1_A * 1_{-A}(x) 1_I * 1_{-I}(x) = (N+u)^3 \sum_{\xi} |\widehat{1_A}(\xi)|^2 |\widehat{1_I}(\xi)|^2 \geq \frac{|A|^2 |I|^2}{N+u},\ \ \ \ \ (2)

in the last step isolating the {\xi = 0} term. Since {A} is a Sidon set we have

\displaystyle (N+u)1_{A} * 1_{-A}(x) \leq 1,

for {0 < |x| \leq u} (note this is not true for other values of {x \in Z}). Thus

\displaystyle (N+u)^2\sum_{x \in Z} 1_A * 1_{-A}(x) 1_I * 1_{-I}(x) \leq |A| u + u^2.\ \ \ \ \ (3)

Combining (2) and (3) and choosing the optimal choice of {u = \lfloor N^{3/4} \rfloor} gives the desired result. {\spadesuit}

Green remarks in his paper that above proof only assumes that {r_{A-A}(x) \leq 1} for {0 < x < N^{3/4}} instead of the full Sidon set property. Combining Example 1 and Theorem 1, we find

\displaystyle N^{1/2} \leq r_2(N) \leq N^{1/2} + N^{1/4} + 1.

It is a famous question of Erdős to improve these bounds (in particular the lower bound), which has been stuck for over fifty years (modulo a improvement by Cilleruelo to {N^{1/2} + N^{1/4} + 1/2}). It seems likely that both the lower and upper bound are not optimal and that the truth lies somewhere between {N^{1/2} + C} and {N^{1/2} + O(N^{\epsilon})} for any {C\geq 1} and {\epsilon > 0}. One can see this blog post of Gowers and the comments within for interesting discussion of this problem.

We now consider general {k} and start with a lower bound.

Proposition 1 (Random sets): For {N} large enough,

\displaystyle r(N) \geq \frac{1}{2}8^{-1/(2k-1)} N^{1/(2k-1)} . \ \ \ \spadesuit

Proof: We use the alteration method, which is discussed in Alon and Spencer’s book on the Probabilistic Method. Choose {A \subset [N]} where each element is chosen independently at random with probability

\displaystyle p = 8^{-1/(2k-1)}\frac{N^{1/(2k-1)}}{N}.

The number of solutions to (1) in {[N]} is at most {N^{2k-1}}, so the expected number of solutions to (1) is at most {p^{2k}N^{2k-1}}. By Markov,

\displaystyle \mathbb{P}(n(A) > 2p^{2k}N^{2k-1}) < \frac{1}{2}.

On the other hand, by say Chebyshev,

\displaystyle \mathbb{P}(|A| < \frac{1}{2} p N) < \frac{1}{2},

for {N} large enough. By the union bound, we may choose an {A} such that {|A| \geq pn/2} and {n(A) \leq 2p^{2k} N^{2k-1}}. By our choice of {p}, we have {n(A) \leq |A|/2}. For each nontrivial solution of (1), we delete an element (the so-called alteration) of {A} that renders it no longer a solution and call the resulting set {A'}. Thus {n(A') = 0} and so {r(N) \geq |A'| \geq pN/2}, as desired. {\spadesuit}

Ruzsa remarks that the probabilistic construction above probably never gives the correct order of magnitude for {r(N)}, where (8) is replaced by a suitable linear equation. We already saw this above, as Example one gives a Sidon set of size {N^{1/2}} while the random construction gives one of size about {N^{1/3}}. Moreover, with an explicit construction, Bose and Chowla showed the following improvement to Proposition 1:

\displaystyle r(N) \geq (1 + o(1)) N^{1 / k},

(see this survey of Kevin O’ Bryant for more information). Timmons showed that

\displaystyle R(N) \geq (2^{1 - 1/k} + o(1))N^{1/k},

by pasting together two sets from the aforementioned Bose-Chowla construction.

We now turn to upper bounds. We let {T_k(A)} be the number of solutions in {A} to (1) For any set with {n(A) = 0}, we have that {T_k(A) \leq k!|A|^k} and so by Cauchy Schwarz

\displaystyle |A|^{2k} \leq |kA| T_k(A) \leq k N k! |A|^k.\ \ \ \ \ (4)

It follows that

\displaystyle r(N) \leq (k! k )^{1/k} N^{1/k},

and thus

\displaystyle r(N) \asymp_k N^{1/k}.

This argument fails to bound {R(N)}, as was observed by Ruzsa, who provided an alternative argument, which we recall below. The basic idea is to still use (4), and to show that {T_k(A)} is still small for sets with {N(A) = 0}. We will need the following lemma.

Lemma 1 (Hölder): Let {f} be a function on the torus. Then

\displaystyle \int |f|^{2k-2} \leq \left( \int |f|^{2k} \right)^{1 - 1/(k-1)} \left(\int |f|^2 \right)^{1/(k-1)}.

To prove this, apply Hölder’s inequality to {|f|^{2k} = |f|^{2k - 2/(k-1)} |f|^{2/(k-1)}} with powers {(k-2)/(k-1)} and {k-1}. Note that one can always interpolate an upper bound for {\int |f|^p} in terms of {\int |f|^u} and {\int |f|^v} as long as {u < p < v}.

Proposition 2: We have

\displaystyle R(N) \leq (1 + o(1) )k^{2 - 1/k} N^{1/k}. \ \ \ \spadesuit

Proof: Suppose that {N(A) = 0} (though we will not make use of this until the end). By (4), it is enough to show

\displaystyle T_k(A) \leq (1 + o(1))k^{2k-2} |A|^k.

Note that

\displaystyle T_k(A) = \int |f|^{2k} , \ \ \ f(x) = \sum_{a\in A} e^{2 \pi i a}.

By Lemma 1 and Parseval, it is enough to show

\displaystyle T_k(A) \leq (1 + o(1))k^2 |A| \int |f|^{2k-2}.

The idea is that {N(A) = 0} allows us to eliminate a variable at a cost of only {k^2 |A|}, as opposed to the trivial bound of {|A|^2}.

We let

\displaystyle s(n) = \#\{(x_1 , \ldots , x_k) \in A^k : x_1 + \ldots + x_k = n, \ \ x_i \ \text{distinct} \},


\displaystyle \sigma(n) = \#\{(x_1 , \ldots , x_k) \in A^k : x_1 + \ldots + x_k = n\}.

Note we have

\displaystyle T_k(A) = \sum_n \sigma(n)^2.

By the triangle inequality in {\ell^2}, it is enough to show

\displaystyle \sum_n (\sigma(n) - s(n))^2 \leq k^4 \int |f|^{2k-2} , \ \ \ \sum_n s(n)^2 \leq k^2 |A| \int |f|^{2k-2} .\ \ \ \ \ (5)

Note the second term is the larger of the two for {|A|} large.

We dispose of the first inequality. Note {\sigma(n) - s(n)} is the number of solutions to {x_1 + \ldots + x_k = n} with some {x_i = x_j}. There are at most {k^2} choices for the indices and then after relabelling we find

\displaystyle 2x_1 + x_2 + \ldots + x_{k-1} = n.


\displaystyle \sum_n (\sigma(n) - s(n))^2 \leq k^4 \int |f(2x)|^2 |f(x)|^{2k-4} \leq k^4 \int |f|^{2k-2},

by Hölder’s inequality.

Now we finally use that {N(A) = 0}. Suppose we have a solution to (1) with distinct {x_i} and distinct {y_j}. Since {N(A) = 0}, this implies {x_i = y_j} for some {1 \leq i , j \leq k}. Thus

\displaystyle \sum_n s(n)^2 \leq k^2 |A| \int |f|^{2k-2},

since we have {k^2} choices for {i,j} and {|A|} choices for {x_i}. {\spadesuit}

The best bound for {r(N)} (for {k} large) is due to Green and is of the form

\displaystyle r(N) \leq (1 + o_N(1)) (1 + o_k(1)) \frac{k}{2e} N^{1/k},\ \ \ \ \ (6)

an improvement to the constant in (4). On the other hand, the best bound for {R(N)} is in a recent paper of Schoen and Shkredov:

\displaystyle R(N) \leq 16 k^{3/2} N^{1/k}, \ \ \ N \ \text{large enough}\ \ \ \ \ (7)

adopting ideas from Timmons as well as Ruzsa’s original work. Schoen and Shkredov implicitly mention the following question.

Question 1: Let {A \subset \mathbb{Z}} such that {N(A) = 0}. Does there exist a large {B \subset A} (say {|B| \geq |A| / 2}) such that {B} has no solutions to

\displaystyle x_1 + \ldots + x_{\ell} = y_1 + \ldots + y_{\ell},\ x_i, y_j \ \text{distinct} \ \ \ 1 \leq \ell \leq k.\ \ \ \ \ (8)

An affirmative answer would allow one to apply (5) and then (4) to {B} and eventually get improved bounds for {R(N)} comparable to what is known for {r(N)} (i.e. the same as (6) up to the absolute constant). Schoen and Shkredov are not able to solve Question 1, but are able to show that (8) holds for many {\ell} and then apply a suitable version of Erdős-Ko-Rado to obtain (7). Underlying their argument is the simple observation that a solution to (1) with {k = m} can be added to a solution with {k = n} to create a new solution with {k = m + n}.

We illustrate part of the argument in the {k=4} case, which is significantly simpler than the general case and already handled by Timmons.

Proposition 3: One has

\displaystyle R_4(N) \leq (1 + o(1)) 2^{1/8} 4^{5/8} N^{1/4}.

Proof: We let {A \subset [N]} with {N(A) = 0}. By the first part of (5) and Lemma 1, we have

\displaystyle T_4(A) \leq (1 + o(1))\sum_n s(n)^2.

By the second part of (5),

\displaystyle \sum_n s(n)^2 \leq 16 |A| \int |f|^{6} \leq 16 |A| (\int |f|^8 )^{1/2} (\int |f|^4)^{1/2},

and so

\displaystyle T_4(A) \leq 16^2 |A|^2 \int |f|^4.

Suppose first that {A} has no solutions to (8) with {\ell = 2}. Then

\displaystyle \int |f|^4 \leq 4 |A|^2 .

Indeed there are at most {2|A|^2} solutions to {a+b= c+d} with either {a=b} or {c=d} and also {\leq 2|A|^2} trivial solutions. It follows that

\displaystyle T_4(A) \leq 4^5 |A|^4.

Combining this with (4) gives the desired result. Now suppose {A} does have a solution to (8) with {\ell =2}, say

\displaystyle x_1 + x_2 = y_1 + y_2. \ \ \ \ \ (9)

Delete these four elements from {A} to create {A'}. Now we claim {A'} does not have a solution to (8) with {\ell = 2}. Indeed if there were, say

\displaystyle a_1 + a_2 = b_1 + b_2,

then adding this to (9) yields

\displaystyle a_1 + a_2 + x_1 + x_2 = b_1 + b_2 + y_1 + y_2,

which contradicts {N(A) = 0}. Thus we may apply the above argument to {A'} to obtain

\displaystyle T_4(A') \leq (1+o(1)) 4^5 |A'|^4,

and the Proposition follows from (4). {\spadesuit}

Exponential Sums over Small Subgroups

The purpose of the post is to recall a theorem of Bourgain and Konyagin that shows cancellation in exponential sums over multiplicative subgroups of {\mathbb{F}_p}, incorporating the point–plane bound incidence bound due to Rudnev. It is notoriously hard to find cancellation in short exponential sums in {\mathbb{F}_p}, for instance improving the Burgess bound is a fundamental open problem in number theory (see this previous blog post for discussion). Bourgain and Konyagin were able to leverage the sum–product phenomenon to show cancellation in certain sums with as few as {p^{\delta}} terms, improving upon the previous best of { p^{1/4+\epsilon}} due to Konyagin (incidentally the Burgess bound relies on a simpler sum–product type bound).

Let {H \leq \mathbb{F}_p^{\times}} be a multiplicative subgroup. We define the Fourier transform of {H} via

\displaystyle \widehat{1_H}(\xi) : = \frac{1}{p} \sum_{h \in H} e_p(-\xi h) , \ \ \ e_p(\theta) : = e^{2 \pi i \theta / p}. \ \ \ \ \ (1)

In (1) and what follows we adopt the notation of Chapter 4 in Tao and Vu (see also my previous blog post for some basic discussion on the discrete Fourier transform).

For any {\xi \in \mathbb{F}_p}, we have the trivial bound

\displaystyle |\widehat{1_H}(\xi)| \leq \mathbb{P}(H) : = |H| / p,

which is obtained at the zero frequency. On the other hand, {1_H} has multiplicative structure and we expect it cannot correlate with an additive character in light of the sum–product phenomenon. This was verified by Bourgain and Konyagin (see also these notes of Green).

Theorem 1 (Bourgain-Glibichuk-Konyagin): Let {H \leq \mathbb{F}_p^{\times}} of size at least {p^{\delta}}. Then

\displaystyle \sup_{\xi \in \mathbb{F}_p^{\times}}|\widehat{1_H}(\xi)| \leq \mathbb{P}(H) p^{-\epsilon}. \ \ \ \spadesuit

Theorem 1 should be compared to the famous Gauss sum estimate (see for instance this previous blog post), but applies to much smaller multiplicative subgroups. The proof relies on three ideas. The first is that if {|\widehat{1_H}(\xi) |} is large for one {\xi \in \mathbb{F}_p^{\times}}, then it is large for many {\xi \in \mathbb{F}_p^{\times}}. Indeed it follows from (1) that

\displaystyle \widehat{1_H}(h \xi) = \widehat{1_H}(\xi) , \ \ h \in H \ \ \ \ \ (2)

We define the spectrum (see Chapter 4 of Tao and Vu for detailed discussion, as well as these notes of Green) via

\displaystyle {\rm Spec}_{\alpha}(H) : = \{\xi \in \mathbb{F}_p : |\widehat{1_H}(\xi) | \geq \alpha \mathbb{P}(H) \}.

By Parseval’s identity of the form

\displaystyle \sum_{\xi \in \mathbb{F}_p} |\widehat{1_H}(\xi)|^2 = \mathbb{P}(H),

and (2), we find

\displaystyle |H| \leq | {\rm Spec}_{\alpha}(H)| \leq \mathbb{P}(H)^{-1} \alpha^{-2}.\ \ \ \ \ (3)

If {|H| = p^{\delta}}, this gives

\displaystyle \alpha \leq p^{1/2 - \delta},

which is only useful for {\delta > 1/2} (for instance this works quite well for Gauss sums). Thus we need new ideas to handle the case {\delta \leq 1/2}. Note this is in alignment with the principle that basic Fourier techniques intrinsically have a “square root barrier.”

The second idea we will use is that {H} has little additive structure in the following form. Recall the additive energy of {A} and {B} is defined via

\displaystyle E^+(A,B) = \{(a,a' , b , b' ) \in A^2 \times B^2 : a + b = a' + b' \}.

Note that

\displaystyle E^+(A,B) = \sum_x r_{A- B}(x)^2 , \ \ \ r_{A-B}(x) : = \#\{(a,b) \in A \times B : x = a -b \}.

Proposition 1 (Sum–Product): Let {H \leq \mathbb{F}_p^{\times}} and {|H|^2 |A| \leq p^2}. Then

\displaystyle E^+(H ,A) \leq |H| |A|^{3/2} \ \ \ \spadesuit

Proposition 1 should be compared to the trivial bound {E^+(H,A) \leq |H| |A|^2}.

Proof: We will use Rudnev’s point–plane incidence bound (Theorem 3 in this paper). To do so, we note that {E^+(H,A)} counts the number of solutions to

\displaystyle h + a = h' + a' , \ \ \ a , a' \in A , \ h, h' \in H.

Since {H} is a multiplicative subgroup, {|H|^2 E^+(H,A)} is the number of solutions to

\displaystyle hh_1 + a = h' h_1' + a' , \ \ \ a , a' \in A , \ h, h' , h_1 , h_1' \in H.

This is precisely the number of incidences between the point set {H \times H \times A} and planes of the form {h x + a = h' y + z}. Thus by Rudnev’s point–plane incidence bound of the form (note there is a condition on the number of maximum colinnear planes which is trivially satisfied in our cartesian product set–up)

\displaystyle I(P , \Pi) \ll n^{3/2} , \ \ \ |P| = |\Pi| = n,

we find

\displaystyle |H|^2 E^+(H,A) \ll |H|^3 |A|^{3/2} . \ \ \ \spadesuit

We now move onto the third idea and general principle is that {{\rm Spec}_{\alpha}(H)} is additively structured. The following Lemma is due to Bourgain and can be found in Lemma 4.37 in Tao and Vu.

Lemma 1 (Additive Structure in Spectrum): Let {A \subset \mathbb{F}_p} and {0 < \alpha \leq 1}. Then for any {S \subset {\rm Spec}_{\alpha}(A)}, one has

\displaystyle \#\{ (\xi_1 , \xi_2) \in S \times S : \xi_1 - \xi_2 \in {\rm Spec}_{\alpha^2/2}(A) \} \geq \frac{\alpha^2}{2} |S|^2. \ \ \ \spadesuit

Lemma 1 roughly asserts that the spectrum is closed under addition. For example, consider the example {A = \{1 , \ldots , K\}} where {K = o(p)}. Here {{\rm Spec}_{\alpha}(A)} is an interval of length {\asymp \alpha^{-1}} (there are more sophisticated examples, see this paper of Green).

Proof of Lemma 1: We set {f = 1_A}. By assumption we have

\displaystyle \alpha \mathbb{P}(A) |S| \leq \sum_{\xi \in S} |\widehat{f}(\xi)| = \sum_{\xi \in S} c(\xi) \widehat{f}(\xi) = \frac{1}{p} \sum_{\xi \in S} \sum_{a \in A} c(\xi) e_p(\xi a),

for some {c(\xi) \in \mathbb{C}} of modulus 1. By changing the order of summation and Cauchy Schwarz,

\displaystyle \mathbb{P}(A)^2 \alpha^2 |S|^2 \leq \frac{|A|}{p^2} \sum_{a\in A} \sum_{\xi , \xi' \in S} c(\xi) c(\xi') e((\xi - \xi') a) \leq \mathbb{P}(A) \sum_{\xi , \xi' \in S} |\widehat{f}(\xi - \xi')|.

Lemma 1 follows from pigeon holing. {\spadesuit}

Suppose, for the sake of discussion, that for all {\alpha}, {S \subset {\rm Spec}_{\alpha}(A)} does not have additive structure in the strong form form {r_{S-S}(x) \ll 1} for all {x}. Then we conclude

\displaystyle \{ (\xi_1 , \xi_2) \in S \times S : \xi_1 - \xi_2 \in T \} \ll |T|,\ \ \ \ \ (4)

and so by Lemma 1 we have a significant growth from {{\rm Spec}_{\alpha}(A)} to {{\rm Spec}_{\alpha^2/2}(A)}. But then applying this again to {{\rm Spec}_{\alpha^2/2}(A)} we have significant growth to {{\rm Spec}_{\alpha^4/8}(A)} and repeating this procedure will eventually contradict the trivial bound

\displaystyle |{\rm Spec}_{\alpha}(A)| \leq p.

When {H} is a multiplicative subgroup, we can show a weaker version of (4) using that {{\rm Spec}_{\alpha}(H)} is a union of cosets of {H} via (2) and Proposition 1. We turn to the details.

Proof of Theorem 1: Fix {0 < \alpha \leq 1} be chosen small enough so that {S_{\alpha} : = {\rm spec}_{\alpha}(H)} is nonempty. By Lemma 1,

\displaystyle \#\{ (\xi_1 , \xi_2) \in S_{\alpha} \times S_{\alpha} : \xi_1 - \xi_2 \in {\rm Spec}_{\alpha^2/2}(H) \} \geq \frac{\alpha^2}{2} |S_{\alpha}|^2,

that is

\displaystyle \frac{\alpha^4}{4} |S_{\alpha}|^4\leq \left( \sum_{z \in {\rm Spec}_{\alpha^2/2}((A)(H) } r_{S_{\alpha} - S_{\alpha}}(z) \right)^2 \leq | {\rm Spec}_{\alpha^2/2}(H)| E^+(S_{\alpha} , S_{\alpha}).\ \ \ \ \ (5)

Now we use Proposition 1 to provide an upper bound for {E^+(S_{\alpha} , S_{\alpha})}. By (2), {S'_{\alpha} : = S_{\alpha} \setminus \{0\}} is a union of cosets of {H}, say {S'_{\alpha} = \cup_{x \in C} Hx}. Thus by the triangle inequality in {\ell^2} and Proposition 1,

\displaystyle E^+(S'_{\alpha} )^{1/2} \leq \sum_{x \in C} E^+(S'_{\alpha} , Hx)^{1/2} = \sum_{x \in C} E^+(S'_{\alpha}x^{-1} , H)^{1/2} \ll \frac{|S_{\alpha}|}{|H|}|H|^{1/2} |S'_{\alpha}|^{3/4}.

Combining with (5), we find

\displaystyle \alpha^4 |H| | {\rm Spec}_{\alpha}(H)|^{1/2} \ll | {\rm Spec}_{\alpha^2/2}(H)| .

By (3), we find that {|{\rm Spec}_{\alpha}(H)| > |H| + 1} and so

\displaystyle \alpha^4 p^{\delta / 2} \leq \alpha^4 |H|^{1/2} \ll \frac{| {\rm Spec}_{\alpha^2/2}(H)|}{|{\rm Spec}_{\alpha}(H)|} .\ \ \ \ \ (6)

Now we let {1 > \alpha_1 > \alpha_2 > \ldots > \alpha_{J+1} > 0}, where {\alpha_{i+1} = \alpha_i^2 / 2} and {\alpha_1 = p^{-\epsilon}}. Thus {{\rm Spec}_{\alpha_1}(H)} contains more than 0 or immediately conclude Theorem 1. Thus (6) holds for all {\alpha_i} since {{\rm Spec}_{\alpha_i}(H)} increases in size as {i} increases. We have

\displaystyle \prod_{i = 1}^{J} \frac{| {\rm Spec}_{\alpha_{i+1}}(H)|}{|{\rm Spec}_{\alpha_i}(H)|} = \frac{| {\rm Spec}_{\alpha_{J+1}}(H)|}{|{\rm Spec}_{\alpha_1}(H)|} \leq p ,

and so there is a {1 \leq j \leq J} such that

\displaystyle \frac{| {\rm Spec}_{\alpha_{j+1}}(H)|}{|{\rm Spec}_{\alpha_j}(H)|} \leq p^{1/J}.

Combining with (6), we find

\displaystyle p^{- 2^J \epsilon + \delta /2} \frac{1}{2^{J}} \ll \alpha_j^4 p^{\delta/2} \ll p^{1/J}.

Choosing {2^J \epsilon \asymp 1}, we find that

\displaystyle p^{\delta / 2} \ll \epsilon^{-1} p^{1 / \log \epsilon^{-1}},

which is a contradiction for {p} large as long as {\delta \ll \log^{-1} \epsilon^{-1}}. {\spadesuit}

As we saw in the proof, the sum–product phenomenon asserts that {H} has little additive structure which is in tension with the general property that {{\rm Spec}_{\alpha}(A)} is additively structured.


Entropy and Sumsets: An example

The following post is a result of a discussion with Imre Ruzsa. Motivated by the following easy inequality in additive combinatorics

\displaystyle A+2 \cdot A \subset A+A+A , \ \ q \cdot A := \{qa : a \in A\},

I asked if the following was true for a finitely valued random variable {X}:

\displaystyle H(X+2 \cdot X) \leq H(X+X+X), \ H(X) := -\sum_{x \in X} \mathbb{P}(X = x) \log_2 \mathbb{P}(X = x).\ \ \ \ \ (1)

Here all sums are of independent copies of the random variables. The idea is that one might expect {X+X} to be a bit more uniform than {2 \cdot X}.

First Imre provided a counterexample to the question

\displaystyle H(X+ 2\cdot Y) \leq H(X+Y+Y).

I find this example is particularly elegant. Let {X} be uniform on {\{0,1\}} and {Y} be uniform on {\{0 , \ldots , n\}}. Then {X+2 \cdot Y} is uniform on {\{0 , \ldots , 2n+1\}}, while the support of {X + Y + Y} is {\{0 , \ldots , 2n+1\}} but is not uniform (there is concentration in the middle thanks to the distribution of {Y+Y}).

We then seriously questioned the validity (1). After some discussion, Imre eventually said something about higher dimensional concentration that made me think one should check (1) for the “Gaussian.” The reason Gaussian is in quotes is that it is not finitely valued as assumed in (1), so strictly speaking we cannot check it for the Gaussian. To see if there was hope, I looked at the differential entropy of a real valued random variable {G} with density {p} defined via

\displaystyle H(G) := -\int_{-\infty}^{\infty} p(x) \log p(x) dx.

Let us take {G} to be the Gaussian with mean zero (this is irrelevant for entropy) and variance 1. Recall some basic properties of variance:

\displaystyle {\rm Var}(aG) = a^2 {\rm Var}(G) , \ \ {\rm Var}(G+G) = 2 {\rm Var}(G),

where {a \in \mathbb{R}} and {G+G} is understood to be the sum of two independent copies of {G}. Thus

\displaystyle {\rm Var}(G + 2 \cdot G) = 5 , \ \ {\rm Var}(G + G +G ) = 3.

So we indeed see that (1) is not true for the Gaussian. To construct a finitely valued random variable that does not satisfy (1), we can convolve a Bernoulli random variable with itself until (1) is not satisfied (assuming that going from discrete to continuous does not destroy (1) which is not obvious without checking as {2 \cdot X} has a strange support condition, for instance the same argument would prove H(2 \cdot G) \geq H(G+G) which is clearly not true for discrete random variables). Anyways, I wrote some quick python code to check this and found that for {X = B + B + B} where {B} is the random variable of a fair coin flip, we have

\displaystyle H(X+2 \cdot X) \approx 2.984 , \ \ H(X+X+X) \approx 2.623.

Here {X+ 2 \cdot X} and {X+X+X} are supported on {\{0 , \ldots , 9\}} and so their entropies are bounded by the entropy of uniform distribution on 10 elements which is

\displaystyle \log_2 10 \approx 3.322.

Sometimes entropy analogs of sumset inequalities hold and sometimes they do not (see this paper of Ruzsa or this paper of Tao, or a host of work by Madiman and coauthors).

Sum of a set and its dilate

Let {A, B \subset \mathbb{Z}} be finite and nonempty. One of the first things we learn in additive combinatorics is

\displaystyle |A+B| \geq |A| +|B| - 1. \ \ \ \ \ (1)

There are a host of results improving on (1) when {B} has certain structure. For instance, all of the following have an short and elementary proof. Here {q \cdot A := \{qa : a \in A\}}.

  • (1) {|A+2\cdot A| \geq 2 |A| - 3}
  • (2) {|A+A -A| \gg |A|^2}, for {A} convex,
  • (3) {|A+AA| \geq |A|^2 + |A| - 1},
  • (4) {|A + A^{-1}| \geq |A|^2}.

Let’s see one way to prove (1). After translation and dilation, we may suppose that {A} contains both even and odd numbers. Let {A_0} be the even numbers in {A} and {A_1} be the odd numbers in {A}. By (1),

\displaystyle |A+2 \cdot A| = |A_0 + 2 \cdot A| + |A_1 + 2 \cdot A| \geq |A_0| + |A| - 1 + |A_1| + |A| - 1 = 3 |A| - 2.

We remark that there is another proof of this in which one picks {2 |A| - 3} elements of {A+2 \cdot A} in increasing order.

Antal Balog and I considered the question of improving (1) when {B = q \cdot A }. We proved the following.

Theorem 1: Let {q\geq 1} be an integer and {A \subset \mathbb{Z}} be finite. Then

\displaystyle |A+q \cdot A| \geq (q + 1) |A| - O_q(1). \ \ \ \spadesuit

This is best possible up to the constant as can be seen by taking {A} to be an arithmetic progression. Below in the fold, we will a proof in the case where {q} is a prime. This case is technically easier, and was first handled by Cilleruelo, Hamidoune, and Serra. Already the case {q=3} is non–trivial, though a proof of {|A+3 \cdot A| \geq 4 |A| - 4} had appeared several times in the literature. We adopt the proof from our paper, which can be modified to handle the general case.

We partition {A} into residue classes modulo {q} as follows:

\displaystyle A = \displaystyle \bigcup_{j = 1}^s A_j , \ \ A_j = a_j + q \cdot A_j^{\prime}, \ \ A_j \neq \emptyset ,\ \ 0 \leq a_j < q.\ \ \ \ \ (2)

We say {A} is fully–distributed mod {q} (FD) if {s=q} or equivalently {A} intersects every residue class modulo {q}. The key fact is that not only the sets {A_i} are disjoint, but so are the sets {A_i + q \cdot A} and we have

\displaystyle A + q \cdot A= \coprod_{i =1}^s A_i + q \cdot A.\ \ \ \ \ (3)

As long as {|A| \geq 2}, we may assume that {s \geq 2} since we may replace {A} with {\frac{1}{q} \cdot (A - x)}. This is actually a key point: Theorem 1 is translation and dilation invariant. We first show that Theorem 1 is true for random sets.

Lemma (FD): Suppose {A} is FD. Then

\displaystyle |A+q \cdot A| \geq (q+1)|A| - q. \ \ \ \spadesuit

Proof: By (1) and (3), we have

\displaystyle |A+q \cdot A| = \sum_{j=1}^q |A_j + q \cdot A| \geq \sum_{j=1}^q \left(|A_j |+ |A| - 1 \right) = (q + 1)|A| - q. \ \ \ \spadesuit

Now we are in a position to prove Theorem 1.

Proof of Theorem 1: We prove

\displaystyle |A+q \cdot A| \geq \frac{m}{q} |A| - C_m,\ \ \ \ \ (4)

by induction on {m}. For {m = 2q}, this follows from (1) with {C_m = 1}. We now assume (4) holds for some {m < q(q+1)} and show it for {m+1}.

Suppose that there is some {1 \leq i \leq s} such that {|A_i| \leq q^{-1} |A|}. Then by (1), (4) and {m < q(q+1)}, we have

\displaystyle |A+q \cdot A| \geq |A_i + q \cdot A| + |(A \setminus A_i) + q \cdot (A \setminus A_i)|

\displaystyle \geq |A_i| + |A| - 1 + \frac{m}{q} (|A| - |A_i|) - C_m \geq \frac{m+1}{q} |A| - C_m - 1.

Suppose now that {A_i'}, as defined in (2) is FD for all {1 \leq i \leq s}. By (3) and translation and dilation invariance of {|X+ q \cdot X|}, we have

\displaystyle |A+q \cdot A| \geq \sum_{j=1}^s |A_j + q \cdot A_j| = \sum_{j=1}^s |A'_j + q \cdot A'_j| \geq (q+1) |A| - q^2.

Thus we may suppose that

\displaystyle |A_i| \geq \frac{|A|}{q} , \ \ 1 \leq i \leq s,\ \ \ \ \ (5)

and there is some {k} such that {A'_k} is not FD. We then apply the following lemma.

Lemma (not FD): Suppose {A_k'} is not FD. Then

\displaystyle |A_j+ q \cdot A| \geq |A_j + q \cdot A_j| + \min_{1 \leq m \leq s} |A_m|. \ \ \ \spadesuit

Let’s use Lemma (not FD) to finish the proof. By Lemma (not FD), (4) twice and (5) we have

\displaystyle |A+q \cdot A| \geq |A_k + q \cdot A| + |(A \setminus A_k) + q \cdot (A \setminus A_k)|

\displaystyle \geq |A_k + q \cdot A_k| + \min_{1 \leq m \leq s} |A_m| + \frac{m}{q} (|A| - |A_k|) - C_m

\displaystyle \geq \frac{m}{q} |A_k| - C_m + \frac{|A|}{q} + \frac{m}{q} (|A| - |A_k|) - C_m

\displaystyle = \frac{m+1}{q} |A| - 2 C_m.

This completes the proof modulo Lemma (not FD).

Proof of Lemma (not FD): Suppose

\displaystyle |A_k + q \cdot A| < |A_k + q \cdot A_k| + \displaystyle \min_{1 \leq m \leq s} |Q_m|.

Then for any {1\leq m \leq s}, we have

\displaystyle |A_m^{\prime}| = |A_m| > |(A_k+ q \cdot A_m )\setminus (A_k+ q \cdot A_k) | = |(a_m - a_k + A_k^{\prime} + q \cdot A_m^{\prime}) \setminus (A_k^{\prime} + q \cdot A_k^{\prime})|.

It follows that for every {x \in A_k^{\prime}} there is a {y \in A_m^{\prime}} such that {a_m-a_k+x+qy\in A'_k + q\cdot A'_k}, and so there is an {x'\in A'_k} such that

\displaystyle a_m - a_k + x \equiv x' \pmod q.

We may repeat this argument with {x'} in place of {x} and so on to get that for every {x \in A'_k} and {w \in \mathbb{Z}}, there is {z \in A'_k} such that

\displaystyle w(a_m - a_k) + x = z \pmod q.

Since {q} is prime, we get that {A'_k} is FD, as desired. {\spadesuit}

Since this work, I worked out the general case of sum of many dilates. After that, Antal Balog and I worked on the analogous problem in vector spaces. Both papers contain several open problems and one of them appears on my problems page. Further, there is a paper of Breuillard and Green on contraction maps, for which sum of dilates is a special case. Also, feel free to see my short video I made on the topic, which highlights some of the problems.

One natural generalization that remains open is

\displaystyle |A+q \cdot A| \geq (|q| + 2d -1)|A| - O_{d , q}(1),

where {A \subset \mathbb{Z}^d} and not contained in any {d-1} dimensional subspace. The cases {d=2} and {d=3} are known to hold.

Breaking the 6/5 threshold for sums and products modulo a prime

Ilya Shkredov and I just released our preprint, “Breaking the 6/5 threshold for sums and products modulo a prime.”

Since this paper of Roche–Newton, Rudnev, and Shkredov, the best bound for the sum–product problem in {\mathbb{F}_p} was given by the following.

Theorem (Sum–Product): Let {A \subset \mathbb{F}_p} of size at most {p^{5/8}}. Then

\displaystyle |A+A| + |AA| \gg |A|^{6/5}. \ \ \spadesuit

The major breakthrough was a point–plane incidence theorem of Rudnev. Let us recall their proof, which has been simplified since their original.

We consider the set of points and lines

\displaystyle P = (A+A) \times (AA), \ \ \ \mathcal{L} = \{(u,v) \in \mathbb{F}_p^2 : v = a(u - c)\}_{a , c \in A}.

Note that each line contains the points {(b + c , ab)} and so there are at least {|A|^3} incidences. On the other hand, one may apply a point–line incidence, for instance the cartesian version found in Theorem 4 of this paper of Stevens and Zeeuw to obtain an upper bound for the number of incidences between {P} and {\mathcal{L}}. Combining these two bounds gives Theorem 1.

We are able to make the following improvement.

Theorem 1: Let {A \subset \mathbb{F}_p} of size at most {p^{3/5}}. Then

\displaystyle |A+A| + |AA| \gg |A|^{6/5 + c}, \ \ \ c = 4/305. \ \ \spadesuit

We can improve this application of a Szemerédi–Trotter type bound in the specific case of the sum–product problem. The basic idea is to evaluate the above proof and obtain a bound for the {\tau}–rich lines, instead of incidences. We then obtain the bound for the higher order energy:

\displaystyle d_4^+(A) \lesssim |AA|^2 |A|^{-2} , \ \ d_4^+(A) := E_4^+(A,B) |A|^{-1} |B|^{-3}.\ \ \ \ \ (1)

A simple application of Cauchy–Schwarz gives

\displaystyle |A+A|^{4/3} d_4^+(A)^{-1/3} \leq |A+A|. \ \ \ \ \ (2)

Combining (1) and (2) recovers Theorem 1. We are able to make an improvement to (2) and in turn to the sum–product problem.