Minimum number of additive tuples in groups of prime order

For a prime number $p$ and a sequence of integers $a_0,\dots,a_k\in \{0,1,\dots,p\}$, let $s(a_0,\dots,a_k)$ be the minimum number of $(k+1)$-tuples $(x_0,\dots,x_k)\in A_0\times\dots\times A_k$ with $x_0=x_1+\dots + x_k$, over subsets $A_0,\dots,A_k\subseteq\mathbb{Z}_p$ of sizes $a_0,\dots,a_k$ respectively. An elegant argument of Lev (independently rediscovered by Samotij and Sudakov) shows that there exists an extremal configuration with all sets $A_i$ being intervals of appropriate length, and that the same conclusion also holds for the related problem, reposed by Bajnok, when $a_0=\dots=a_k=:a$ and $A_0=\dots=A_k$, provided $k$ is not equal 1 modulo $p$. By applying basic Fourier analysis, we show for Bajnok's problem that if $p\ge 13$ and $a\in\{3,\dots,p-3\}$ are fixed while $k\equiv 1\pmod p$ tends to infinity, then the extremal configuration alternates between at least two affine non-equivalent sets.


Introduction
Let Γ be a given finite Abelian group, with the group operation written additively.
Corollary 2. For every k 2 with k ≡ 1 (mod p) and a ∈ [0, p], there is t ∈ Z p such that s k (a) = s k ([t, t + a − 1]).
Unfortunately, if k 3, then there may be sets A different from APs that attain equality in Corollary 2 with s k (|A|) > 0 (which is in contrast to the case k = 2). For example, our (non-exhaustive) search showed that this happens already for p = 17, when Also, already the case k = 2 of the more general Theorem 1 exhibits extra solutions. Of course, by analysing the proof of Theorem 1 or Corollary 2 one can write a necessary and sufficient condition for the cases of equality. We do this in Section 2; in some cases this condition can be simplified.
However, by using basic Fourier analysis on Z p , we can describe the extremal sets for Corollary 2 when k ≡ 1 (mod p) is sufficiently large.
Theorem 3. Let a prime p 7 and an integer a ∈ [3, p − 3] be fixed, and let k ≡ 1 (mod p) be sufficiently large. Then there exists t ∈ Z p for which the only s k (a)-extremal sets are ξ · [t, t + a − 1] for all non-zero ξ ∈ Z p . Problem 4. Find a 'good' description of all extremal families for Corollary 2 (or perhaps Theorem 1) for k 3.
While Corollary 2 provides an example of an s k (a)-extremal set for k ≡ 1 (mod p), the case k ≡ 1 (mod p) of the s k (a)-problem turns out to be somewhat special. Here, translating a set A has no effect on the quantity s k (A). More generally, let A be the group of all invertible affine transformations of Z p , that is, it consists of maps x → ξ · x + η, x ∈ Z p , for ξ, η ∈ Z p with ξ = 0. Then s k (α(A)) = s k (A), for every k ≡ 1 (mod p) and α ∈ A. (2) Let us call two subsets A, By (2), we need to consider sets only up to this equivalence. Trivially, any two subsets of Z p of size a are equivalent if a 2 or a p − 2. Again using Fourier analysis on Z p , we show the following result.
Theorem 5. Let a prime p 7 and an integer a ∈ [3, p − 3] be fixed, and let k ≡ 1 (mod p) be sufficiently large. Then the following statements hold for the s k (a)-problem. (b) I is the unique extremal set for infinitely many k; (c) s k (a) < s k (I ) for infinitely many k, provided there are at least three nonequivalent a-subsets of Z p .
It is not hard to see that there are at least three non-equivalent a-subsets of Z p if and only if p 13 and a ∈ [3, p − 3], or p 11 and a ∈ [4, p − 4]. Thus Theorem 5 characterises pairs (p, a) for which there exists an a-subset A which is s k (a)-extremal for all large k ≡ 1 (mod p). Problem 7. Given a ∈ [3, p − 3], find a 'good' description of all a-subsets of Z p that are s k (a)-extremal for at least one (resp. infinitely many) values of k ≡ 1 (mod p).
Problem 8. Is it true that for every a ∈ [3, p − 3] there is k 0 such that for all k k 0 with k ≡ 1 (mod p), any two s k (a)-extremal sets are affine equivalent?

Proof of Theorem 1
Here we prove Theorem 1 by adopting the proof of Samotij and Sudakov [7].
Let A 1 , . . . , A k be subsets of Z p . Define σ(x; A 1 , . . . , A k ) as the number of k-tuples (x 1 , . . . , x k ) ∈ A 1 × · · · × A k with x = x 1 + · · · + x k . Also, for an integer r 0, let These notions are related to our problem because of the following easy identity: Let an interval mean an arithmetic progression with difference 1, i.e. a subset I of Z p of form {x, x + 1, . . . , x + y}. Its centre is x + y/2 ∈ Z p ; it is unique if I is proper (that is, 0 < |I| < p). Note the following easy properties of the sets N r : 1. These sets are nested: 2. If each A i is an interval with centre c i , then N r (A 1 , . . . , A k ) is an interval with centre c 1 + · · · + c k .
We will also need the following result of Pollard [6, Theorem 1].
Theorem 9. Let p be a prime, k 1, and A 1 , . . . , A k be subsets of Z p of sizes a 1 , . . . , a k .
Then for every integer r 1, we have Proof of Theorem 1. Let A 0 , . . . , A k be some extremal sets for the s(a 0 , . . . , a k )-problem. We can assume that 0 < a 0 < p, because s(A 0 , . . . , A k ) is 0 if a 0 = 0 and k i=1 a i if a 0 = p, regardless of the choice of the sets A i . Since p − a 0 , all integers r r 0 + 1.
The nested intervals , with t independent of r, which has as small as possible intersection with each N r -interval above given their sizes, that is, This and Pollard's theorem give the following chain of inequalities: giving the required. Let us write a necessary and sufficient condition for equality in Theorem 1 in the case a 0 , . . . , a k ∈ [1, p − 1]. Let r 0 0 be defined by (5)- (6). Then, by (4), a sequence A 0 , . . . , A k ⊆ Z p of sets of sizes respectively a 0 , . . . , a k is extremal if and only if Let us now concentrate on the case k = 2, trying to simplify the above condition. We can assume that no a i is equal to 0 or p (otherwise the choice of the other two sets has no effect on s(A 0 , A 1 , A 2 ) and every triple of sets of sizes a 0 , a 1 and a 2 is extremal). Also, as in [7], let us exclude the case s(a 0 , a 1 , a 2 ) = 0, as then there are in general many extremal configurations. Note that s(a 0 , a 1 , a 2 ) = 0 if and only if r 0 = 0; also, by the Cauchy-Davenport theorem (the special case k = 2 and r = 1 of Theorem 9), this is equivalent to a 1 + a 2 − 1 p − a 0 . Assume by symmetry that a 1 a 2 . Note that (5) implies that r 0 a 1 .
The condition in (10) states that we have equality in Pollard's theorem. A result of Nazarewicz, O'Brien, O'Neill and Staples [5,Theorem 3] characterises when this happens (for k = 2), which in our notation is the following.
Theorem 10. For k = 2 and 1 r 0 a 1 a 2 < p, we have equality in (10) if and only if at least one of the following conditions holds: 3. a 1 = a 2 = r 0 + 1 and A 2 = g − A 1 for some g ∈ Z p , 4. A 1 and A 2 are arithmetic progressions with the same difference.
Let us try to write more explicitly each of these four cases, when combined with (8) and (9).
Next, suppose that a 1 + a 2 p + r 0 . Then, for any two sets A 1 and A 2 of sizes a 1 and a 2 , we have N r 0 (A 1 , A 2 ) = Z p ; thus (9) holds automatically. Similarly to the previous case, there does not seem to be a nice characterisation of (8 Next, suppose that we are in the third case. The primality of p implies that g ∈ Z p satisfying A 2 = g −A 1 is unique and thus N r 0 +1 (A 1 , A 2 ) = {g}. Therefore (8) is equivalent to A 0 g. Also, note that if I 1 and I 2 are intervals of size r 0 + 1, then n r 0 (I 1 , I 2 ) = 3. By the definition of r 0 , we have p − 2 a 0 p − 1. Thus we can choose any integer , and then let A 0 be obtained from Z p by removing g and at most one further element of N r 0 (A 1 , A 2 ). Here, A 0 is always an AP (as a subset of Z p of size a 0 p − 2) but A 1 and A 2 need not be.
Finally, let us show that if A 1 and A 2 are arithmetic progressions with the same difference d and we are not in Case 1 nor 2 of Theorem 10, then A 0 is also an arithmetic progression whose difference is d. By (1), it is enough to prove this when A 1 = [a 1 ] and A 2 = [a 2 ] (and d = 1). Since a 1 + a 2 p − 1 + r 0 and r 0 + 1 a 1 a 2 , we have that have sizes respectively a 1 + a 2 − 2r 0 + 1 < p and a 1 + a 2 − 2r 0 − 1 > 0. We see that N r 0 +1 (A 1 , A 2 ) is obtained from the proper interval N r 0 (A 1 , A 2 ) by removing its two endpoints. Thus A 0 , which is sandwiched between the complements of these two intervals by (8) Parseval's identity states that The convolution of two functions f, g : Z p → C is given by It is not hard to show that the Fourier transform of a convolution equals the product of Fourier transforms, i.e.
the electronic journal of combinatorics 26(1) (2019), #P1.30 We write f * k for the convolution of f with itself k times. (So, for example, f * 2 = f * f .) Denote by 1 A the indicator function of A ⊆ Z p which assumes value 1 on A and 0 on Z p \ A. We will call 1 A (0) = |A| the trivial Fourier coefficient of A. Since the Fourier transform behaves very nicely with respect to convolution, it is not surprising that our parameter of interest, s k (A), can be written as a simple function of the Fourier coefficients of 1 A . Indeed, let A ⊆ Z p and x ∈ Z p . Then the number of tuples (a 1 , . . . , a k ) ∈ A k such that a 1 + . . . + a k = x (which is σ(x; A, . . . , A) in the notation of Section 2) is precisely . The function s k (A) counts such a tuple if and only if its sum x also lies in A. Thus, Since every set A ⊆ Z p of size a has the same trivial Fourier coefficient (namely 1 A (0) = a), let us re-write (13) as Thus we need to minimise F (A) (which is a real number for any A) over a-subsets A ⊆ Z p .
To do this when k is sufficiently large, we will consider the largest in absolute value non- In order to prove Theorems 3 and 5, we will make some preliminary observations about these special sets. The set of a-subsets which are affine equivalent to I is precisely the set of a-APs. Next we will show that F (I) = 2 Note that (−1) γ(a−1)(k−1) equals (−1) γ if both a, k are even and 1 otherwise. To see (15), let γ ∈ {1, . . . , p−1 and e −πi(a−1)s equals 1 if (a − 1)s is even, and −1 if (a − 1)s is odd. Note that, since p is an odd prime, (a − 1)s is odd if and only if a and k are both even. So (16) is real, and the fact that 1 I (p − γ) = 1 I (γ) implies that the corresponding term for p − γ is the same as for γ. This gives (15). A very similar calculation to (16) shows that Given r > 0 and 0 θ < 2π, we write arg(re θi ) := θ.
Proposition 11. Suppose that p 7 is prime and a ∈ Since ω m+1 , ω −m lie on the unit circle, the argument of ω m+1 + ω −m is either π/p or π + π/p. But the bounds on m imply that it has positive real part, so arg(ω m+1 + ω −m ) = π/p. By looking at the non-degenerate parallelogram in the complex plane with vertices 0, 1 I − (1), ω m+1 + ω −m , 1 I (1), we see that the argument of 1 I (1) lies strictly between that of 1 I − (1) and ω m+1 + ω −m , i.e. strictly between 0 and π/p, giving the required. We say that an a-subset A is a punctured interval if A = I + t or A = −I + t for some t ∈ Z p . That is, A can be obtained from an interval of length a + 1 by removing a penultimate point.
Lemma 12. Let p 7 be prime and let a ∈ {3, . . . , p − 3}. Then the sets I, I ⊆ Z p are not affine equivalent. Thus no punctured interval is affine equivalent to an interval.
Proof. Suppose on the contrary that there is α ∈ A with α(I ) = I. Let a reflection mean an affine map R c with c ∈ Z p that maps x to −x + c. Clearly, I = [a] is invariant under the reflection R := R a−1 . Thus I is invariant under the map R := α −1 • R • α. As is easy to see, R is also some reflection and thus preserves the cyclic distances in Z p . So R has to fix a, the unique element of I with both distance-1 neighbours lying outside of I . Furthermore, R has to fix a − 2, the unique element of I at distance 2 from a. However, no reflection can fix two distinct elements of Z p , a contradiction.
We remark that the previous lemma can also be deduced from Proposition 11. Indeed, for any A ⊆ Z p , the multiset of Fourier coefficients of A is the same as that of x · A for x ∈ Z p \ {0}, and translating a subset changes the argument of Fourier coefficients by an integer multiple of 2π/p. Thus for every subset which is affine equivalent to I, the argument of each of its Fourier coefficients is an integer multiple of π/p. Proof. Given D ∈ Zp a , we claim that there is some D pri ∈ Zp a with the following properties: • D pri is affine equivalent to D; • ρ(D) = |1 D pri (1)|; and the electronic journal of combinatorics 26(1) (2019), #P1.30 Call such a D pri a primary image of D. Indeed, suppose that ρ(D) = |1 D (γ)| for some non-zero γ ∈ Z p , and let 1 D (γ) = r e θ i for some r > 0 and 0 θ < 2π. (Note that we have r > 0 since p is prime.) Choose ∈ {0, . . . , p − 1} and −π/p < φ π/p such that θ = 2π /p + φ. Let D pri := γ · D + . Then as required.
Let D ⊆ Z p have size a and write 1 D (1) = re θi . Assume by the above that −π/p < θ π/p. For all j ∈ Z p , let where (z) denotes the real part of z ∈ C. Given any a-subset E of Z p , we have Informally speaking, the main idea of the proof is that if we fix the direction e iθ , then the projection length is maximised if we take a distinct elements j ∈ Z p with the a largest values of h(j), that is, if we take some interval (with the runner-up being a punctured interval).
Let us provide a formal statement and proof of this now.
Claim 14. Let I a be the set of length-a intervals in Z p . The case when −π/p < θ < 0 is almost identical except now j : We can now prove part (i) of the lemma. Suppose A ∈ Zp with equality in the first inequality if and only if D ∈ M 2 (D). Since C is a punctured interval, it is not affine equivalent to an interval. So the first part of the lemma implies that |1 C (1)| m 2 (a). Thus we have equality everywhere and so D ∈ M 2 (D). Therefore B is the affine image of a punctured interval, as required. Further, if B is a punctured interval, then D is a punctured interval if and only if γ = ±1. This completes the proof of (ii).
We will now prove Theorem 3.
Let t be optimal. To prove the theorem, we will show that F (ξ · (I + t)) < F (A) (and so s k (ξ · (I + t)) < s k (A)) for any a-subset A ⊆ Z p which is not a dilation of I + t.
We will first show that F (I+t) < F (A) for any a-subset A which is not affine equivalent to an interval. By Lemma 13(i), we have that |1 I+t (±1)| = m 1 (a) and ρ(A) m 2 (a).
The remaining case is when A = ζ · (I + v) for some non-optimal v ∈ Z p and non-zero ζ ∈ Z p . Since s k (A) = s k (I + v), we may assume that ζ = 1. Note that cos(θ t ) cos(π − π/p) < cos(π − 2π/p) cos(θ v ). Thus where the last inequality uses the fact that k is sufficiently large. Thus F (I +t) < F (I +v), as required.
Finally, using similar techniques, we prove Theorem 5. Suppose now that at least one of a, k is odd. Let A be an a-set not equivalent to I. Again by Lemma 13, we have 2(m 1 (a)) k+1 − (p − 1)(m 2 (a)) k+1 > 0.
So the interval I and its affine images have in fact the largest number of additive (k + 1)tuples among all a-subsets of Z p . In particular, s k (a) < s k (I).
Suppose that there is some A ∈ Zp a which is not affine equivalent to I or I . (If there is no such A, then the unique extremal sets are affine images of I for all k > k 0 (a, p), giving the required.) Write ρ := re θi = 1 I (1). Then by Lemma 13(ii), we have r = m 2 (a), and ρ(A) m 3 (a). Given k 2, let s ∈ N be such that k = sp + 1. Then Proposition 11 implies that there is an even integer ∈ N for which c := pθ − π ∈ (−π, π) \ {0}. Let ε := 1 3 min{|c|, π − |c|} > 0. Given an integer t, say that s ∈ N is t-good if sc ∈ ((t − 1 2 )π + ε, (t + 1 2 )π − ε). This real interval has length π − 2ε > |c| > 0, so must contain at least one integer multiple of c. In other words, for all t ∈ Z \ {0} with the same sign as c, there exists a t-good integer s > 0. As spθ ≡ sc (mod 2π), the sign of cos(spθ) is (−1) t . Moreover, Lemma 13 implies that m 2 (a) > m 3 (a), m 2 (a). Thus, when k = sp + 1 > k 0 (a, p), the absolute value of 2m 2 (a) k+1 cos(spθ) is greater than the right-hand side of (21). Thus, for large |t|, we have F (A) > F (I ) if t is even and F (A) < F (I ) if t is odd, implying the theorem by (14).