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Multiplicatively irreducibility of small perturbations of shifted kk-th powers

Chi Hoi Yip School of Mathematics
Georgia Institute of Technology
Atlanta, GA 30332
United States
[email protected]
Abstract.

Motivated by a conjecture of Erdős on the additively irreducibility of small perturbations of the set of squares, recently Hajdu and Sárközy studied a multiplicative analogue of the conjecture for shifted kk-th powers. They conjectured that for each k2k\geq 2, if one changes o(X1/k)o(X^{1/k}) elements of Mk={xk+1:x}M_{k}^{\prime}=\{x^{k}+1:x\in\mathbb{N}\} up to XX, then the resulting set cannot be written as a product set ABAB nontrivially. In this paper, we confirm a more general version of their conjecture for k3k\geq 3.

Key words and phrases:
multiplicative decomposition, shifted powers
2020 Mathematics Subject Classification:
11P70, 11B30, 11D45

1. Introduction

Let \mathbb{N} be the set of positive integers. In this paper, we study questions related to additive decompositions and multiplicative decompositions. A set SS\subset\mathbb{N} is said to be multiplicatively reducible if it has a multiplicative decomposition S=AB={ab:aA,bB}S=AB=\{ab:a\in A,b\in B\}, where A,BA,B are subsets of \mathbb{N} with size at least 22. Similarly, SS is additively reducible if it can be written as a sumset A+B={a+b:aA,bB}A+B=\{a+b:a\in A,b\in B\}, where A,BA,B are subset of \mathbb{N} with size at least 22. There is a large amount of literature on the study of additive dnd multiplicative decompositions for sets with different arithmetic structures; we refer to a nice survey by Elsholtz [7].

The set of perfect squares is additively irreducible simply because the gap between consecutive squares tends to infinity. Erdős conjectured that all small perturbations of the set of squares are still additively irreducible.

Conjecture 1.1 (Erdős).

If k2k\geq 2 and we change o(X1/2)o(X^{1/2}) elements of the set of squares up to XX (deleting some of its elements and adding some positive integers), then the new set RR is always additively irreducible.

This conjecture is still open, with the best-known progress due to Sárközy and Szemerédi [18], where they showed that 1.1 holds if one replaces o(X1/2)o(X^{1/2}) with o(X1/2/2(3+ϵ)logX/loglogX)o(X^{1/2}/2^{(3+\epsilon)\log X/\log\log X}) for any ϵ>0\epsilon>0. We also refer to related results by Bienvenu [2] and Leonetti [15] with a probabilistic flavor. It appears that the finite field analogue of 1.1 is even more challenging (if true); we refer to [12, 16, 19, 21] for some partial results.

Recently, Hajdu and Sárközy studied the multiplicative decompositions of polynomial sequences with integer coefficients in a series of three papers [9, 10, 11]. In particular, in the first two papers, they provided a simple proof of the following result: for each k2k\geq 2, if one changes finitely many elements of the set Mk={xk+1:x}M_{k}^{\prime}=\{x^{k}+1:x\in\mathbb{N}\}, then the new set remains multiplicatively irreducible. We also refer to the study of finite field analogues of the same result in [13, 17].

In the third paper of the series [11], Hajdu and Sárközy studied the following multiplicative analogue of 1.1.

Conjecture 1.2 (Hajdu and Sárközy).

If k2k\geq 2 and we change o(X1/k)o(X^{1/k}) elements of the set {xk+1:x}\{x^{k}+1:x\in\mathbb{N}\} up to XX, then the new set RR is always multiplicatively irreducible.

1.2 is best possible in the sense that if one changes a positive proportion of elements from {xk+1:x}\{x^{k}+1:x\in\mathbb{N}\} (equivalently, changes at most cX1/kcX^{1/k} elements of the set {xk+1:x}\{x^{k}+1:x\in\mathbb{N}\} up to XX for some constant c>0c>0), then the resulting set could be multiplicatively reducible. For example, let m2m\geq 2 be an integer and set Rm=ABR_{m}=AB, where A={1,m}A=\{1,m\} and B={xk+1:x}B=\{x^{k}+1:x\in\mathbb{N}\}; note that RmR_{m} is obtained by adding (m1/k+o(1))Xk(m^{-1/k}+o(1))X^{k} elements to {xk+1:x}\{x^{k}+1:x\in\mathbb{N}\} up to XX and m1/k0m^{-1/k}\to 0 as mm\to\infty.

Based on a skillful and sophisticated argument involving various tools from Diophantine approximation, Diophantine equations, extremal graph theory, and multiplicative number theory, Hajdu and Sárközy [11, Theorem 2.1] proved a weaker version of 1.2. More precisely, they proved that 1.2 holds if o(X1/k)o(X^{1/k}) is replaced by

o(X1/kexp((log2+ϵ)logXloglogX))o\bigg{(}X^{1/k}\exp\bigg{(}-(\log 2+\epsilon)\frac{\log X}{\log\log X}\bigg{)}\bigg{)}

for any ϵ>0\epsilon>0. They remarked that both 1.1 and 1.2 “seem to be beyond reach in their original form”.

In this paper, we prove a more general version of 1.2 for k3k\geq 3.

Theorem 1.3.

Let k,nk,n be integers with k3k\geq 3 and n0n\neq 0. If we change o(X1/k)o(X^{1/k}) elements of the set {xk+n:x}\{x^{k}+n:x\in\mathbb{N}\}\cap\mathbb{N} up to XX, then the new set RR is always multiplicative irreducible.

One key ingredient in our proof is the connection of multiplicative decompositions of the set MkM_{k}^{\prime} and the bipartite variants of Diophantine tuples, first considered by the author [20] as an attempt to make some of the results in [9, 10] effective. There are many well-studied generalizations and variants of Diophantine tuples; see the recent book of Dujella [6] for an overview. The relevant variant in our setting is the following bipartite variant: for each k2k\geq 2 and each nonzero integer nn, we call a pair of sets (A,B)(A,B) a bipartite Diophantine tuple with property BDk(n)BD_{k}(n), if A,BA,B are two subsets of \mathbb{N} with size at least 22, such that ab+nab+n is a kk-th power for each aAa\in A and bBb\in B. While this concept was only formally introduced by the author [20] recently, the same objects have been for example studied by Gyarmati [8], Bugeaud and Dujella [3], and Bugeaud and Gyarmati [4], since two decades ago. The connection is the following: if we can give a good absolute upper bound on min{|A|,|B|}\min\{|A|,|B|\} among all bipartite Diophantine tuples (A,B)(A,B) with property BDk(n)BD_{k}(n), then it seems plausible that the following Kövari–Sós–Turán theorem [14] from extremal graph theory can be used to make some partial progress on 1.2.

Lemma 1.4 (Kövari–Sós–Turán theorem).

Let GG be a bipartite graph with vertex classes UU and VV such that |U|=m|U|=m and |V|=n|V|=n. Assume that there do not exist a set XUX\subset U with size ss and a set YVY\subset V with size tt, such that xx and yy are adjacent for all xXx\in X and yYy\in Y. Then the number of edges of GG is at most (s1)1/t(nt+1)m11/t+(t1)m.(s-1)^{1/t}(n-t+1)m^{1-1/t}+(t-1)m.

Our proof techniques cannot handle the case k=2k=2 in 1.2. In fact, it is an open question to show unconditionally that if (A,B)(A,B) is a bipartite Diophantine tuple with property BD2(1)BD_{2}(1), then min{|A|,|B|}\min\{|A|,|B|\} is bounded by an absolute constant [1, 4]; note that this would follow easily if we assume the uniformity conjecture [5]. As for the case for k3k\geq 3, the same question was partially addressed by the author [20]; in particular, it was shown in [20, Theorem 2.2] that if k3k\geq 3 is fixed, and |n||n|\to\infty, then for any bipartite Diophantine tuple (A,B)(A,B) with property BDk(n)BD_{k}(n), we have min{|A|,|B|}klog|n|\min\{|A|,|B|\}\ll_{k}\log|n|. Clearly, such a result is not strong enough for the application to 1.2. Instead, we realize that studying a “local version” of bipartite Diophantine tuples is sufficient to prove 1.2 for k3k\geq 3.

2. Forbidden local structures

In this section, we prove some refined “local estimates” on bipartite Diophantine tuples with some carefully chosen parameters. We list several results from [20] and deduce some useful corollaries. We first introduce the following constants defined in [20]:

s3=6,s4=4,s5=3, and sk=2 for k6;\displaystyle s_{3}=6,\quad s_{4}=4,\quad s_{5}=3,\quad\text{ and }\quad s_{k}=2\quad\text{ for }k\geq 6;
t3=15399938,t4=343,t5=9723,t6=294, and tk=k2+k4k26k+6 for k7.t_{3}=\frac{15399}{938},\quad t_{4}=\frac{34}{3},\quad t_{5}=\frac{97}{23},\quad t_{6}=\frac{29}{4},\text{ and }\quad t_{k}=\frac{k^{2}+k-4}{k^{2}-6k+6}\quad\text{ for }k\geq 7.

The following proposition is one of the key results in [20]. Its proof is based on a combination of an explicit version of the bounds of the number of solutions of Thue inequalities, repeated applications of gap principles, and some combinatorial arguments.

Proposition 2.1 ([20, Proposition 4.1]).

Let k3k\geq 3 and let nn be a nonzero integer. Let A,BA,B\subset\mathbb{N} such that A={a1,a2,,a}A=\{a_{1},a_{2},\ldots,a_{\ell}\} and B={b1,b2,,bm}B=\{b_{1},b_{2},\ldots,b_{m}\} with a1<a2<<aa_{1}<a_{2}<\cdots<a_{\ell} and b1<b2<<bmb_{1}<b_{2}<\cdots<b_{m}, and AB+n{xk:x}AB+n\subset\{x^{k}:x\in\mathbb{N}\}. If k>3k>3, further assume that msk+1m\geq s_{k}+1, 2\ell\geq 2, and a2bmska_{2}\leq b_{m-s_{k}}; if k=3k=3, further assume that m7m\geq 7, 3\ell\geq 3, and a3bm6a_{3}\leq b_{m-6}. Then at most sks_{k} elements in BB are at least 2|n|tk2|n|^{t_{k}}.

For our purpose, we shall use the following immediate corollary of Proposition 2.1. Note that tk17t_{k}\leq 17 and sk6s_{k}\leq 6 for all k3k\geq 3.

Corollary 2.2.

Let k3k\geq 3 and nn be a nonzero integer. There do not exist integers a1,a2,a3a_{1},a_{2},a_{3} and b1,b2,,b7b_{1},b_{2},\ldots,b_{7} with a1<a2<a3b1<b2<<b7a_{1}<a_{2}<a_{3}\leq b_{1}<b_{2}<\ldots<b_{7} and 2|n|17b12|n|^{17}\leq b_{1}, such that aibj+na_{i}b_{j}+n is a kk-th power for all 1i31\leq i\leq 3 and 1j71\leq j\leq 7.

The following lemma is based on a simple gap principle.

Lemma 2.3 ([20, Lemma 3.6]).

Let k3k\geq 3 and let nn be a nonzero integer. Let a,b,c,da,b,c,d are positive integers such that a<ba<b, c<dc<d, and ac2|n|ac\geq 2|n|. Suppose further that ac+n,bc+n,ad+n,bd+nac+n,bc+n,ad+n,bd+n are kk-th powers. Then bdkk(ac)k1/(4k1|n|k)bd\geq k^{k}(ac)^{k-1}/(4^{k-1}|n|^{k}).

Next, we deduce two corollaries of 2.3.

Corollary 2.4.

Let k3k\geq 3 and let nn be a nonzero integer. If X>46(k1)|n|6kX>4^{6(k-1)}|n|^{6k}, then there do not exist integers a1,a2,b1,b2a_{1},a_{2},b_{1},b_{2} with X1/3<a1<a2X1/2<b1<b2XX^{1/3}<a_{1}<a_{2}\leq X^{1/2}<b_{1}<b_{2}\leq X such that aibj+na_{i}b_{j}+n are kk-th powers for all 1i,j21\leq i,j\leq 2.

Proof.

Suppose otherwise that there do exist a1,a2,b1,b2a_{1},a_{2},b_{1},b_{2} with the required property. Note that we have a1b1X5/6>2|n|a_{1}b_{1}\geq X^{5/6}>2|n|. Then by Lemma 2.3, we have a2b2kk(a1b1)k1/(4k1|n|k)a_{2}b_{2}\geq k^{k}(a_{1}b_{1})^{k-1}/(4^{k-1}|n|^{k}). It follows from the assumptions X>46(k1)|n|6kX>4^{6(k-1)}|n|^{6k} and X1/3<a1<a2X1/2<b1<b2XX^{1/3}<a_{1}<a_{2}\leq X^{1/2}<b_{1}<b_{2}\leq X that

b2kk(a1b1)k14k1|n|ka214k1|n|ka12b12a214k1|n|kX1/6b12b12>X,b_{2}\geq\frac{k^{k}(a_{1}b_{1})^{k-1}}{4^{k-1}|n|^{k}a_{2}}\geq\frac{1}{4^{k-1}|n|^{k}}\cdot\frac{a_{1}^{2}b_{1}^{2}}{a_{2}}\geq\frac{1}{4^{k-1}|n|^{k}}\cdot X^{1/6}b_{1}^{2}\geq b_{1}^{2}>X,

violating the assumption that b2Xb_{2}\leq X. ∎

Corollary 2.5.

Let k3k\geq 3 and let nn be a nonzero integer. Let a1,a2a_{1},a_{2} be positive integers with a1<a2a_{1}<a_{2}. Then for XX sufficiently large, the number of positive integers bXb\leq X such that a1b+na_{1}b+n and a2b+na_{2}b+n are both kk-th powers are at most 2loglogX2\log\log X.

Proof.

Suppose that b1<b2<<bm<Xb_{1}<b_{2}<\cdots<b_{m}<X are positive integers such that a1bi+na_{1}b_{i}+n and a2bi+na_{2}b_{i}+n are both kk-th powers for each 1im1\leq i\leq m. For each 1im11\leq i\leq m-1, applying 2.3 to a1,a2,bi,bi+1a_{1},a_{2},b_{i},b_{i+1}, we get

bi+1kk(a1bi)k14k1|n|ka2Cbi2,b_{i+1}\geq\frac{k^{k}(a_{1}b_{i})^{k-1}}{4^{k-1}|n|^{k}a_{2}}\geq Cb_{i}^{2},

where CC is a positive constant depending on k,n,a1,a2k,n,a_{1},a_{2}. It follows that for each 1im11\leq i\leq m-1, we have logbi+12logbi+logC\log b_{i+1}\geq 2\log b_{i}+\log C and thus logbi+1+logC2(logbi+C)\log b_{i+1}+\log C\geq 2(\log b_{i}+C). Choose the smallest jj such that logbj+logC1\log b_{j}+\log C\geq 1; note that if such jj does not exist, then we have me/Cm\leq e/C and we are already done. Since bjjb_{j}\geq j, we have je/Cj\leq e/C. It follows that

logX+logClogbm+logC2mj(logbj+logC)2mj2me/C\log X+\log C\geq\log b_{m}+\log C\geq 2^{m-j}(\log b_{j}+\log C)\geq 2^{m-j}\geq 2^{m-e/C}

and thus m2loglogXm\leq 2\log\log X for XX sufficiently large. ∎

3. Proof of the main result

Our proof is inspired by several arguments used in [11, 20]. In the following, we use the following standard notation: for each set AA\subset\mathbb{N} and each positive real number XX, we write A(X)=|{xA:xX}|A(X)=|\{x\in A:x\leq X\}|.

Proof of Theorem 1.3.

Write M={xk+n:x}M=\{x^{k}+n:x\in\mathbb{N}\}\cap\mathbb{N}. We can write R=QSR=Q\cup S, where QMQ\subset M and SM=S\cap M=\emptyset. By the assumption on RR, we have Q(X)=(1o(1))X1/kQ(X)=(1-o(1))X^{1/k} and S(X)=o(X1/k)S(X)=o(X^{1/k}). For the sake of contradiction, suppose that RR is multiplicatively reducible. Then we can write R=ABR=A\cdot B, where A,BA,B are subsets of \mathbb{N} with size at least 22.

Claim 3.1.

Let X0max{4|n|34,46(k1)|n|6k}X_{0}\geq\max\{4|n|^{34},4^{6(k-1)}|n|^{6k}\} be a sufficiently large number such that Q(X)12X1/kQ(X)\geq\frac{1}{2}X^{1/k} whenever XX0X\geq X_{0}. If XX0X\geq X_{0}, then one of the following holds:

  1. (1)

    max{A(X),B(X)}140X1/k\max\{A(X),B(X)\}\geq\frac{1}{40}X^{1/k};

  2. (2)

    max{A(X1/2),B(X1/2)}140X1/2k\max\{A(X^{1/2}),B(X^{1/2})\}\geq\frac{1}{40}X^{1/2k};

  3. (3)

    max{A(X1/3),B(X1/3)}140X1/3k\max\{A(X^{1/3}),B(X^{1/3})\}\geq\frac{1}{40}X^{1/3k}.

Proof of claim.

Let XX0X\geq X_{0} be fixed. Write

A1=A[1,X1/3],A2=A(X1/3,X1/2],A3=A(X1/2,X],A_{1}=A\cap[1,X^{1/3}],\quad A_{2}=A\cap(X^{1/3},X^{1/2}],\quad A_{3}=A\cap(X^{1/2},X],

and similarly define sets B1,B2,B3B_{1},B_{2},B_{3}. For each integer xR[1,X]x\in R\cap[1,X], it can be written as x=abx=ab for some aAa\in A and bBb\in B with 1a,bx1\leq a,b\leq x; note that we always have either axa\leq\sqrt{x} or bxb\leq\sqrt{x}. Thus,

Q[1,X]\displaystyle Q\cap[1,X] R[1,X]\displaystyle\subset R\cap[1,X]
((A1A2)(B1B2B3))((A1A2A3)(B1B2))\displaystyle\subset((A_{1}\cup A_{2})\cdot(B_{1}\cup B_{2}\cup B_{3}))\cup((A_{1}\cup A_{2}\cup A_{3})\cdot(B_{1}\cup B_{2}))
=((A1A2)(B1B2))(A1B3)(A2B3)(B1A3)(B2A3).\displaystyle=((A_{1}\cup A_{2})\cdot(B_{1}\cup B_{2}))\cup(A_{1}\cdot B_{3})\cup(A_{2}\cdot B_{3})\cup(B_{1}\cdot A_{3})\cup(B_{2}\cdot A_{3}).

Observe that if abQab\in Q, then abnab-n is a kk-th power. Next, we consider the contribution of the five sets on the right-hand side of the above equation to Q[1,X]Q\cap[1,X].

  1. (1)

    Clearly, the number of pairs (a,b)(A1A2)×(B1B2)(a,b)\in(A_{1}\cup A_{2})\times(B_{1}\cup B_{2}) such that abQab\in Q is at most |A1A2||B1B2|=A(X1/2)B(X1/2)|A_{1}\cup A_{2}||B_{1}\cup B_{2}|=A(X^{1/2})B(X^{1/2}).

  2. (2)

    Build a bipartite graph GG with vertex classes A1A_{1} and B3B_{3} such that for each aA1a\in A_{1}, and bB3b\in B_{3}, there is an edge between aa and bb if and only if abnab-n is a kk-th power. Then the number of pairs (a,b)A1×B3(a,b)\in A_{1}\times B_{3} such that abQab\in Q is at most the number of edges of GG. Since X04|n|34X_{0}\geq 4|n|^{34}, we have minbB3bXX0=2|n|17\min_{b\in B_{3}}b\geq\sqrt{X}\geq\sqrt{X_{0}}=2|n|^{17}. Also note that we have maxaAaX1/3<X1/2minbB3b\max_{a\in A}a\leq X^{1/3}<X^{1/2}\leq\min_{b\in B_{3}}b. Thus, by Corollary 2.2, there do not exist three distinct elements a1,a2,a3Aa_{1},a_{2},a_{3}\in A, and seven distinct elements b1,b2,,b7Bb_{1},b_{2},\cdots,b_{7}\in B such that aibjna_{i}b_{j}-n is a kk-th power for each i{1,2}i\in\{1,2\} and 1j71\leq j\leq 7. Thus, applying the Kövari–Sós–Turán theorem (Lemma 1.4) with U=B,V=A,s=7,t=3U=B,V=A,s=7,t=3, the number of edges of GG is at most

    63|A1||B3|11/3+2|B3|2A(X1/3)B(X)2/3+2B(X).\sqrt[3]{6}|A_{1}||B_{3}|^{1-1/3}+2|B_{3}|\leq 2A(X^{1/3})B(X)^{2/3}+2B(X).
  3. (3)

    Since X046(k1)|n|6kX_{0}\geq 4^{6(k-1)}|n|^{6k}, Corollary 2.4 implies that there do not exist two distinct elements a1,a2A2a_{1},a_{2}\in A_{2}, and two distinct elements b1,b2B3b_{1},b_{2}\in B_{3} such that aibjna_{i}b_{j}-n is a kk-th power for each 1i,j21\leq i,j\leq 2. Thus, similar to the analysis in (2), the Kövari–Sós–Turán theorem implies that the number of pairs (a,b)A2×B3(a,b)\in A_{2}\times B_{3} such that abQab\in Q is at most 2|A2||B3|11/2+|B3|2A(X1/2)B(X)1/2+B(X).\sqrt{2}|A_{2}||B_{3}|^{1-1/2}+|B_{3}|\leq 2A(X^{1/2})B(X)^{1/2}+B(X).

  4. (4)

    Similar to (2), the number of pairs (b,a)B1×A3(b,a)\in B_{1}\times A_{3} such that abQab\in Q is at most 2B(X1/3)A(X)2/3+2A(X)2B(X^{1/3})A(X)^{2/3}+2A(X).

  5. (5)

    Similar to (3), the number of pairs (b,a)B2×A3(b,a)\in B_{2}\times A_{3} such that abQab\in Q is at most 2B(X1/2)A(X)1/2+A(X)2B(X^{1/2})A(X)^{1/2}+A(X).

It follows that

Q(X)\displaystyle Q(X) A(X1/2)B(X1/2)+2A(X1/3)B(X)2/3+2B(X)+2A(X1/2)B(X)1/2+B(X)\displaystyle\leq A(X^{1/2})B(X^{1/2})+2A(X^{1/3})B(X)^{2/3}+2B(X)+2A(X^{1/2})B(X)^{1/2}+B(X)
+2B(X1/3)A(X)2/3+2A(X)+2B(X1/2)A(X)1/2+A(X).\displaystyle\quad+2B(X^{1/3})A(X)^{2/3}+2A(X)+2B(X^{1/2})A(X)^{1/2}+A(X). (1)

Suppose the statement of the claim is false; then inequality (1) implies that

Q(X)\displaystyle Q(X) (140+240+240+240+140+240+240+240+140)X1/k<12X1/k,\displaystyle\leq\bigg{(}\frac{1}{40}+\frac{2}{40}+\frac{2}{40}+\frac{2}{40}+\frac{1}{40}+\frac{2}{40}+\frac{2}{40}+\frac{2}{40}+\frac{1}{40}\bigg{)}X^{1/k}<\frac{1}{2}X^{1/k},

contradicting the assumption that Q(X)12X1/kQ(X)\geq\frac{1}{2}X^{1/k}. This finishes the proof of claim. ∎

By 3.1, there is an increasing sequence (Xj)j=1(X_{j})_{j=1}^{\infty} with XjX_{j}\to\infty, such that

max{A(Xj),B(Xj)}140Xj1/k\max\{A(X_{j}),B(X_{j})\}\geq\frac{1}{40}X_{j}^{1/k}

for each jj\in\mathbb{N}. Without loss of generality, by passing to a subsequence, we may assume that B(Xj)140Xj1/kB(X_{j})\geq\frac{1}{40}X_{j}^{1/k} for each jj\in\mathbb{N}. Since |A|2|A|\geq 2, so we can pick two fixed elements a1,a2Aa_{1},a_{2}\in A with a1<a2a_{1}<a_{2}. For each bB[1,X]b\in B\cap[1,X], we have a1b<a2ba2Xa_{1}b<a_{2}b\leq a_{2}X. Thus, the number of bB[1,X]b\in B\cap[1,X] with aibSa_{i}b\in S for some i{1,2}i\in\{1,2\} is at most S(a2X)=o((a2X)1/k)=o(X1/k)S(a_{2}X)=o((a_{2}X)^{1/k})=o(X^{1/k}). It follows that the number of bB[1,Xj]b\in B\cap[1,X_{j}] with aibQa_{i}b\in Q (in particular, aibna_{i}b-n is a kk-th power) for i{1,2}i\in\{1,2\} is at least

B(Xj)o(Xj1/k)140Xj1/ko(Xj1/k)=(140o(1))Xj1/k>180Xj1/k>2loglogXjB(X_{j})-o(X_{j}^{1/k})\geq\frac{1}{40}X_{j}^{1/k}-o(X_{j}^{1/k})=(\frac{1}{40}-o(1))X_{j}^{1/k}>\frac{1}{80}X_{j}^{1/k}>2\log\log X_{j}

for all jj large enough. However, this contradicts 2.5. ∎

Acknowledgments

The author thanks Ernie Croot, Greg Martin, and Jiaxi Nie for helpful discussions.

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