分类: Number theory

  • From Sum-Free Sets to Strongly 2-Primitive Sets: Localization, Patching, and the Search for Stability

    Two extremal problems can share a proof architecture without sharing a proof. Benjamin Bedert’s 2025 breakthrough on large sum-free subsets is additive and Fourier-analytic. A July 2026 manuscript on strongly $2$-primitive sets is multiplicative and hypergraph-theoretic. Their common language is not a transferable sieve, but a four-step design: localize, build a strong object in each local block, prevent leakage between blocks, and add the gains. That comparison points toward the right stability question – and also reveals why the most naive version of stability is false.

    Status and scope. Bedert’s theorem is the February 2025 arXiv preprint cited below. The $27/2$ theorem is contained in a five-page manuscript by Przemek Chojecki supplied for this post in July 2026; no public preprint or peer-reviewed version was found, and the Erdős Problems page for #793 still labels the problem open at the time of writing. The written proof was checked here for internal coherence, but that is not independent peer review. The factor-graph and normalized-stability results later in this article are the additional analysis developed in the accompanying long note. Conditional and open statements are labelled explicitly.

    Here is the headline before the details. The supplied manuscript claims

    $$
    F(n)=\pi(n)+\left(\frac{27}{2}+o(1)\right)
    \frac{n^{2/3}}{(\log n)^2},
    $$

    where $F(n)$ is the largest size of a strongly $2$-primitive subset of $[1,n]$. One might expect every near-extremizer to be close to the construction behind this formula: almost all primes, plus products of three primes near $n^{1/3}$ arranged as a nearly saturated linear $3$-uniform hypergraph. That literal statement is false. There is, however, an exact and useful replacement: after contracting harmless private-prime lifts, every near-extremizer has a canonical prime layer and a small exceptional hub carrying the entire second-order excess. If the residual hub is mostly made of squarefree triples, then pair saturation and $n^{1/3}$-scale stability follow. Proving – or refuting – that triple-core hypothesis is the remaining inverse problem.

    1. Two problems that look similar only from far away

    1.1 The sum-free extraction problem

    A set $B\subset\mathbb Z$ is sum-free if it contains no $x,y,z$, not necessarily distinct, with

    $$x+y=z.$$

    For a finite set $A$ of integers, write

    $$
    S(A)=\max\{|B|:B\subseteq A,\ B\text{ is sum-free}\},
    $$

    and, for $N$-element sets of positive integers,

    $$
    S(N)=\min_{|A|=N}S(A).
    $$

    This is an extraction problem. An adversary gives us an arbitrary host set $A$, and we must find a large structured subset inside it. Erdős’s middle-third argument gives $S(A)\ge |A|/3$. Alon and Kleitman improved this to $(N+1)/3$, and Jean Bourgain proved $S(N)\ge (N+2)/3$ in 1997. The long-standing qualitative question was whether the additive improvement can tend to infinity:

    $$S(N)\ge \frac N3+\omega(N),\qquad \omega(N)\longrightarrow\infty.$$

    Bedert answered yes, proving that an absolute $c\gt0$ exists such that

    $$
    S(A)\ge \frac{|A|}{3}+c\log\log |A|.
    $$

    1.2 The strongly $2$-primitive packing problem

    A set $A\subseteq[1,n]$ is strongly $2$-primitive when

    $$
    a\nmid bc
    \qquad
    (a,b,c\in A,\ a\notin\{b,c\}),
    $$

    where $b=c$ is allowed. The word strongly matters. Under a more recent convention, a $2$-primitive set only forbids witnesses $b,c$ that are distinct. For example, $\{4,5,6\}$ passes that weaker test but fails the strong one because $4\mid6^2$.

    Now define

    $$
    F(n)=\max\{|A|:A\subseteq[1,n],\ A\text{ is strongly }2\text{-primitive}\}.
    $$

    This is a packing problem. The host interval is fixed, and we directly construct the largest possible forbidden-divisibility family. The all-primes set gives the leading term $\pi(n)$. In 1938 Erdős proved upper and lower bounds of the form

    $$
    \pi(n)+c_1\frac{n^{2/3}}{(\log n)^2}
    \le F(n)\le
    \pi(n)+c_2\frac{n^{2/3}}{(\log n)^2},
    $$

    and later asked whether the second-order term has an asymptotic constant. This is the modern Erdős Problem #793. The supplied 2026 manuscript proposes that the constant is $27/2$; it is important not to say that Erdős himself conjectured this numerical value.

    Comparison of the sum-free extraction problem and the strongly 2-primitive packing problem
    Figure 1. The quantifiers already separate the two questions. Sum-free theory extracts a subset from an arbitrary host; the multiplicative problem packs a forbidden-divisibility family into a fixed interval.

    1.3 Equality versus order

    The relation $a\nmid bc$ is not ordinary product-freeness. The set $\{6,10,15\}$, for instance, has no internal equality $xy=z$, yet $6\mid10\cdot15$. Prime valuations expose the real geometry:

    $$
    a\mid bc
    \quad\Longleftrightarrow\quad
    v_p(a)\le v_p(b)+v_p(c)
    \quad\text{for every prime }p.
    $$

    Thus the forbidden relation is a coordinatewise domination inequality in the divisor lattice. For squarefree integers, if $E_a=\{p:p\mid a\}$, it becomes

    $$E_a\subseteq E_b\cup E_c.$$

    So the natural combinatorial object is a $2$-cover-free family, not the solution set of a linear equation. Fourier characters are superb at detecting equations such as $x+y=z$. They do not come with an evident contractive projection that detects the partial order $\nu(a)\le\nu(b)+\nu(c)$. This is the first reason Bedert’s proof cannot simply be copied into the multiplicative setting.

    2. The $27/2$ upper bound: every element needs a private factor

    Set

    $$
    y=n^{1/3},\qquad
    M=\frac{y}{\log n},\qquad
    \Sigma_n=M^2=\frac{n^{2/3}}{(\log n)^2}.
    $$

    The upper bound begins with a small lemma that contains more information than the inequality it proves.

    Private-factor lemma. Let $\mathcal B$ be a set of positive integers, and choose for every $a\in A$ a factorization $a=u_av_a$ with $u_a,v_a\in\mathcal B$. If $A$ is strongly $2$-primitive, then $|A|\le|\mathcal B|$.

    Think of $E_a=\{u_a,v_a\}$ as a two-element multiset. If $u_a\ne v_a$ and neither coordinate is private to $a$, another chosen pair contains $u_a$ and another contains $v_a$; the corresponding two elements have product divisible by $a$. If $u_a=v_a=x$, failure of privacy would give another pair containing $x$ twice, hence another element equal to $x^2=a$. Therefore every $a$ has a private coordinate, and these coordinates are distinct. That is the injection $A\hookrightarrow\mathcal B$.

    The manuscript chooses the multiplicative $2$-basis

    $$
    \begin{aligned}
    \mathcal B_0&=[1,n^{3/5}],\\
    \mathcal B_1&=\{p\text{ prime}:n^{3/5}\lt p\le n\},\\
    \mathcal B_2&=\{pq:p,q\le y\text{ prime}\},\\
    \mathcal B_3&=\{qr:y\lt q\le n^{2/5},\ r\le n/q^2,\ q,r\text{ prime}\}.
    \end{aligned}
    $$

    Every $m\le n$ can be factored into two elements of $\mathcal B=\mathcal B_0\cup\mathcal B_1\cup\mathcal B_2\cup\mathcal B_3$. The case split is elementary but carefully tuned. Small $m$ can be balanced into two factors below $n^{3/5}$, a prime factor above $n^{2/5}$ can be split off, and the remaining difficult case groups two of the three largest prime factors into an element of $\mathcal B_2$ or $\mathcal B_3$.

    The main prime layer comes from $\mathcal B_1$. The genuinely second-order counts are

    $$
    |\mathcal B_2|
    =\binom{\pi(y)+1}{2}
    =\left(\frac92+o(1)\right)\Sigma_n,
    $$

    because $\pi(n^{1/3})\sim3n^{1/3}/\log n=3M$, and

    $$
    |\mathcal B_3|
    =\sum_{y\lt q\le n^{2/5}}\pi\!\left(\frac n{q^2}\right)
    =(9+o(1))\Sigma_n.
    $$

    The contribution of $\mathcal B_0$ and all relevant overlaps is $o(\Sigma_n)$. Hence

    $$
    |\mathcal B|
    =\pi(n)+\left(\frac92+9+o(1)\right)\Sigma_n
    =\pi(n)+\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The private-factor injection then gives the upper bound. A caution that becomes crucial for stability: $\mathcal B_2$ and $\mathcal B_3$ are labels in an upper-bound certificate. Near-saturation of these labels does not immediately say that the original elements of $A$ are themselves products of two or three primes.

    3. The lower bound: turn properly coloured edges into prime triples

    The lower bound lives in a linear $3$-uniform hypergraph. Let $\mathcal H$ be a family of triples of distinct primes such that any two triples share at most one prime and every edge product is at most $n$. Define

    $$
    A_{\mathcal H}
    =\{p\le n:p\text{ prime and }p\notin V(\mathcal H)\}
    \cup
    \left\{\prod_{p\in E}p:E\in\mathcal H\right\}.
    $$

    Then

    $$|A_{\mathcal H}|=\pi(n)-|V(\mathcal H)|+|\mathcal H|.$$

    Why is this strongly $2$-primitive? A target triple product has three distinct prime coordinates. Each other hyperedge supplies at most one of them, so two other elements supply at most two. The same argument still works when the two witnesses coincide. Primes outside the vertex set remain singleton elements and cannot divide any other chosen element.

    3.1 Logarithmic prime bins

    Fix a small mesh $h\gt0$ and divide primes near $y=n^{1/3}$ into bins

    $$
    P_i=\{p\text{ prime}:ye^{ih}\lt p\le ye^{(i+1)h}\},
    \qquad
    \Delta_i=e^{(i+1)h}-e^{ih}.
    $$

    For a cell satisfying

    $$i\le j,\qquad i+2j\le-4,$$

    put $k=-i-j-3$. Then $i\le j\lt k$, and any $p\in P_i$, $q\in P_j$, $r\in P_k$ obeys $pqr\le n$.

    If $i\lt j$, properly edge-colour the complete bipartite graph between $P_i$ and $P_j$, using primes of $P_k$ as colours. When $i=j$, do the same with the complete graph on $P_i$. The lower pair $\{p,q\}$, coloured by $r$, becomes the hyperedge $\{p,q,r\}$.

    A complete bipartite graph between two prime bins properly edge-coloured by a third prime bin
    Figure 2. Within a cell, every colour class is a matching, so two produced triples never share a lower pair. Across cells, the sorted signature $(i,j,k)$ has fixed sum $-3$; two shared bin indices force the third and hence force the same cell.

    The proper colouring is the local no-interference mechanism. The constant-sum signature is the global one. Together they make the union over all cells a linear hypergraph.

    3.2 The cell weight

    For any fixed finite collection of bins, the prime number theorem gives

    $$|P_i|=(3+o(1))M\Delta_i.$$

    Off the diagonal, a cell contributes approximately $9M^2\Delta_i\Delta_j$ edges. A diagonal cell contributes half as many unordered pairs. The exact geometric-series identity is

    $$
    \sum_{\substack{i\lt j\\i+2j\le-4}}\Delta_i\Delta_j
    +\frac12\sum_{i\le-2}\Delta_i^2
    =e^{-h}+\frac12e^{-2h}.
    $$

    Letting $h\to0$, the weight tends to $3/2$, so

    $$
    |\mathcal H|
    =\left(9\cdot\frac32-o(1)\right)\Sigma_n
    =\left(\frac{27}{2}-o(1)\right)\Sigma_n.
    $$

    The number of vertices used is only $o(\Sigma_n)$. Replacing those primes by the triple products therefore yields the matching lower bound.

    The logarithmic feasible pair region split into the 9 over 2 and 9 counting regimes
    Figure 3. The same constant is visible in the canonical feasible pair space. The labels $9/2$ and $9$ are prime-counting contributions, not Euclidean areas of the drawing.

    4. From Erdős’s middle third to Bedert’s $c\log\log N$

    We now return to the additive problem. Let $\mathbb T=\mathbb R/\mathbb Z$, and let $\phi=\mathbf1_{(1/3,2/3)}$. The middle third of the circle is sum-free: two points in it cannot add, modulo $1$, to another point in it. Therefore, for every $x\in\mathbb T$,

    $$A_x=\{a\in A:ax\pmod1\in(1/3,2/3)\}$$

    is sum-free. Averaging $|A_x|$ over $x$ gives $|A|/3$. The problem is to force a positive fluctuation above that mean.

    After a harmless normalization, the relevant Fourier series has the form

    $$
    F_A(x)=\sum_{a\in A}\sum_{m\ge1}
    \frac{\chi(m)}m\cos(2\pi max),
    $$

    where $\chi$ is the non-principal real character modulo $3$. A lower bound for $\|F_A\|_1$ gives a one-sided large value and hence a large sum-free subset.

    4.1 Where Bourgain’s sieve loses a logarithm

    The historical correction is worth making explicit. The relevant predecessor is Jean Bourgain’s 1997 paper, not “John Bogan, 1993.” Also, the Littlewood $L^1$ conjecture had already been proved in 1981, independently by McGehee-Pigno-Smith and Konyagin. Bourgain’s argument combines Fourier analysis with a Möbius operation that filters harmonic indices.

    Schematically, if $R_Q$ denotes the positive $Q$-rough integers, then

    $$
    \sum_{d\mid\prod_{p\le Q}p}
    \frac{\mu(d)\chi(d)}d F_A(dx)
    =\sum_{a\in A}\cos(2\pi ax)+\mathcal R_Q(x).
    $$

    The left side costs

    $$
    \prod_{p\le Q}\left(1+\frac1p\right)\asymp\log Q
    $$

    under the triangle inequality. If $Q$ is large enough to make the rough remainder negligible directly, this factor consumes the logarithm delivered by a Littlewood-type lower bound. This explains why merely “using the Littlewood theorem harder” does not produce the desired unbounded gain.

    4.2 Bedert’s split: medium primes are sieved, small primes are projected

    Bedert chooses

    $$Q_1=(\log N)^{1/2},\qquad Q=(\log N)^{20}.$$

    He Möbius-sieves only the medium primes $Q_1\le p\le Q$. Their reciprocal sum is $O(1)$, so the norm loss is a constant rather than a logarithm. The small primes $p\le Q_1$ are handled by a different operation.

    For every small prime, record the exact valuation $\nu_p(a)$ and the unit residue after removing that prime power. Chinese remaindering packages all this data into a joint fibre $A(r,\nu)$. Fourier projection to a residue class,

    $$
    \operatorname{Proj}(H;\rho\bmod q)(x)
    =\sum_{m\equiv\rho\ (q)}\widehat H(m)e(mx),
    $$

    is an $L^1$-contraction:

    $$
    \|\operatorname{Proj}(H;\rho\bmod q)\|_1\le\|H\|_1.
    $$

    At the main valuation level, the projection removes small-prime harmonic contamination without paying the full Möbius product. After truncation, one sees a large exponential sum plus an $L^2$-small error, to which a robust McGehee-Pigno-Smith test function applies.

    At a general valuation level, lower fibres can still leak into the projected channel. Bedert proves an isolation-or-descent alternative: either the desired $L^1$ lower bound is already present, or a large fibre has a strictly lower fibre losing at most a controlled polylogarithmic factor. Iterating and reversing this descent produces a chain

    $$
    \nu^{(1)}\prec\nu^{(2)}\prec\cdots\prec\nu^{(J)},
    \qquad
    J\gg\frac{\log N}{\log\log N},
    $$

    whose fibre sizes grow geometrically.

    4.3 The non-Archimedean no-leakage lemma

    Associate to these levels the nested moduli

    $$
    q_i=\prod_{p\le Q_1}p^{\nu_p^{(i)}+1},
    \qquad q_1\mid q_2\mid\cdots\mid q_J.
    $$

    Let $g_i$ be the normalized indicator of the $i$-th residue block and put $Q_i(x)=\exp(-|\widehat g_i(x)|)$. The non-Archimedean MPS test function is assembled from

    $$
    \Phi_J=
    \widehat g_J+\widehat g_{J-1}Q_J+\cdots+
    \widehat g_1Q_2\cdots Q_J.
    $$

    The decisive observation is that $|\widehat g_i|$ is $1/q_i$-periodic, hence $\widehat Q_i$ is supported on $q_i\mathbb Z$. Multiplying an earlier block by a later $Q_k$ shifts its frequencies only by multiples of $q_k$. Since $q_i\mid q_k$, the earlier block cannot escape its residue class modulo $q_i$. Main inner products contribute one controlled unit per block, while geometric fibre growth makes the cross terms summable.

    That is Bedert’s genuine no-leakage mechanism. It is more precise than the slogan “use many congruence classes and patch them.”

    One further qualification prevents a common misstatement. Bedert obtains a suitable $F_4$-isomorphic model $B$ of the original set and proves the required $L^1$ bound there. The norm $\|F_A\|_1$ is not claimed to be invariant under an $F_4$-isomorphism. What is preserved is the four-term additive information needed for sum-freeness, so $S(A)=S(B)$.

    Bedert’s inverse output is also stronger than the numerical lower bound. If $S(A)\le N/3+C$, his results force low additive dimension, a dense $F_4$-model, large additive energy in every substantial subset, and a “99% Structure Theorem” decomposing all but a small exceptional set into large small-doubling pieces. This is a genuine inverse theorem for host sets resisting sum-free extraction, but not an edit-distance classification by one canonical extremizer.

    5. The real bridge: localize, build, prevent leakage, sum

    Parallel flowcharts for Bedert's p-adic Fourier proof and the logarithmic hypergraph construction
    Figure 4. The analogy is architectural. Bedert preserves Fourier residue lanes through nested moduli; the multiplicative construction preserves low codegrees through matching colours and unique scale signatures.

    The dictionary is now clean:

    Role Bedert’s sum-free proof Strongly $2$-primitive proof
    Underlying relation Linear equation in an abelian group Coordinatewise domination of prime valuations
    Local coordinates Exact small-prime valuations and unit residues Logarithmic sizes of prime factors
    Local block Joint $p$-adic fibre and MPS block Scale cell and properly coloured graph
    No interference Nested moduli preserve Fourier support Colour matchings and unique cell signatures preserve linearity
    Accumulation One controlled inner product per fibre One triple per admissible lower pair
    Inverse output Additive dimension, energy, small-doubling pieces Private factors and a cover-free exponent core

    This is a substantial connection, but it is a connection of proof design. Bedert’s Fourier projection has no direct analogue for the divisor partial order. The most plausible transfer is therefore not a line-by-line proof but a research program: find a multiplicative localization, an isolation-or-descent alternative, and a no-leakage invariant that survives repeated prime powers and larger supports.

    6. The first stability guess is false

    The lower construction suggests a tempting statement: every set with

    $$
    |A|\ge
    \pi(n)+\left(\frac{27}{2}-o(1)\right)\Sigma_n
    $$

    should differ in only $o(\Sigma_n)$ places from almost all primes plus a near-optimal linear family of triple products. Two examples show why this is too rigid.

    6.1 A private-prime lift

    Start with a near-optimal linear triple family $\mathcal H_n$ using odd primes below $n/2$. Keep primes above $n/2$, replace every unused prime $p\le n/2$ by $2p$, and retain the triple products. Each $2p$ has the private prime divisor $p$; no other chosen element contains that prime. The triple products are still protected by linearity. The new set remains strongly $2$-primitive and has the same second-order asymptotic size.

    But essentially every prime below $n/2$ has been replaced by a composite. The symmetric-difference distance from the all-primes model is

    $$
    \Theta(\pi(n/2))=\Theta\!\left(\frac n{\log n}\right),
    $$

    and

    $$
    \frac{\pi(n/2)}{\Sigma_n}
    \asymp n^{1/3}\log n\longrightarrow\infty.
    $$

    So literal edit-distance stability fails by much more than the scale of the second-order term.

    6.2 A nonlinear sunflower at almost no cost

    Even if we insist on squarefree triple products, linearity need not hold in the original representation. Add a sunflower

    $$\{\{2,3,r\}:r\in R\}$$

    for many private petals $r$. Remove the singleton primes $2,3,r$ and add the products $6r$. The net cardinality loss is only $2$, while the support family contains $\asymp n/\log n$ edges sharing the pair $\{2,3\}$. It is $2$-cover-free because every target has its own private petal, but any linear subfamily contains at most one sunflower edge.

    The two examples have the same cause: degree-one prime coordinates carry a huge amount of neutral decoration. Stability can only become true after quotienting out this freedom.

    7. Near equality in the upper bound is an exact star forest

    Fix the chosen factorizations $a=u_av_a$ from the multiplicative basis proof. Make a graph $G_A$ with vertex set $\mathcal B$, one edge $\{u_a,v_a\}$ for each $a\in A$, and allow loops.

    Exact factor-graph theorem. Every nonloop edge has a degree-one endpoint, and every loop is an isolated component. Thus $G_A$ is a disjoint union of stars and isolated loops. If $U$ is the number of unused vertices and $c(G_A)$ is the number of nonloop star components, then

    $$|\mathcal B|-|A|=U+c(G_A).$$

    The proof is the private-factor lemma read without discarding information. If a nonloop edge $\{u,v\}$ had both endpoints in other edges, the corresponding two elements would have product divisible by $uv=a$. If a loop at $u$ met another edge, then $u^2$ would divide the square of the other element.

    A factor graph decomposed into stars, an isolated loop and unused vertices
    Figure 5. The upper-bound defect is counted exactly: an unused basis coordinate costs one, and each nonloop star component costs one.

    More quantitatively,

    $$
    \bigl|\{b\in\mathcal B\setminus\mathcal B_0:
    \deg_{G_A}(b)\ne1\}\bigr|
    \le |\mathcal B|-|A|.
    $$

    If

    $$
    |A|\ge
    \pi(n)+\left(\frac{27}{2}-\delta_n\right)\Sigma_n,
    $$

    then the defect $D_n=|\mathcal B|-|A|$ is at most $(\delta_n+o(1))\Sigma_n$. Hence almost every secondary coordinate in both $\mathcal B_2$ and $\mathcal B_3$ occurs in exactly one chosen factor pair. This is a rigorous stability theorem for the upper-bound certificate. It is not yet a classification of the original integers.

    8. Private-prime compression gives the correct normal form

    Suppose a prime $p$ divides $a\in A$ and divides no other element of $A$. Replacing $a$ by $p$ preserves the cardinality and strong $2$-primitivity. The new target $p$ cannot divide a product of two other elements because neither contains $p$; for any unchanged target, replacing $a$ by a divisor only decreases the product available to cover it.

    For primes $p\gt n^{3/5}$, the basis factorization can be chosen to expose $p$ in every multiple. Thus a degree-one large-prime coordinate is genuinely a globally private prime and can be contracted. Contract all such coordinates simultaneously.

    Normalized stability modulo private-prime lifts. From every strongly $2$-primitive $A\subseteq[1,n]$ satisfying the near-extremal bound above, private-prime contractions produce a strongly $2$-primitive set $\widetilde A$ of the same size with

    $$\widetilde A=(\mathbb P\cap[1,n]\setminus V)\cup C,\qquad \operatorname{supp}(c)\subseteq V\quad(c\in C).$$

    where

    $$|V|\le(\delta_n+o(1))\Sigma_n,\qquad |C|-|V|=|A|-\pi(n).$$

    In particular, if $\delta_n=o(1)$, then

    $$
    |V|=o(\Sigma_n),
    \qquad
    |C|=\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The geometry is now transparent. Before compression, the leading $\pi(n)$ coordinates may be represented by arbitrary private multiples. After compression, almost every prime appears literally. Every residual composite is supported entirely on the small exceptional prime hub $V$, and its excess over the missing primes is exactly the second-order gain.

    Private-prime lifts contracted to a normal form with singleton primes and a small exceptional hub
    Figure 6. Literal stability fails on the left. The quotient by private-prime contractions produces the exact normal form on the right.

    There is one more unconditional consequence. At most $|V|$ members of $C$ have support of size at most two. Indeed, any one- or two-prime residual element must have a valuation coordinate in which it is a strict global maximum; assign the element to such a prime. Two elements cannot receive the same prime. Therefore, in a normalized near-extremizer, all but $o(\Sigma_n)$ residual composites have at least three distinct prime divisors.

    This is close to the hoped-for triple picture, but it does not prove that the typical element has exactly three prime factors, that it is squarefree, or that its prime factors lie near $n^{1/3}$.

    9. Conditional stability when the residual core is made of triples

    Now impose an additional hypothesis:

    Squarefree-triple hypothesis. All but $o(\Sigma_n)$ elements of $C$ are products of three distinct primes.

    Let $\mathcal H$ be the support family of those triples. Strong $2$-primitivity says precisely that it is $2$-cover-free:

    $$
    E\nsubseteq F\cup G
    \qquad(E,F,G\in\mathcal H,\ E\notin\{F,G\}),
    $$

    with $F=G$ allowed.

    9.1 Pruning to a linear core

    If two triples $E,F$ share a pair and $x$ is the third vertex of $E$, then $x$ has degree one. Otherwise another edge $G$ containing $x$, together with $F$, would cover $E$. Delete such a petal edge whenever a repeated pair occurs. Every deletion removes at least one vertex as well as one edge, so the excess $|\mathcal H|-|V(\mathcal H)|$ does not decrease.

    The result is a linear subfamily $\mathcal L$ with

    $$
    |\mathcal L|-|V(\mathcal L)|
    \ge |\mathcal H|-|V(\mathcal H)|,
    $$

    and only $o(\Sigma_n)$ edges are lost in the normalized near-extremal setting. Notice the nuance: linearity is recovered after pruning private petals; it is not asserted for every cover-free representation.

    9.2 Saturating the feasible pair space

    Define

    $$
    \mathcal D_n=
    \{(p,q):p\lt q\text{ prime and }pq^2\le n\}.
    $$

    Sort an edge of $\mathcal L$ as $p\lt q\lt r$ and map it to $(p,q)$. Linearity makes this map injective. Since $r\gt q$ and $pqr\le n$, the image lies in $\mathcal D_n$. Direct counting gives

    $$
    |\mathcal D_n|
    =\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The range $q\le n^{1/3}$ contributes $(9/2+o(1))\Sigma_n$; the range $n^{1/3}\lt q\le n^{2/5}$ contributes $(9+o(1))\Sigma_n$; the tail is negligible. Since the linear core already has $(27/2-o(1))\Sigma_n$ edges, its lower-pair map misses only $o(\Sigma_n)$ feasible pairs.

    Moreover, for every fixed $\varepsilon\gt0$, all but $o(\Sigma_n)$ edges satisfy

    $$
    n^{1/3-\varepsilon}
    \le p,q,r\le
    n^{1/3+\varepsilon}.
    $$

    This is the desired scale stability. It does not imply uniqueness. Different one-factorizations or different proper edge-colourings can change $\Theta(\Sigma_n)$ triples while preserving the same pair occupancy and extremal count. The stable object is the saturated pair space, not an individual colouring.

    10. The remaining inverse problem is weighted and cover-free

    After normalization, write every $c\in C$ as an exponent vector

    $$
    \nu(c)=(v_p(c))_{p\in V}\in\mathbb Z_{\ge0}^{V}.
    $$

    Strong $2$-primitivity becomes the weighted cover-free condition

    $$
    \nu(c)\nleq\nu(c_1)+\nu(c_2)
    \quad\text{coordinatewise}
    $$

    whenever $c\notin\{c_1,c_2\}$. We know that

    $$
    |C|=\left(\frac{27}{2}+o(1)\right)\Sigma_n,
    \qquad |V|=o(\Sigma_n),
    $$

    and that almost every vector has support at least three. What remains unknown is the sharper assertion

    $$
    \#\{c\in C:c\text{ is not a squarefree product of three primes}\}
    =o(\Sigma_n).
    $$

    This gap may conceal genuine alternative near-extremizers. If a triple $pqr$ has product slack, one can contemplate replacing it by $2pqr$ while deleting the singleton prime $2$. For a linear support family, the underlying three private coordinates still prevent coverage. It is not known whether a positive density of such bounded-hub multiplier layers can coexist globally near the $27/2$ threshold, or whether cross-layer collisions force them to be negligible.

    Open normalized stability problem. Classify weighted $2$-cover-free exponent families under the product constraint $c\le n$ whose excess is $(27/2-o(1))\Sigma_n$. Decide whether the residual core is predominantly squarefree of support three, or whether bounded-hub multiplier layers produce genuinely different normalized near-extremizers.

    10.1 What a Bedert-style descent would need

    The additive proof suggests three design requirements, not three ready-made lemmas.

    1. A local order projection. One needs to isolate a valuation or support layer while controlling contamination from lower exponent vectors. A divisor-lattice zeta or Möbius transform is a possible language, but no analogue of Bedert’s $L^1$-contractive residue projection is presently available.
    2. An isolation-or-descent alternative. If the secondary basis slots do not have predominantly prime cofactors, the argument should descend to a structured hub of repeated factors with a quantitative gain or a summable loss.
    3. A no-leakage invariant. Low codegree and scale signatures work for squarefree triples. A general invariant must survive repeated exponents and supports of size at least four.

    The star-forest theorem is already a first inverse statement of this kind: near equality forces almost every basis coordinate to be a leaf attached to a small collection of hubs. The hard step is to turn that certificate-level hub structure into a classification of the original weighted exponent vectors.

    11. What is proved, conditional, and open

    • Published/preprint additive result: Bedert proves $S(A)\ge |A|/3+c\log\log|A|$, together with inverse information involving additive dimension, dense $F_4$-models, energy and a $99\%$ decomposition into large small-doubling pieces.
    • Supplied 2026 manuscript: the claimed $27/2$ asymptotic follows from the multiplicative-basis upper bound and the logarithmic-cell hypergraph construction described above. The status caveat at the beginning remains in force.
    • Unconditional stability developed in the accompanying note: literal edit stability is false; the upper-bound factor graph is an exact star forest; private-prime compression gives the normal form $(\mathbb P\setminus V)\cup C$; and almost every residual composite has at least three distinct prime divisors.
    • Conditional stability: if almost all residual composites are squarefree triples, pruning gives a near-maximal linear core whose lower pairs saturate $\mathcal D_n$, and almost every prime factor lies at scale $n^{1/3+o(1)}$.
    • Open: prove that the normalized residual core is predominantly squarefree of support three, or construct a different near-extremal weighted cover-free core.

    The conceptual moral is simple but useful. Bedert’s $p$-adic fibres and the multiplicative logarithmic cells are not the same mathematical object. Yet both proofs win by choosing coordinates in which local constructions can be made strong and then finding an exact invariant that prevents those constructions from interfering. In the stability problem, the same philosophy says to quotient the neutral directions first. Once private-prime lifts are removed, the true obstruction becomes visible: a weighted cover-free family on a tiny prime hub. That is the right object for the next theorem.

    References and source trail

    1. P. Erdős, On sequences of integers no one of which divides the product of two others and on some related problems, 1938.
    2. P. Erdős, On some applications of graph theory to number-theoretic problems, 1969; see also Erdős Problem #793.
    3. O. Carruth McGehee, L. Pigno and B. Smith, Hardy’s inequality and the $L^1$-norm of exponential sums, Annals of Mathematics 113 (1981), 613-618.
    4. J. Bourgain, Estimates related to sumfree subsets of sets of integers, Israel Journal of Mathematics 97 (1997), 71-92.
    5. B. Bedert, Large sum-free subsets of sets of integers via $L^1$-estimates for trigonometric series, arXiv:2502.08624v1, 12 February 2025.
    6. P. Chojecki, The Second Term for Strongly 2-Primitive Sets, user-supplied five-page manuscript, July 2026; no public identifier located at the time of writing.
  • Ramanujan-Nagell theorem:平方数与 2^n 相差 7 的有限性

    旧博客原文

    原题:The Ramanujan-Nagell Theorem: Understanding the Proof

    The Ramanujan-Nagell Theorem: Understanding the Proof


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Ramanujan-Nagell theorem 研究的是一个看起来非常小的指数丢番图方程:

    $$x^2+7=2^n.$$

    它的结论是整数解只有有限个,而且正整数解恰好为

    $$(x,n)=(1,3),(3,4),(5,5),(11,7),(181,15).$$

    Ramanujan-Nagell theorem:平方数与 2^n 相差 7 的有限性
    Ramanujan-Nagell 方程把初等同余、二次域分解和 Lucas 序列约束压缩在同一个指数丢番图问题中。

    1. 初等筛选

    先看奇偶性。若 $x$ 为偶数,则左边 $x^2+7$ 为奇数,不可能等于 $2^n$;所以 $x$ 必为奇数。再看模 $8$,奇数平方恒为 $1$,所以 $x^2+7\equiv0\pmod 8$,这只说明 $n\ge3$,但已经排除了很多无意义情形。

    2. 代数数论中的分解

    真正有力的观察是把方程写成

    $$(x+\sqrt{-7})(x-\sqrt{-7})=2^n.$$

    在 $\mathbb Q(\sqrt{-7})$ 的整数环中,$2$ 可以分解成两个共轭因子。由于这个二次域的类数很小,理想层面的分解可以被提升为元素层面的约束,于是 $x+\sqrt{-7}$ 必须接近某个基本元素的 $n$ 次幂。

    3. Lucas 序列的出现

    把共轭相减,得到的不是任意等式,而是一个 Lucas 型序列项必须等于很小的数:

    $$\frac{\alpha^n-\bar\alpha^n}{\alpha-\bar\alpha}=\pm 1\quad\text{or}\quad \pm 7.$$

    这种递推序列增长很快,同时在模意义下有强限制。少数小 $n$ 需要直接检查,大 $n$ 则被递推结构和同余条件排除。

    4. 为什么这条定理有代表性

    这类问题的典型形状是:初等同余给出粗过滤,二次域分解给出结构,最后用 Lucas 序列或线性形式估计把无限可能压成有限检查。Ramanujan-Nagell 方程的漂亮之处在于,所有这些工具都集中在一个非常短的公式里。

  • Diophantine approximation:Dirichlet 定理、抽屉原理与最佳逼近

    旧博客原文

    原题:Diophantine approximation

    I explain some general ideal in the theory of diophantine approximation, some of them is original by myself, begin with a toy model, then consider the application on folklore Swirsing-Schmidt conjecture.

    \tableofcontents

    1. Dirichlet theorem, the toy model

    The very basic theorem in the theory of Diophantine approximation is the well known Dirichlet approximation theorem, the statement is following.

    Theorem 1 (Dirichlet theorem) for all {\alpha} is a irrational number, we have infinity rational number {\frac{q}{p}} such that:

    \displaystyle |\alpha-\frac{q}{p}|<\frac{1}{p^2} \ \ \ \ \ (1)

    Remark 1 It is easy to see the condition of irrational is crucial. There is a best constant version of it, said, instead of {1}, the best constant in the suitable sense for the theorem 1 should be {\frac{1}{\sqrt{5}}} and arrive by {\frac{\sqrt{5}+1}{2}} at least. The strategy of the proof of the best constant version involve the Frey sequences.

    Now we begin to explain the strategies to attack the problem.

    \paragraph{Argument 1, boxes principle} We begin with a easiest one, i.e. by the argument of box principle, the box principle is following,

    Theorem 2 (Boxes principle) Given {n\in {\mathbb N}} and two finite sets {A={a_1,a_2,...,a_n,a_{n+1}}}, set {B={b_1,...,b_{n}}}, if we have a map:

    \displaystyle f:A\longrightarrow B \ \ \ \ \ (2)

    Then there exists a element {b_k\in B} such that there exist at least two element {a_i,a_j\in A}, {f(a_i)=f(a_j)=b_k}.

    Proof: The proof is trivial. \Box

    Now consider, {\forall N\in {\mathbb N}}, the sequences {x,2x,...,Nx}, then {\{ix\}\in [0,1], \forall i\in \{1,2,...,n\}}. Divide {[0,1]} in an average way to {N} part: {[\frac{k-1}{N},\frac{k}{N}]}. Then the linear structure involve (which, in fact play a crucial role in the approach). And the key point is to look at {\{nx\}} and integers.

    \paragraph{Argument 2, continue fractional} We know, for irrational number {x}, {x} have a infinite long continue fractional:

    \displaystyle x=q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k+...}}}} \ \ \ \ \ (3)

    Then

    \displaystyle |x-q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}|\sim \frac{1}{(q_1q_2...q_{k-1})^2q_k} \ \ \ \ \ (4)

    And we have,

    \displaystyle \frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}=\frac{a_n}{b_n}, (a_n,b_n)=1 \ \ \ \ \ (5)

    Then {b_n=O(q_1...q_k)}.

    \paragraph{Argument 3, Bohr set argument} We begin with some kind of Bohr set:

    \displaystyle B_p=I-\cup_{q\in \{0,1,...,p-1\}}(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2}) \ \ \ \ \ (6)

    The key point is the shift of Bohr set, on the vertical line i.e. {|B_p\cap B_{p+1}|} is very slow, and can be explained by

    \displaystyle \frac{k}{p+1}+\frac{1}{(p+1)^2}>\frac{k}{p}-\frac{1}{p^2} \ \ \ \ \ (7)

    So:

    \displaystyle \frac{1}{p^2}+\frac{1}{(p+1)^2}>\frac{k}{p(p+1)} \ \ \ \ \ (8)

    in {|B_p \cap B_{p+1}|\sim \frac{1}{p(p+1)}} But in fact they are not really independent, as the number of Bohr sets increase, then you can calculate the correlation, thanks to the harmonic sires increasing very slowly, wwe can get something non trivial by this argument, but it seems not enough to cover the whole theorem 1.

    \paragraph{Argument 4, mountain bootstrap argument} This argument is more clever than 3, although both two arguments try to gain the property we want in 1 from investigate the whole space {[0,1]} but not {x}, this argument is more clever.

    Now I explain the main argument, it is nothing but sphere packing, with the set of balls

    \displaystyle \Omega=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall p\in {\mathbb N}, 1\leq q\leq p-1 \} \ \ \ \ \ (9)

    and define its subset

    \displaystyle \Omega_l=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall 1\leq p\leq l, 1\leq q\leq p-1 \} \ \ \ \ \ (10)

    Then {\Omega_l\subset \Omega}, and {\Omega =\cup_{l\in {\mathbb N}}\Omega_l}. If we can proof,

    Lemma 3 For all {l\in {\mathbb N}}, there is a subset {A_l} of {\Omega-\Omega_l} such that {\cup_{i\in A_l}B_i=[0,1]}.

    Remark 2 If we can proof 3, it is easy to see the theorem 1 follows.

    Proof: The proof follows very standard in analysis, may be complex analysis? Key point is we start with a ball {B_{p,q}}, whatever it is, this is not important, the important thing is we can take some ball {B_{p',q'}} with the center of {B_{p',q'}} in {B_{p,q}}, then try to consider {B_{p',q'}\cup B_{p,q}} to extension {B_{p,q}} and then we find the boudary is also larger then we can extension again, step by step just like mountain bootstrap argument. So we involve in two possible ending,

    1. The extension process could extension {B{p,q}} to whole space.
    2. we can not use the extension argument to extension to the whole space.

    If we are in the first situation, then we are safe, there is nothing need proof. If we are in second case, anyway we take a ball {B_{p,q}=(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2})}. Then try to find good ball {B_{p',q'}} to approximate {B_{p,q}}, but this is difficult… \Box

    Remark 3 Argument 1 is too clever to be true in generalization, argument 2 is standard, by the power of renormalization. argument 3 and argument 4 have gap… I remember I have got a proof similar to argument 4 here many years ago, but I forgot how to get it…

    2. Schimidt conjecture

    The Schimidt conjecture could be look as the generalization of Dirchlet approximation theorem 1 to algebraic number version, to do this, we need define the height of a algebraic number.

    Definition 4 We say a number {\alpha\in {\mathbb C}} is a {k-}order algebraic number if and only is the minimal polynomial of {\alpha}, {f(x)=a_nx^n+...+a_1x+a_0, a_n\neq 0} have degree {deg(f)=n, f\in {\mathbb Z}[x]}.

    Definition 5 (Height) Now we define the height of a {k-}th order algebraic number as {H(\alpha):=\max\{\|a_n\|_h,\|a_{n-1}\|_h,...,\|a_0\|_h\}}, Where

    \displaystyle h(a_i)=\|a_i\|_{\infty} \ \ \ \ \ (11)

    Now we state the conjecture:

    Theorem 6 (Swiring-Schimidt conjecture) For all transendental number {x\in {\mathbb C}}, there is infinitely {\alpha} are {k-}th algebraic number such that:

    \displaystyle |x-\alpha|<\frac{c_k}{H(\alpha)^{k+1}} \ \ \ \ \ (12)

    Where {c_k} is a constant only related to {k} but not {x}.

    I point out the conjecture is very related to the map:

    \displaystyle F:(x_1,...,x_n) \longrightarrow (\sigma_1(x_1,...,x_n),\sigma_2(x_1,...,x_n),...,\sigma_n(x_1,...,x_n)) \ \ \ \ \ (13)

    Where {\sigma_k(x_1,...,x_n)=\sum_{1\leq i_1<...<i_k\leq n}\Pi_{j=1}^kx_{i_1}x_{i_2}...x_{i_k}} is the {k-}th symmetric sum.

    Remark 4 {F} is a map {{\mathbb C}^n\rightarrow {\mathbb C}^n}, what we consider is its inverse, {G=F^{-1}}, but {G} is not smooth, it occur singularity when {x_i=x_j} for some {i\neq j}. And the map, as we know, the singularity depend on the quantity {\Pi_{1\leq i< j\leq n}(x_i-x_j)}.

    Remark 5 I then say something about the geometric behaviour of the map {G}, as we know, what we have in mind is consider the map {G} as a distortion {{\mathbb C}^n\rightarrow {\mathbb C}^n}, Then {H(\alpha)} is just the pullback of the canonical metric on {{\mathbb C}}(morally) to {{\mathbb C}}.


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Diophantine approximation 的基本问题是:一个无理数 $\alpha$ 能被有理数 $p/q$ 逼近到什么程度?Dirichlet 定理给出最基本的答案,而连分数告诉我们哪些分母是真正的最佳逼近。

    Diophantine approximation:Dirichlet 定理、抽屉原理与最佳逼近
    Dirichlet 定理的抽屉原理证明依赖小数部分的差分线性结构。

    1. Dirichlet 定理

    对任意无理数 $\alpha$,存在无穷多个有理数 $p/q$,使得

    $$\left|\alpha-\frac pq\right|<\frac1{q^2}.$$

    一个有限版本是:给定 $Q$,存在 $1\le q\le Q$ 和整数 $p$,使

    $$|q\alpha-p|<\frac1Q.$$

    2. 抽屉原理证明

    看 $Q+1$ 个数的小数部分

    $$0,\{\alpha\},\{2\alpha\},\ldots,\{Q\alpha\}.$$

    把 $[0,1]$ 分成 $Q$ 个长度 $1/Q$ 的区间。两个小数部分落在同一区间,于是它们的差给出某个 $q\alpha$ 距离整数小于 $1/Q$。这就是 Dirichlet 定理最干净的证明。

    3. 线性结构在哪里

    抽屉原理本身只是计数,但这里真正起作用的是线性结构:两个点 $\{a\alpha\}$ 和 $\{b\alpha\}$ 接近,差就变成 $\{(a-b)\alpha\}$ 接近整数。没有这个差分结构,抽屉原理不会自动给出有理逼近。

    4. 连分数与最佳逼近

    连分数展开

    $$\alpha=[a_0;a_1,a_2,\ldots]$$

    给出 convergents $p_k/q_k$。它们满足

    $$\left|\alpha-\frac{p_k}{q_k}\right|<\frac1{q_kq_{k+1}}.$$

    这些分母 $q_k$ 是最佳逼近的自然尺度。若 $a_k$ 有界,则 $\alpha$ 是 badly approximable;若某些 $a_k$ 很大,就会出现异常好的逼近。

    5. 更高维和 Schmidt 猜想的方向

    高维 Diophantine approximation 会把一个数的逼近问题变成向量、线性形式或流形上的逼近问题。此时抽屉原理仍然给出基准结果,但最佳常数、例外集维数和代数数逼近会变得更深。许多问题最后会进入 geometry of numbers、dynamical systems on homogeneous spaces 或 Schmidt game 的语言。

  • 短区间中 Mobius 函数与 nil-sequence 的相关估计

    旧博客原文

    原题:The correlation of Mobius function and nil-sequences in short interval

    I wish to establish the following estimate:

    Conjecture :(correlation of Mobius function and nil-sequences in short interval)

    \lambda(n) is the liouville function we wish the following estimate is true.

    \int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH).

    Where we have H\to \infty as x\to \infty, \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\} is a compact space.

    I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

    There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

    Theorem (multiplication function in short interval)

    f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. H\to \infty as x\to infty, then we have the following result,

    \int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH).

    And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

    Theorem(correlation of multiplication function and nil-sequences in long interval)

    f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. g(n)=a_n^m+...+a_1n+a_0 is a polynomial function then we have the following result,

    \int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X).


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    希望建立的估计可以粗略写成:对 Liouville 函数或 Mobius 函数 $\lambda(n)$,以及复杂度受控的 nil-sequence $F(g^n x)$,在短区间 $I=[X,X+H]$ 上有

    $$\frac1H\sum_{n\in I}\lambda(n)F(g^n x)=o(1).$$

    这里 $H=H(X)\to\infty$,但 $H$ 可以远小于 $X$。这类估计如果成立,会把短区间乘法函数理论和 Sarnak/Chowla 型问题连接起来。

    短区间中 Mobius 函数与 nil-sequence 的相关估计
    短区间中的 Mobius-nilsequence 相关估计试图在局部窗口内捕捉乘法函数的随机性。

    1. 为什么是 nil-sequence

    nil-sequence 是低复杂度动力系统轨道的模型。多项式相位 $e(P(n))$ 是最基本例子,更高阶 nilmanifold 上的轨道则对应高阶 Fourier 分析中的结构部分。若 Mobius 与所有这类低复杂度序列正交,就说明它在动力系统意义下表现得像随机噪声。

    2. 短区间困难

    长区间中可以使用 Bourgain-Sarnak-Ziegler criterion、Vinogradov 型估计和 nilsequence equidistribution。短区间的问题更硬,因为平均长度不够,许多全局消去无法直接使用。

    Matomaki-Radziwill 的定理说明,乘法函数在几乎所有短区间中仍有平均消去。这给出一个强烈信号:如果 nil-sequence 的结构在短窗口上足够规则,那么相关和也应当消失。

    3. 与 Chowla 的关系

    对数平均 Chowla 猜想可以通过 Fourier 展开和结构分解,转化为乘法函数与低复杂度序列的相关估计。这里的短区间版本相当于把“全局随机性”压缩到局部窗口中观察。

    4. 可能路线

    一个可行框架是:先用短区间乘法函数定理处理非结构部分,再对 nil-orbit 做定量 equidistribution 分解,最后用 BSZ 型准则控制剩余相关。核心瓶颈是所有常数都必须对短区间长度 $H$ 有足够好的依赖。

  • 代数数的 Diophantine approximation:Liouville、Roth 与 Vandermonde 约束

    旧博客原文

    原题:Diophantine approximation of algebraic number

    An important theorem in Diophantine approximation is the theorem of Liuoville:

    **Liuoville Theorem** If x is a algebraic number of degree n over the rational number then there exists a constant c(x) > 0 such that:\left|x-{\frac {p}{q}}\right|>{\frac {c(x)}{q^{{n}}}}

    holds for every integer p,q\in N^* where q>0.

    This theorem explain a phenomenon, the approximation of algebraic number by rational number could not be very well. Which was generated later to **Thue–Siegel–Roth theorem**, them could be used to proof a lots of constant is not algebraic, i.e. transcendentals .

    My questions is in another direction, now let us not just consider one root  \alpha_1 of a integer polynomial P(x)=a_mx^m+...+a_1x+a_0 but consider all roots of it, i.e. \{\alpha_1,...,\alpha_m\}, which is based on a observation : If we define

    \sigma_k(P(x))=\sum_{1\leq \alpha_{i_1}<\alpha_{i_2}<...<\alpha_{i_k}\leq m}\alpha_{i_1}\alpha_{i_2}...\alpha_{i_k}

    By **Vieta theorem** we know \sigma_k(n)\in \mathbb Q for all k\in N^*, this will lead to some restriction and in fact destroy the uniformly distribution of (\alpha_1,...,\alpha_m)\in [0,1]^m. In fact the most important one is the determination of Vandermon Determinant:
    V(P(x))=\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j).

    We know \Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\in \mathbb Q so when \Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\neq 0 we could use this to proof a nontrivial estimate for \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}.
    \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).

    by combine the A-G inequality and \Pi_{1\leq \alpha_i<\alpha_j\leq n}(\alpha_i-\alpha_j)=\lambda\neq 0.While by continue fractional expansion we only know a trivial estimate of type \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n}).

    my question is the following:
    Is there still have a nontrivial estimate for \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z} (which could be slight weaker), if we don’t have the whole power of **Vieta theorem**? more precisely:

    **problem 1**

    if we have \sigma_k((\alpha_1,...,\alpha_m))=\lambda_k\in \mathbb Q for all k\in \{1,2,...,m'\} where m'<m, is there still some nontrivial estimate of,

    \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}

    hold for all n\in N^*?

    One reason to consider this could be true is that although \{\alpha_1,...,\alpha_m\} is not roots of a integer polynomial but we could image in some suitable metric space X the gromov-hausdorff distance of tuple (\alpha_1,...,\alpha_m) and a tuple come form roots of integer polynomial is small . And it seems reasonable to image this type of asymptotic quality is continue with the G-H distance on X.

    Another problem is what happen when V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0. More precisely,

    **problem 2**

    What happen when V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0 , is this result,

    \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).

    still true?

    Let us go a litter further, if these two problem both have a satisfied answer, what is the higher dimensional case?

    **problem 3**

    Given m\in \mathbb N^*. If tuple (y_1,...,y_k) is very closed to the zero set of a variety in \mathbb Z[x_1,...,x_m] in \mathbb (Z^{m})^k in the sense a lots of symmetric sum of y_1,...,y_k belong to \mathbb Q^m, will this lead to some good estimate for

    \sum_{1\leq s\leq k}||y_sn||_{\mathbb R^m/\mathbb Z^m}?

    I think these type of result should be investigated very well, Iappreciate to any pointer with useful comments and answer, both on given some strategy to solve these problems or given some reference about these problems.


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    代数数不能被有理数“过分好”地逼近。Liouville theorem 是这件事的第一层形式,Thue-Siegel-Roth theorem 则给出几乎最优的结论。

    代数数的 Diophantine approximation:Liouville、Roth 与 Vandermonde 约束
    Liouville 和 Roth 定理说明代数数不能被有理数过分好地逼近。

    1. Liouville theorem

    若 $\alpha$ 是次数 $d$ 的代数数,则存在 $C(\alpha)>0$,使对所有有理数 $p/q$,

    $$\left|\alpha-\frac pq\right|\ge \frac{C(\alpha)}{q^d}.$$

    证明的核心是把 $\alpha$ 代入整数多项式 $P$,再估计 $P(p/q)$ 不可能是太小的非零有理数。

    2. Roth theorem

    Roth theorem 大幅加强 Liouville:若 $\alpha$ 是无理代数数,则对任意 $\varepsilon>0$,

    $$\left|\alpha-\frac pq\right|<\frac1{q^{2+\varepsilon}}$$

    只有有限多个有理解。换句话说,代数数的有理逼近指数不能超过 $2$ 太多。

    3. 多个根的约束

    若考虑一个整数多项式的所有根 $\alpha_1,\ldots,\alpha_d$,Vieta 定理给出系数与根的对称函数之间的整数关系。Vandermonde determinant

    $$\prod_{i

    又控制根之间不能全部过分靠近。

    4. 从单点逼近到整体结构

    单个根的有理逼近只看一个 $\alpha$;所有根一起看时,还会出现判别式、对称多项式和高度的约束。整体代数结构比单个连分数展开更刚性。

    5. 一个自然问题

    如果没有完整 Vieta 结构,只知道某些弱的对称约束,是否仍能推出非平凡逼近下界?这类问题位于 Diophantine approximation、代数高度和几何不等式之间。

  • Van der Corput trick:从差分到均匀分布

    旧博客原文

    原题:Van der curpurt trick

    There is the statement of Van der carport theorem:

    Given a sequences \{x_n\}_{n=1}^{\infty} in S_1, if \forall k\in N^*, \{x_{n+k}-x_n\} is uniformly distributed, then \{x_n\}_{n=1}^{\infty} is uniformly distributed.

    I do not know how to establish this theorem with no extra condition, but this result is true at least for polynomial flow.

    |\sum_{n=1}^Ne^{2\pi imQ(n)}|= \sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}

    = \sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}=\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}} \leq \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}

    = \sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}

    \leq ....\leq

    \sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1)

     

    This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.

    Uniformly distribution result of F_p:
    Given Q(n)=a_kn^k+...+a_1n+a_0, \{Q(0),Q(1),...,Q(p-1)\} coverages
     to a uniformly distribution in \{0,1,...,p-1\}
     as p \to \infty.

    This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

    S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)=o(n)

    Maybe with the help of B-Z-S theorem.

    The standard estimate of Mobius function is:

    \sum_{n\leq X:n\equiv a~(mod~q)} \mu^2(n)=\frac{6}{\pi^2} \prod_{p|q} \left(1-\frac{1}{p^2} \right)\frac{X}{q}+E(X,q,a)

    The error term O_{\varepsilon}\left(\sqrt{X/q} +q^{\frac{1}{2}+\varepsilon}\right) is true for q\leq X^{\frac{2}{3}-\varepsilon}.

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Van der Corput trick 的核心是:不要直接估计一个振荡平均,而是估计它与平移后的相关。若所有非零差分序列都足够均匀,那么原序列本身也应当均匀。

    Van der Corput trick:从差分到均匀分布
    Van der Corput trick 用差分相关替代原始平均,在多项式相位问题中会降低次数。

    1. 均匀分布版本

    设 $(a_n)$ 是 $\mathbb T^d$ 中的序列。一个典型命题是:若对每个 $h\ne0$,差分序列

    $$a_{n+h}-a_n$$

    都在 $\mathbb T^d$ 中均匀分布,那么 $a_n$ 也均匀分布。证明通常通过 Weyl criterion,把问题化成指数和估计。

    2. 指数和不等式

    对复数序列 $u_n$,Van der Corput 不等式给出

    $$\left|\frac1N\sum_{n\le N}u_n\right|^2
    \lesssim \frac1H+\frac1H\sum_{1\le h\le H}\left|\frac1N\sum_{n\le N-h}u_{n+h}\overline{u_n}\right|.$$

    右侧出现的就是相关项。若相关项都小,则原平均小。

    3. 多项式相位

    当 $u_n=e(P(n))$ 且 $P$ 是次数 $k$ 的多项式时,差分 $P(n+h)-P(n)$ 的次数降为 $k-1$。因此可以用归纳证明 Weyl 型均匀分布结论。

    4. 与乘法函数相关

    在 Mobius 或 Liouville 与低复杂度序列的相关估计中,Van der Corput trick 常用于把一个序列的复杂度下降一层,再配合 Bourgain-Sarnak-Ziegler criterion。它的角色不是给出最终消去,而是把问题改写成更适合结构分析的形式。

  • Baragar-Bourgain-Gamburd-Sarnak 猜想:Markov triples 的模 p 连通性

    旧博客原文

    原题:Baragar-Bourgain-Gamburd-Sarnak conjecture

    M is the markov triple (x,y,z):

    x^2+y^2+z^2=xyz and (x,y,x)\in \mathbb Z^3  \ \ \ \  (*).

    It is easy to see:

    R_1: (x,y,z)\to (3yz-x,y,z).

    map markov triple to markov triple.

    This is also true for R_2,R_3. and the transform R_1,R_2,R_3 and permutation a classical result of markov claim that all solution of  (*) could be generated from (1,1,1). I get a similar result for a similar algebraic equation 1 half years ago when consider a Q version of problem about 1-form given by Xu Bin.

    Now  we know the graph with root (1,1,1) and with node generate by transform R_1\cup R_2 \cup R_3 \cup S_3 is connected.

    The B-B-G-S conjecture is is the connected property still true for prime p surfficed  large?


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Markov triples 是方程

    $$x^2+y^2+z^2=3xyz$$

    的正整数解。经典 Markov 定理说,从根解 $(1,1,1)$ 出发,反复使用 Vieta involution

    $$(x,y,z)\mapsto (x,y,3xy-z)$$

    以及坐标置换,可以生成所有正整数解。

    Baragar-Bourgain-Gamburd-Sarnak 猜想:Markov triples 的模 p 连通性
    Markov triples 的 Vieta involution 生成解图;模 p 后的问题变成有限域曲面上的连通性问题。

    1. 解图

    把每个解看作图的一个顶点,若两个解由一次 Vieta involution 或置换相连,就连一条边。整数正解形成一棵以 $(1,1,1)$ 为根的巨大图。

    2. 模 p 的问题

    把方程放到有限域 $\mathbb F_p$ 上,得到有限集合

    $$X_p=\{(x,y,z)\in\mathbb F_p^3:x^2+y^2+z^2=3xyz\}.$$

    同样的 involution 仍然作用在 $X_p$ 上。Baragar-Bourgain-Gamburd-Sarnak 方向的问题是:当 $p$ 足够大时,这个作用图是否在主要部分上连通,甚至是否具有 expansion 性质?

    3. 为什么这不是普通图论

    图的边来自代数自同构,因此顶点集合有强烈的代数几何结构。连通性问题本质上是在问:这些简单变换能否在有限域曲面上产生足够大的轨道。

    4. 可能工具

    这类问题常混合使用代数几何、有限群展开、谱间隙和 sum-product 现象。若能证明没有大的不变子集,就能排除图分裂成多个大块的可能。

  • 用互不相交闭区间覆盖半开区间:Ostrowski 表示与有效均匀分布

    旧博客原文

    原题:Covering a non-closed interval by disjoint closed intervals

    this note will talk about the Ostrowski representation and approximation by continue fraction.

    As well-known,by the Weyl criterion,\{n\alpha\} is uniformly distribution in [0,1] iff \alpha\in R-Q.

    i.e. we have:\forall 0\leq a\leq b\leq 1,we have:

    \lim_{N\to \infty}|\{1\leq n\leq N|\{n\alpha\}\in [a,b]\}|=(b-a)N+o(N).

    but this will not give the effective version.i.e. we do not the the more information about the decay of o(N).

    we will give a approach of effective version of \alpha with smooth condition by give another proof of the uniformly distribution (in fact to to decomposition the interval [a,b] in to a finite sums of special intervals).and get the result:

    D_N=\int_{M}D_N(\theta)d\mu=\int_Msup_{0<a<b<1}|\sum_{n=1}^{N}\chi_{(a,b)}(\{\theta n \})-N(b-a)|d\mu\sim O(log N)

    if the term in the continuous fraction of \alpha have a up bound.this is so called \alpha is smooth.


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    讨论无理旋转 $x\mapsto x+\alpha$ 的均匀分布时,Weyl criterion 告诉我们

    $$\frac1N\sum_{n\le N}e(k n\alpha)\to0,\qquad k\ne0.$$

    但这个判别法本身并不给出非常几何的有效误差。若想知道轨道落入某个区间 $I$ 的次数和 $N|I|$ 相差多少,就需要把区间和时间长度一起分解得更细。

    用互不相交闭区间覆盖半开区间:Ostrowski 表示与有效均匀分布
    Ostrowski 表示把轨道长度分解成连分数分母尺度,半开区间也可以拆成有限个互不相交的闭区间来控制端点误差。

    1. 半开区间为什么麻烦

    区间若不是闭的,端点会造成计数上的小麻烦;但在动力系统里,端点通常只贡献有限误差。因此可以把半开区间分解成有限个互不相交的闭区间,再把端点误差单独处理。

    2. 连分数与最佳逼近

    设 $\alpha=[a_0;a_1,a_2,\ldots]$,其收敛分母为 $q_j$。最佳逼近性质说明,长度约为 $\|q_j\alpha\|$ 的小区间正好对应旋转轨道的自然尺度。

    Ostrowski 表示把整数 $N$ 写成

    $$N=\sum_j b_j q_j,$$

    其中系数 $b_j$ 由连分数项控制。于是前 $N$ 次轨道可以分成若干个以 $q_j$ 为长度的块。

    3. 有效均匀分布

    若 $\alpha$ 的连分数项有一致上界,也就是 bounded type,那么每个尺度的坏误差都不会积累得太快。这样可以得到 Denjoy-Koksma 型估计:

    $$\left|\sum_{n

    这就是原来想要的“smooth”或有效版本:不仅知道趋于均匀,还知道偏差怎样增长。

  • Bourgain-Sarnak-Ziegler criterion:Mobius 正交性的有限检验

    旧博客原文

    原题:Bourgain-Sarnak-Ziegler Criterion

    img_0516.jpgimg_0517.jpgBourgain-Sarnak-Ziegler定理可以视为Vingrodov均值定理的有限版本。


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Bourgain-Sarnak-Ziegler criterion 可以看成 Vinogradov 均值思想的有限版本:若一个有界序列在不同素数倍采样下彼此近似正交,那么它与 Mobius 函数也应当正交。

    Bourgain-Sarnak-Ziegler criterion:Mobius 正交性的有限检验
    BSZ criterion 用不同素数倍采样的相关消失来推出 Mobius 正交性。

    1. 判别法的形式

    设 $a_n$ 是有界序列。若对不同素数 $p\ne q$,有

    $$\sum_{n\le N}a_{pn}\overline{a_{qn}}=o(N)$$

    并且这个估计对一批素数足够一致,那么可以推出

    $$\sum_{n\le N}\mu(n)a_n=o(N).$$

    2. 为什么素数倍相关重要

    Mobius 函数的困难在于它携带素因子结构。BSZ criterion 的想法是:不直接分析 $\mu(n)$,而是检查序列 $a_n$ 对不同素数尺度是否产生相关。如果所有这些相关都小,Mobius 就没有可利用的结构与之耦合。

    3. 与 Sarnak 猜想

    在动力系统中,常取 $a_n=f(T^n x)$。于是 Mobius disjointness 变成

    $$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to0.$$

    BSZ 把这个问题转化为比较 $T^p$ 和 $T^q$ 产生的两个轨道序列。

    4. 有限版本的意义

    称它为 Vinogradov 均值定理的有限版本,是因为它同样通过“多重相关消失”来控制原始振荡和。它特别适合低复杂度系统,例如 nilsequence、skew product 或 substitution dynamics。

  • 随机矩阵与 zeta 函数的 moment 估计

    旧博客原文

    原题:随机矩阵与zeta函数的moment估计

    旧站归档中的这篇正文原本为空。


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Riemann zeta 函数在临界线上的矩估计是解析数论的中心问题之一。随机矩阵理论给出一个非常强的模型:$\zeta(1/2+it)$ 的局部统计像酉矩阵特征多项式在单位圆上的取值。

    随机矩阵与 zeta 函数的 moment 估计
    随机酉矩阵特征多项式为 zeta 函数临界线矩估计提供了主项指数和常数结构的模型。

    1. zeta 矩

    第 $2k$ 阶矩通常写成

    $$M_k(T)=\int_0^T|\zeta(1/2+it)|^{2k}\,dt.$$

    猜想主项形如

    $$M_k(T)\sim C_k T(\log T)^{k^2}.$$

    2. CUE 模型

    取 $U\in U(N)$ 为 Haar 随机酉矩阵,考虑特征多项式

    $$Z_U(\theta)=\det(I-e^{-i\theta}U).$$

    Keating-Snaith 模型把 $N$ 与 $\log T$ 对应,并用 $\mathbb E|Z_U(\theta)|^{2k}$ 预测 zeta 矩中的 $k^2$ 指数。

    3. 算术因子

    随机矩阵只解释对称性和谱统计部分。zeta 函数还包含 Euler product,因此常数 $C_k$ 应分解成随机矩阵因子和算术 Euler product 因子。

    4. 为什么这个模型有效

    零点统计、特征多项式矩和 logarithmic correlation 都显示出同一类结构。随机矩阵不是证明本身,但它给出正确主项、正确指数和很多低阶项的预测,是组织 zeta 矩问题的有效语言。