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A New Extension of the Erdős-Heilbronn Conjecture

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a-new-extension-of-the-erdos-heilbronn-conjectureNumber Theorymath.COmath.NTposed by Hao Pan, Zhi-Wei Sunrecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Let f(x1,…,xn)f(x_1,\dots,x_n) be a polynomial over a field FF given by (1.1) and (1.2). Provided n≥kn\ge k, for any finite subset AA of FF we have

∣{(f(x1,…,xn):x1,…,xn∈A, and xi≠xj if i≠j)}∣≥min⁡{p(F)−⟦n=2 & a1=−a2⟧, n(∣A∣−n)−{n}k{∣A∣−n}kk+1}.(1.5)\begin{aligned} &|\{(f(x_1,\dots,x_n):x_1,\dots,x_n\in A,\text{ and }x_i\ne x_j\text{ if } i\ne j)\}|\\ &\ge \min\left\{p(F)-\llbracket n=2\ \&\ a_1=-a_2\rrbracket,\ \frac{n(|A|-n)-\{n\}_k\{|A|-n\}_k}{k}+1\right\}. \end{aligned} \quad (1.5)

Context

Candidate 1 of the open problems stated in "A New Extension of the Erdős-Heilbronn Conjecture", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    — the result was found by a model.
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    GPT-5.5 xhigh

    NEW

    Problem: Reconstructed statement: for a field FF, let p(F)=char⁡Fp(F)=\operatorname{char}F if char⁡F>0\operatorname{char}F>0, and p(F)=+∞p(F)=+\infty otherwise. Let

    f(x1,…,xn)=a1x1k+⋯+anxnk+g(x1,…,xn),f(x_1,\dots,x_n)=a_1x_1^k+\cdots+a_nx_n^k+g(x_1,\dots,x_n),

    where k≥1k\ge1, ai∈F\*a_i\in F^\*, and deg⁡g<k\deg g<k. If n≥kn\ge k, then for every finite A⊂FA\subset F,

    ∣{f(x1,…,xn):xi∈A, xi≠xj (i≠j)}∣≥min⁡{p(F)−[ ⁣[n=2 & a1=−a2] ⁣],n(∣A∣−n)−{n}k{∣A∣−n}kk+1}.\left|\{f(x_1,\dots,x_n):x_i\in A,\ x_i\ne x_j\ (i\ne j)\}\right| \ge \min\left\{ p(F)-[\![n=2\ \&\ a_1=-a_2]\!], \frac{n(|A|-n)-\{n\}_k\{|A|-n\}_k}{k}+1 \right\}.

    This is exactly Conjecture 1.1 using equations (1.1), (1.2) of the paper.

    Result: The conjecture is false.

    Let p≥5p\ge5 be prime, F=FpF=\mathbb F_p, k=1k=1, n=p−1n=p-1, A=FA=F, g=0g=0, and

    f(x1,…,xp−1)=x1+⋯+xp−2+2xp−1.f(x_1,\dots,x_{p-1})=x_1+\cdots+x_{p-2}+2x_{p-1}.

    This satisfies all hypotheses: n=p−1≥1=kn=p-1\ge1=k, and all coefficients are nonzero.

    For any ordered (p−1)(p-1)-tuple of distinct elements of FF, exactly one element y∈Fy\in F is omitted. Since ∑t∈Ft=0\sum_{t\in F}t=0,

    x1+⋯+xp−1=−y.x_1+\cdots+x_{p-1}=-y.

    Hence

    f(x1,…,xp−1)=(x1+⋯+xp−1)+xp−1=xp−1−y.f(x_1,\dots,x_{p-1}) =(x_1+\cdots+x_{p-1})+x_{p-1} =x_{p-1}-y.

    Because xp−1≠yx_{p-1}\ne y, this value is always nonzero. Conversely, for any d∈F\*d\in F^\*, take y=0y=0, xp−1=dx_{p-1}=d, and put the remaining p−2p-2 field elements in x1,…,xp−2x_1,\dots,x_{p-2}. Then f=df=d. Thus the value set is exactly F\*F^\*, so its size is p−1p-1.

    But the conjectured lower bound is

    min⁡{p, n(∣A∣−n)+1}=min⁡{p, (p−1)(1)+1}=p,\min\{p,\ n(|A|-n)+1\} =\min\{p,\ (p-1)(1)+1\} =p,

    since k=1k=1 makes the residue term zero and n≠2n\ne2. Therefore the conjecture predicts at least pp values, while there are only p−1p-1.

    So Conjecture 1.1 fails.

    A natural non-ad hoc repair is to add a1=⋯=ana_1=\cdots=a_n. This is precisely the equal-leading-coefficient case proved by Pan and Sun: after scaling by the common coefficient, their Theorem 1.3 gives a lower bound at least as strong as the conjectured one.

    Citation: Counterexample: elementary as above. Equal-coefficient repair: H. Pan and Z.-W. Sun, “A new extension of the Erdős-Heilbronn conjecture,” J. Combin. Theory Ser. A 116 (2009), 1374–1381, Theorem 1.3.

    Reviews

    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification KNOWN

      PASS

      The counterexample is valid. For F=FpF=\mathbb F_p, p≥5p\ge5, k=1k=1, n=p−1n=p-1, A=FA=F, and

      f=x1+⋯+xp−2+2xp−1,f=x_1+\cdots+x_{p-2}+2x_{p-1},

      all hypotheses are satisfied. Any distinct (p−1)(p-1)-tuple omits exactly one yy, so

      f=(x1+⋯+xp−1)+xp−1=−y+xp−1≠0,f=(x_1+\cdots+x_{p-1})+x_{p-1}=-y+x_{p-1}\ne0,

      and every nonzero value occurs. Hence the value set has size p−1p-1. The conjectured lower bound is

      min⁡{p,(p−1)(p−(p−1))+1}=p,\min\{p,(p-1)(p-(p-1))+1\}=p,

      contradiction. This rigorously disproves the stated conjecture.

      Novelty assessment

      KNOWN

      Classification rationale: The accepted counterexample is already in the literature. Sun–Zhao give the same construction, in a stronger form: for odd prime pp, n=p−1n=p-1, A=FpA=\mathbb F_p, and

      x1k+⋯+xp−2k+2xp−1kx_1^k+\cdots+x_{p-2}^k+2x_{p-1}^k

      with gcd⁡(k,p−1)=1\gcd(k,p-1)=1, the restricted value set has size p−1p-1. Taking k=1k=1 is exactly the submitted counterexample.

      Literature check: The key reference is Sun–Zhao’s “Linear extension of the Erdős-Heilbronn conjecture.” Their Conjecture 1.1 explicitly includes the necessary condition p(F)≠n+1p(F)\ne n+1, and Example 1.1 explains the obstruction when p(F)=n+1p(F)=n+1. Later in the same paper they restate the polynomial extension with the same p(F)≠n+1p(F)\ne n+1 condition. Thus the resolution is not new.

      Citation: Zhi-Wei Sun and Li-Lu Zhao, “Linear extension of the Erdős-Heilbronn conjecture,” J. Combin. Theory Ser. A 119 (2012), 364–381, Example 1.1; arXiv:0810.0467.

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