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Math-649-A9 · revision r1 · on Math-649

gpt pro 5.4 attempt on Math-649, version 2

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Contents
  1. Round 7 (Problem 30): a distinct-gap improvement of the Singer embedding
  2. 1) Round-7 Objective
  3. 2) Round-6 / Earlier Foundation Used
  4. 3) New Insight / Tool (Round 7)
  5. 4) Attack Plan (Round 7)
  6. 5) Work (Round 7)
  7. 5.1. Distinct cyclic gaps in a perfect difference set
  8. 5.2. Extremal bound for the largest distinct gap
  9. 5.3. Improved Singer embedding
  10. 5.4. Consequence for the second-order term
  11. 6) Adversarial Verification
  12. 7) Final
  13. 8) Completion Estimate
  14. 9) References

Round 7 (Problem 30): a distinct-gap improvement of the Singer embedding

1) Round-7 Objective

Path (A): proof direction (strengthen the explicit lower-bound family). Round 6 globalised the Round 5 Singer subsequence by combining it with primes in short intervals, but the actual subsequence construction from Round 5 still used only the crude fact that a $(q+1)$-point subset of a cycle of length $q^2+q+1$ has a gap of size at least the average, namely $q+1$.

The most promising direction is therefore to sharpen the cyclic geometry of the Singer construction itself. In this round we do exactly that.

The new theorem is: \[ h\!\left(q^2-\left\lfloor \frac q2\right\rfloor+1\right)\ge q+1 \qquad\text{for every prime power }q. \] This strictly improves the Round 5 family \[ h(q^2+1)\ge q+1. \] As a consequence, \[ \limsup_{N\to\infty}\left(h(N)-\sqrt N\right)\ge \frac54, \] improving the Round 5 bound $\ge 1$.

The original Erdős--Turán problem remains open: is it true that for every $\varepsilon>0$, \[ h(N)=\sqrt N+O_\varepsilon(N^\varepsilon)\,? \]

2) Round-6 / Earlier Foundation Used

We rely on the following previously vetted results.
  • (R5 Singer existence theorem) For every prime power $q$ and \[ v=q^2+q+1, \] there exists a perfect difference set \[ D\subset \mathbb Z/v\mathbb Z \qquad\text{with}\qquad |D|=q+1. \]

  • (R5 injective modular-to-integer transfer) If $B\subset \mathbb Z$ is finite, reduction modulo $v$ is injective on $B$, and the image of $B$ is Sidon in $\mathbb Z/v\mathbb Z$, then $B$ is Sidon in $\mathbb Z$.

  • (R5 cut-at-a-gap principle) If a translate of $D\subset \mathbb Z/v\mathbb Z$ lies on an arc of length $L$, then choosing integer representatives on that arc and translating to start at $1$ gives an integer Sidon set inside $\{1,\dots,L+1\}$.

  • (R6 final status) The main Erdős--Turán question remains open; Round 6 produced global lower bounds from the Round 5 family, but did not improve the Singer subsequence itself.

We will strengthen the Round 5 gap argument by exploiting a new structural property of perfect difference sets.

3) New Insight / Tool (Round 7)

The cyclic gaps in a perfect difference set are pairwise distinct.

Let \[ 0\le d_0<d_1<\cdots<d_q\le v-1 \] be representatives of a perfect difference set $D\subset \mathbb Z/v\mathbb Z$, and define the cyclic gaps \[ g_i:=d_{i+1}-d_i\quad(0\le i\le q-1),\qquad g_q:=d_0+v-d_q. \] Each $g_i$ is an ordered difference of two elements of $D$. Since perfect difference sets have all nonzero ordered differences distinct, the $g_i$ must be pairwise distinct.

This single observation upgrades the Round 5 averaging bound: instead of merely knowing that the largest gap is at least the average $(q^2+q+1)/(q+1)$, we now know it is the largest among $q+1$ distinct positive integers summing to $q^2+q+1$. That forces a much larger gap.

4) Attack Plan (Round 7)

The exact gap left after Round 6 is this:
Round 6 globalised the old Singer subsequence, but the intrinsic lower-bound subsequence itself still only came from the weak bound "max gap $\ge$ average gap".

To improve it, we:

  1. prove that the cyclic gaps in a perfect difference set are distinct;
  2. prove a sharp extremal lemma for the largest of $q+1$ distinct positive integers with fixed sum $q^2+q+1$;
  3. cut the Singer cycle at this improved gap;
  4. transfer back to the integer setting using the Round 5 injective transfer lemma.

This yields a strictly shorter ambient interval than in Round 5, and therefore a larger lower bound for the second-order term.

5) Work (Round 7)

5.1. Distinct cyclic gaps in a perfect difference set

Lemma (Distinct-gap lemma): Let $q$ be a prime power, let $v=q^2+q+1$, and let \[ D\subset \mathbb Z/v\mathbb Z \] be a perfect difference set of size $q+1$. Choose representatives \[ 0\le d_0<d_1<\cdots<d_q\le v-1 \] and define cyclic gaps \[ g_i:=d_{i+1}-d_i\quad(0\le i\le q-1),\qquad g_q:=d_0+v-d_q. \] Then:
  1. each $g_i$ is a positive integer;
  2. \[ \sum_{i=0}^q g_i=v; \]
  3. the integers $g_0,\dots,g_q$ are pairwise distinct.
Proof: The first two assertions are immediate from the definition of the cyclic gaps.

For distinctness, note that for $0\le i\le q-1$ the gap $g_i$ is exactly the ordered difference \[ d_{i+1}-d_i \pmod v, \] and \[ g_q\equiv d_0-d_q \pmod v. \] Thus every $g_i$ is a nonzero ordered difference between two elements of $D$.

If $g_i=g_j$ for some $i\neq j$, then two distinct ordered pairs of elements of $D$ would yield the same nonzero ordered difference modulo $v$, contradicting the defining property of a perfect difference set. Therefore the $g_i$ are pairwise distinct. ∎

5.2. Extremal bound for the largest distinct gap

Lemma (Largest distinct summand bound): Let $m\ge 1$, and let \[ 1\le x_1< x_2<\cdots< x_m \] be distinct positive integers with sum \[ x_1+\cdots+x_m=S. \] Then \[ x_m\ge \left\lceil \frac{S+\binom{m}{2}}{m}\right\rceil. \]
Proof: Since the $x_i$ are distinct and all at most $x_m$, the largest possible value of their sum for a fixed $x_m$ is obtained by taking \[ x_m,\ x_m-1,\ \dots,\ x_m-(m-1). \] Hence \[ S\le x_m+(x_m-1)+\cdots+(x_m-(m-1)) =m x_m-\binom{m}{2}. \] Rearranging gives \[ x_m\ge \frac{S+\binom{m}{2}}{m}, \] and therefore \[ x_m\ge \left\lceil \frac{S+\binom{m}{2}}{m}\right\rceil. \] ∎
Corollary (Improved gap lower bound for Singer parameters): With the notation of Lemma Lemma (Distinct-gap lemma), let \[ G:=\max_{0\le i\le q} g_i. \] Then \[ G\ge \left\lfloor \frac{3q}{2}\right\rfloor+1. \]
Proof: Apply Lemma Lemma (Largest distinct summand bound) with \[ m=q+1,\qquad S=v=q^2+q+1. \] Since the $g_i$ are distinct positive integers summing to $v$, we obtain \[ G\ge \left\lceil \frac{q^2+q+1+\binom{q+1}{2}}{q+1}\right\rceil = \left\lceil \frac{q^2+q+1+\frac{q(q+1)}2}{q+1}\right\rceil = \left\lceil \frac{3q}{2}+\frac{1}{q+1}\right\rceil. \] Because $1/(q+1)>0$, this equals $\lfloor 3q/2\rfloor+1$. ∎

5.3. Improved Singer embedding

Theorem (Improved Singer subsequence): For every prime power $q$, there exists an integer Sidon set \[ S\subseteq \left\{1,2,\dots, q^2-\left\lfloor \frac q2\right\rfloor+1\right\} \] with \[ |S|=q+1. \] Equivalently, \[ h\!\left(q^2-\left\lfloor \frac q2\right\rfloor+1\right)\ge q+1. \]
Proof: Let \[ v:=q^2+q+1, \] and let \[ D\subset \mathbb Z/v\mathbb Z \] be a perfect difference set of size $q+1$, guaranteed by the Round 5 Singer existence theorem.

Choose representatives \[ 0\le d_0<d_1<\cdots<d_q\le v-1 \] and define cyclic gaps $g_0,\dots,g_q$ as in Lemma Lemma (Distinct-gap lemma). Let \[ G:=\max_i g_i. \] By Corollary Corollary (Improved gap lower bound for Singer parameters), \[ G\ge \left\lfloor \frac{3q}{2}\right\rfloor+1. \]

Now cut the cycle at a gap of size $G$. That is, translate $D$ so that the point after this largest gap becomes $0$, and then choose integer representatives on the complementary arc. By the Round 5 cut-at-a-gap principle, we obtain a set \[ B\subset \mathbb Z \] whose reduction modulo $v$ is a translate of $D$, and whose diameter satisfies \[ \operatorname{diam}(B)\le v-G. \] Using the bound on $G$ gives \[ \operatorname{diam}(B)\le q^2+q+1-\left(\left\lfloor \frac{3q}{2}\right\rfloor+1\right) = q^2+q-\left\lfloor \frac{3q}{2}\right\rfloor. \] Since \[ q^2+q-\left\lfloor \frac{3q}{2}\right\rfloor = q^2-\left\lfloor \frac q2\right\rfloor < v, \] reduction modulo $v$ is injective on $B$. Moreover the image of $B$ modulo $v$ is a translate of the perfect difference set $D$, hence is Sidon in $\mathbb Z/v\mathbb Z$.

By the Round 5 injective modular-to-integer transfer lemma, $B$ is Sidon in $\mathbb Z$. Finally translate $B$ so that its minimum becomes $1$: \[ S:=B-(\min B)+1. \] Then \[ |S|=|B|=q+1 \] and \[ S\subseteq \{1,2,\dots,\operatorname{diam}(B)+1\} \subseteq \left\{1,2,\dots, q^2-\left\lfloor \frac q2\right\rfloor+1\right\}. \] Thus $S$ is an integer Sidon set of the required size in the claimed interval. ∎

5.4. Consequence for the second-order term

Corollary (Improved limsup): One has \[ \limsup_{N\to\infty}\left(h(N)-\sqrt N\right)\ge \frac54. \]
Proof: Set \[ N_q:=q^2-\left\lfloor \frac q2\right\rfloor+1, \] where $q$ ranges over prime powers. By Theorem Theorem (Improved Singer subsequence), \[ h(N_q)\ge q+1. \] Therefore \[ h(N_q)-\sqrt{N_q}\ge q+1-\sqrt{N_q}. \]

Now \[ \left\lfloor \frac q2\right\rfloor\ge \frac q2-\frac12, \] so \[ N_q\le q^2-\frac q2+\frac32. \] Hence \[ q+1-\sqrt{N_q}\ge q+1-\sqrt{q^2-\frac q2+\frac32}. \] Let \[ \alpha_q:=q-\frac14. \] Then \[ \alpha_q^2=q^2-\frac q2+\frac1{16}, \] so \[ q^2-\frac q2+\frac32=\alpha_q^2+\frac{23}{16}. \] Therefore \[ \sqrt{q^2-\frac q2+\frac32} =\alpha_q+\frac{\frac{23}{16}}{\alpha_q+\sqrt{q^2-\frac q2+\frac32}}. \] As $q\to\infty$, the correction term tends to $0$, so \[ \sqrt{q^2-\frac q2+\frac32}=q-\frac14+o(1). \] Consequently \[ q+1-\sqrt{q^2-\frac q2+\frac32}=\frac54+o(1), \] and thus \[ h(N_q)-\sqrt{N_q}\ge \frac54+o(1). \] Taking the limsup over prime powers $q$ gives \[ \limsup_{N\to\infty}\left(h(N)-\sqrt N\right)\ge \frac54. \] ∎

6) Adversarial Verification

We check the new argument carefully.
  • Are the cyclic gaps really ordered differences in the modular sense? Yes. For $0\le i\le q-1$, one has $g_i=d_{i+1}-d_i$ as an integer and hence modulo $v$. For the wrap-around gap, $g_q=d_0+v-d_q$, so modulo $v$ this is $d_0-d_q$. Thus every cyclic gap is indeed a nonzero ordered difference.

  • Could two equal cyclic gaps come from the same ordered pair? No. The ordered pairs $(d_{i+1},d_i)$ are distinct as $i$ varies cyclically, so equality of two gaps would give two distinct ordered pairs with the same nonzero difference, contradicting perfectness.

  • Is the extremal bound in Lemma Lemma (Largest distinct summand bound) sharp enough? Yes. It is the best possible bound obtainable from only "distinct positive integers with fixed sum", and it is exactly the information needed to improve the Round 5 average-gap argument.

  • Is the new ambient length formula correct? From the cut-at-gap step, the final interval length is \[ \operatorname{diam}(B)+1\le (v-G)+1. \] Since $G\ge \lfloor 3q/2\rfloor+1$ and $v=q^2+q+1$, this becomes \[ v-G+1\le q^2+q+1-\left(\left\lfloor\frac{3q}{2}\right\rfloor+1\right)+1 =q^2-\left\lfloor\frac q2\right\rfloor+1. \] So there is no off-by-one error.

  • Does injectivity of reduction modulo $v$ hold? Yes, because \[ \operatorname{diam}(B)\le q^2-\left\lfloor \frac q2\right\rfloor < q^2+q+1=v. \] Therefore no two distinct elements of $B$ can be congruent modulo $v$.

  • Does the limsup computation really tend to $5/4$? Yes. The parameter \[ N_q=q^2-\left\lfloor \frac q2\right\rfloor+1 \] satisfies \[ N_q=q^2-\frac q2+O(1), \] so \[ \sqrt{N_q}=q-\frac14+o(1), \] which yields \[ q+1-\sqrt{N_q}=\frac54+o(1). \]

7) Final

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 7 proves a genuinely stronger explicit lower-bound family than any previous round: \[ h\!\left(q^2-\left\lfloor \frac q2\right\rfloor+1\right)\ge q+1 \qquad(q\text{ a prime power}), \] which strictly improves the Round 5 family $h(q^2+1)\ge q+1$.

As a consequence, \[ \limsup_{N\to\infty}\left(h(N)-\sqrt N\right)\ge \frac54, \] improving the Round 5 lower bound $\ge 1$.

The original Erdős--Turán problem --- whether for every $\varepsilon>0$, \[ h(N)=\sqrt N+O_\varepsilon(N^\varepsilon) \] --- remains open.

8) Completion Estimate

COMPLETION: 80%

9) References

  • T. F. Bloom (ed.), Erdős Problem #30, https://www.erdosproblems.com/30, accessed 7 March 2026.
  • J. Singer, A theorem in finite projective geometry and some applications to number theory, Trans. Amer. Math. Soc. 43 $1938$, 377--385.