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Math-656-A5 · revision r1 · on Math-656

gpt pro 5.4 attempt on Math-656

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Contents
  1. Erdős Problem #44 --- Round 4
  2. 1) ROUND-4 OBJECTIVE
  3. 2) ROUND-3 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-4)
  5. 4) ATTACK PLAN (ROUND-4)
  6. 5) WORK (ROUND-4)
  7. 5.1\; Prime near a prescribed target
  8. 5.2\; Main theorem: every sufficiently large cutoff is admissible
  9. 5.3\; Reformulation of the advance
  10. 6) ADVERSARIAL VERIFICATION
  11. 7) FINAL
  12. 8) COMPLETION ESTIMATE (MANDATORY)
  13. 9) REFERENCES

Erdős Problem #44 --- Round 4

1) ROUND-4 OBJECTIVE

Path (A): proof. Given the Round-3 status, the most promising direction is not to restart the global problem, but to close the cutoff-quantifier gap left by Round 2 and Round 3. Round 2 proved that every finite Sidon set extends to a Sidon superset of density $(1/\sqrt2-o(1))$ for some cutoff $M$, and Round 3 upgraded this to an infinite perfect-difference completion with the same limsup density. In this round I prove a strictly stronger finite theorem: \[ \textit{for every sufficiently large } M, \textit{ not merely for some } M, \] there is a Sidon extension of size at least $(1/\sqrt2-\varepsilon)\sqrt M$. This does not settle the original constant-$1$ target, but it eliminates the remaining "sporadic $M$" weakness in the Round-2 construction.

2) ROUND-3 FOUNDATION USED

I use the following previously established ingredient, which Round 3 explicitly preserved as valid progress from Round 2.

Round-2 block-extension construction. Fix a finite Sidon set $A\subset [N]$ and a prime power $q$. Put \[ L:=q^2+q+1. \] Whenever $L>N$, the Round-2 Singer-block + forbidden-difference pruning + sum-range separation argument produces a set \[ B\subset [N+L+1,\,N+2L] \] such that $A\cup B$ is Sidon and \[ |A\cup B|\ge q-N+2. \] I use this construction as a black box; I do not re-prove it here.

The unresolved issue after Round 3 was therefore not how to build one dense block, but how to choose the block order $q$ so that the resulting cutoff lands at a prescribed large $M$.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredient is a prime-near-target selection lemma. By the prime number theorem, for every fixed $\delta>0$, every sufficiently large interval \[ [(1-\delta)x,\,x] \] contains a prime. Since a prime is a prime power, this lets us choose the Singer parameter $q$ very close to the target value \[ \sqrt{\frac{M-N}{2}}, \] so that the Round-2 block occupies almost exactly half of $[1,M]$. This upgrades "there exists some $M$" to "every sufficiently large $M$" without changing the constant $1/\sqrt2$.

4) ATTACK PLAN (ROUND-4)

Exact gap left after Round 3. Round 3 supplied infinitely many dense cutoffs, but not a theorem for a given large $M$.

Precise claim to prove now. For every finite Sidon set $A\subset [N]$ and every $\varepsilon>0$, there exists $M_0=M_0(A,\varepsilon)$ such that for every $M\ge M_0$ there exists \[ B\subset \{N+1,\dots,M\} \] with $A\cup B\subset [M]$ Sidon and \[ |A\cup B|\ge \left(\frac1{\sqrt2}-\varepsilon\right)\sqrt M. \]

Why this plan works. The Round-2 construction already gives the right density once $q$ is chosen. The only missing step is to ensure simultaneously that \[ q^2+q+1\le \frac{M-N}{2} \] and \[ q=(1-o(1))\sqrt{\frac M2}. \] A prime in a short relative interval does exactly this.

5) WORK (ROUND-4)

5.1\; Prime near a prescribed target

Lemma (Prime in a short relative interval): For every $\delta\in(0,1)$ there exists $x_0=x_0(\delta)$ such that every real $x\ge x_0$ admits a prime $q$ with \[ (1-\delta)x\le q\le x. \]
Proof: By the prime number theorem, \[ \pi(x)\sim \frac{x}{\log x} \qquad (x\to\infty). \] Hence for fixed $\delta\in(0,1)$, \[ \pi(x)-\pi((1-\delta)x) \sim \frac{x}{\log x}-\frac{(1-\delta)x}{\log((1-\delta)x)} \sim \frac{\delta x}{\log x}>0. \] Therefore $\pi(x)-\pi((1-\delta)x)\ge 1$ for all sufficiently large $x$, which means that the interval $[(1-\delta)x,x]$ contains a prime. ∎

5.2\; Main theorem: every sufficiently large cutoff is admissible

Theorem (Cofinite-cutoff extension at constant $1/\sqrt2$): Let $N\ge 1$ and let $A\subset [N]$ be a Sidon set. Then for every $\varepsilon>0$ there exists $M_0=M_0(A,\varepsilon)$ such that for every integer $M\ge M_0$ there exists a set \[ B\subset \{N+1,\dots,M\} \] for which $A\cup B\subset [M]$ is Sidon and \[ |A\cup B|\ge \left(\frac1{\sqrt2}-\varepsilon\right)\sqrt M. \]
Proof: If $\varepsilon\ge 1/\sqrt2$, then the target lower bound is nonpositive and the statement is trivial. So assume \[ 0<\varepsilon<\frac1{\sqrt2}. \] Choose $\delta\in(0,1/2)$ so small that \begin{equation} \label{wsref-0-65-71-3a-64-65-6c-74-61-2d-63-68-6f-69-63-65} \frac{1-\delta}{\sqrt2}>\frac1{\sqrt2}-\frac\varepsilon2. \end{equation} Apply Lemma Lemma (Prime in a short relative interval) with this $\delta$, and let $x_0=x_0(\delta)$ be the corresponding threshold.

For $M>N$, define \[ X(M):=\sqrt{\frac{M-N}{2}}-1. \] Then \[ \frac{X(M)}{\sqrt M}\longrightarrow \frac1{\sqrt2} \qquad (M\to\infty), \] so by (eq:delta-choice), \[ \frac{(1-\delta)X(M)-N+2}{\sqrt M}\longrightarrow \frac{1-\delta}{\sqrt2}>\frac1{\sqrt2}-\frac\varepsilon2. \] Hence there exists $M_1$ such that for every $M\ge M_1$, \begin{equation} \label{wsref-0-65-71-3a-73-69-7a-65-2d-74-61-72-67-65-74} (1-\delta)X(M)-N+2\ge \left(\frac1{\sqrt2}-\varepsilon\right)\sqrt M. \end{equation} Choose also $M_2$ so large that for every $M\ge M_2$, \begin{equation} \label{wsref-0-65-71-3a-61-75-78-2d-74-68-72-65-73-68-6f-6c-64-73} X(M)\ge x_0, \qquad (1-\delta)X(M)\ge N, \qquad M\ge N+2. \end{equation} Set \[ M_0:=\max\{M_1,M_2\}. \] Now fix any integer $M\ge M_0$. By (eq:aux-thresholds) and Lemma Lemma (Prime in a short relative interval), there exists a prime $q$ such that \[ (1-\delta)X(M)\le q\le X(M). \] In particular $q\ge N$, so with \[ L:=q^2+q+1 \] we have $L>N$.

We next check that the Round-2 block fits inside $[M]$. Since $q\le X(M)$ and $X(M)=\sqrt{(M-N)/2}-1$, writing $Y:=\sqrt{(M-N)/2}$ gives $q\le Y-1$, and therefore \[ L=q^2+q+1\le (Y-1)^2+(Y-1)+1=Y^2-Y+1. \] Because $M\ge N+2$, one has $Y\ge 1$, so $Y^2-Y+1\le Y^2=(M-N)/2$. Hence \[ L\le \frac{M-N}{2}, \qquad\text{so}\qquad N+2L\le M. \]

Now invoke the Round-2 block-extension construction with this prime power $q$. It yields a set \[ B\subset [N+L+1,\,N+2L]\subset [N+1,\,M] \] such that $A\cup B$ is Sidon and \[ |A\cup B|\ge q-N+2. \] Finally, \[ |A\cup B|\ge q-N+2\ge (1-\delta)X(M)-N+2 \] and so by (eq:size-target), \[ |A\cup B|\ge \left(\frac1{\sqrt2}-\varepsilon\right)\sqrt M. \] This proves the theorem. ∎

5.3\; Reformulation of the advance

Theorem thm:cofinite-M strictly strengthens the Round-2 theorem. Round 2 gave, for fixed $(A,\varepsilon)$, existence of at least one suitable cutoff $M$. Round 4 gives every sufficiently large cutoff $M$. Thus the remaining gap to Erdős Problem #44 is no longer a quantifier problem about which $M$ are reachable; it is purely the density constant $1$ versus $1/\sqrt2$.

6) ADVERSARIAL VERIFICATION

  • Did I accidentally use more than Round 2 proved? No. The only imported fact is the explicit Round-2 construction: for each prime power $q$ with $L=q^2+q+1>N$, it produces a block $B\subset [N+L+1,N+2L]$ and the lower bound $|A\cup B|\ge q-N+2$.

  • Does the block really lie inside the prescribed ambient interval $[M]$? Yes. The estimate $L\le (M-N)/2$ gives $N+2L\le M$, so the entire Round-2 block sits inside $[N+1,M]$.

  • Could $q$ be too small, invalidating the Round-2 separation argument? No. By construction $q\ge N$, hence $L=q^2+q+1>N$.

  • Is the prime-selection lemma strong enough? Yes. We only need a prime in a fixed relative interval $[(1-\delta)x,x]$, and the prime number theorem implies this for all sufficiently large $x$.

  • Boundary cases. If $\varepsilon\ge 1/\sqrt2$, the theorem is trivial. If $A=\varnothing$, the same proof works verbatim with any $N\ge 1$ such that $A\subset [N]$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED). This round does not settle the original constant-$1$ conjecture. However, it proves a genuinely stronger theorem than any previous round: \[ \textit{for every finite Sidon set }A\subset [N]\textit{ and every }\varepsilon>0,\textit{ every sufficiently large }M \] admits a Sidon extension inside $[M]$ of size at least \[ \left(\frac1{\sqrt2}-\varepsilon\right)\sqrt M. \] So the Round-2 / Round-3 constant $1/\sqrt2$ is now available with the correct cofinite quantifier in $M$; the only unresolved issue is whether the constant can be pushed all the way from $1/\sqrt2$ to $1$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 78%

9) REFERENCES

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008. (Prime number theorem.)
  • J. Singer, A theorem in finite projective geometry and some applications to number theory, Trans. Amer. Math. Soc. 43 $1938$, 377--385. (Underlying finite-geometric Sidon block used in Round 2.)