gpt pro 5.4 attempt on Math-653

Erdős Problem #39 (Round 4)

1) ROUND-4 OBJECTIVE

Path (C) --- obstruction/correction.

While adversarially checking the Round 3 construction, I found a concrete flaw in the mixed-radix/discrete-log definition: for $p\in P_k$, the digit $x_j(p)$ was defined by \[ g_j^{x_j(p)}\equiv p \pmod{q_j}\qquad (1\le j\le k), \] but this congruence has no solution when $p=q_j$, because then the right-hand side is $0$ mod $q_j$ while the left-hand side lies in $\mathbb F_{q_j}^{\times}$. Thus the Round 2/3 construction was not literally well-defined on the full prime index set as stated.

The most promising Round 4 direction is therefore:

1. identify the exact exceptional indices where the definition fails;
2. prove that this obstruction is only finite;
3. give the minimal corrected statement and show that the Round 3 two-threshold theorem survives unchanged after this repair;
4. isolate the exact point where the current bad-prime counting mechanism becomes critical.

So the goal of this round is not to solve the original Erdős problem outright, but to repair a genuine logical gap and convert the Round 3 progress into a fully rigorous corrected theorem.

2) Round-3 FOUNDATION USED

I use the following specific Round 3 ingredients, but I do not re-prove them unless they need modification after the correction:

• The mixed-radix/discrete-log framework with bases $b_j=4q_j$, prime blocks \[ P_k=\{p\text{ prime}: 2^{c(k-1)^2-3}<p\le 2^{ck^2-3}\}, \] and integers $a_p$ built from digits $x_j(p)$.
• The exact repeated-sum window from Round 3: if a repeated sum occurs then \[ (1-c)k_1^2<k_2^2<\frac{c}{1-c}k_1^2. \]
• The Round 3 "no deletion" theorem for $0<c\le c_0:=\frac{3-\sqrt5}{2}$, conditional on the digit construction being well-defined.
• The Round 3 bad-prime exponent estimate \[ |B_{k}(c)|\le 2^{\gamma(c)k^2+o(k^2)}, \qquad \gamma(c):=\frac{3c-1}{1-c}, \] again conditional on the digit construction being well-defined.
• The Round 2 endpoint argument at $c=\sqrt2-1$, which sharpened the critical exponent comparison by a constant-level estimate.

3) NEW INSIGHT / TOOL (ROUND-4)

There are three genuinely new ingredients in this round.

• Finite-exception repair lemma. The obstruction $p=q_j$ can occur only in finitely many initial blocks. This gives a minimal corrected statement: remove a finite exceptional set $E(c)$ of problematic primes.
• Transfer principle. After removing $E(c)$, every Round 3 argument applies verbatim to the corrected construction, and the counting function remains $x^{c+o(1)}$ because only finitely many indices are lost.
• Sharp barrier theorem for the present counting method. The exponent \[ \gamma(c)=\frac{3c-1}{1-c} \] is the exact exponent produced by the Round 3 union-bound/divisor-counting argument. Hence that route is power-saving if and only if $c<\sqrt2-1$, critical at $c=\sqrt2-1$, and non-power-saving for $c>\sqrt2-1$.

4) ATTACK PLAN (ROUND-4)

The exact gaps left after Round 3 are:

• the construction was not literally defined for indices $p=q_j$ with $j\le k(p)$;
• therefore the Round 3 theorem, as written, needed a correction;
• the obstruction at $c=\sqrt2-1$ was described, but not isolated as a sharp theorem about the present counting method.

I will therefore prove the following claims:

1. The set of problematic primes \[ E(c):=\{p\text{ prime}: p\in P_k\text{ for some }k\text{ and }p=q_j\text{ for some }1\le j\le k\} \] is finite.
2. After replacing the index set $\mathbb P$ by $\mathbb P\setminus E(c)$, the whole Round 3 theory becomes rigorous with no loss in the exponent.
3. The present bad-prime counting route proves $|B_k(c)|=o(|P_k|)$ exactly for $c<\sqrt2-1$ and cannot yield that conclusion by exponent comparison alone when $c>\sqrt2-1$.

5) WORK (ROUND-4)

5.1. The concrete flaw

For each prime $p$, let $k(p)$ be the unique block index with $p\in P_{k(p)}$. In Rounds 2--3, for every $p\in P_k$ and every $1\le j\le k$, the $j$-th digit of $a_p$ was defined by \[ g_j^{x_j(p)}\equiv p \pmod{q_j}, \qquad q_j< x_j(p)<2q_j, \] where $g_j$ is a primitive root mod $q_j$.

This requires $p\not\equiv 0\pmod{q_j}$, i.e. $p\neq q_j$. So the original definition is illegal whenever $p=q_j$ for some $1\le j\le k(p)$. This is not a cosmetic issue: in that case $p$ has no discrete logarithm in $\mathbb F_{q_j}^{\times}$.

Thus the construction from Round 2/3 was not fully well-defined as written.

5.2. The obstruction is finite

Lemma 5.1 (finite-exception lemma).

Fix $0<c<1/2$. If a prime $p$ lies in $P_k$ and also equals $q_j$ for some $1\le j\le k$, then \[ c(k-1)^2<2k+4. \] Consequently the exceptional set \[ E(c):=\{p\text{ prime}: p\in P_k,\ p=q_j\text{ for some }1\le j\le k\} \] is finite.

Proof.

Assume $p\in P_k$ and $p=q_j$ with $1\le j\le k$. Since $p\in P_k$, \[ 2^{c(k-1)^2-3}<p. \] Since $p=q_j$ and $j\le k$, the defining choice of the $q_j$ gives \[ p=q_j\le 2^{2j+1}\le 2^{2k+1}. \] Therefore \[ 2^{c(k-1)^2-3}<2^{2k+1}, \] which is equivalent to \[ c(k-1)^2<2k+4. \] The left-hand side is quadratic in $k$ and the right-hand side is linear, so this can hold only for finitely many $k$. Hence $E(c)$ is finite. $\square$

A convenient explicit threshold is any integer $K(c)$ satisfying \[ c(K(c)-1)^2>2K(c)+4. \] Then Lemma 5.1 implies that for every $k\ge K(c)$ and every $p\in P_k$, one has $p\neq q_j$ for all $1\le j\le k$.

5.3. Minimal corrected statement

Definition 5.2 (corrected prime blocks).

For each $k\ge 2$, define \[ P_k^{\circ}:=P_k\setminus \{q_j:1\le j\le k\}. \] Equivalently, \[ \bigcup_{k\ge2}P_k^{\circ}=\mathbb P\setminus E(c). \] For $p\in P_k^{\circ}$, define the digits $x_j(p)$ for $1\le j\le k$ exactly as in Rounds 2--3: \[ g_j^{x_j(p)}\equiv p\pmod{q_j}, \qquad q_j< x_j(p)<2q_j, \] and set $x_j(p)=0$ for $j>k$. Then define $a_p$ and the corrected sequence \[ A_{q,c}^{\circ}:=\{a_p: p\in \bigcup_{k\ge2}P_k^{\circ}\}. \]

Lemma 5.3 (eventual agreement with the original blocks).

For all $k\ge K(c)$, one has $P_k^{\circ}=P_k$. In particular, \[ |P_k^{\circ}|=|P_k| \qquad (k\ge K(c)). \]

Proof.

By the defining property of $K(c)$, Lemma 5.1 rules out every equality $p=q_j$ with $p\in P_k$ and $1\le j\le k$ once $k\ge K(c)$. Thus no prime is removed from $P_k$ for $k\ge K(c)$. $\square$

This is the minimal correction: only the finitely many illegal indices are removed. A simpler but equivalent variant is to discard all blocks $P_k$ with $k<K(c)$; this differs from the minimal correction by only finitely many indices and leaves all asymptotic statements unchanged.

5.4. Counting function after the repair

Proposition 5.4.

For every fixed $0<c<1/2$, the corrected sequence satisfies \[ A_{q,c}^{\circ}(x)=x^{c+o(1)}. \]

Proof.

Let \[ B_k:=(4q_1)\cdots (4q_k). \] Round 2 already showed that \[ 2^{k^2+2k}<B_k\le 2^{k^2+4k}, \] so if $B_k<x\le B_{k+1}$, then $k^2=\log_2 x+o(\log x)$.

For the lower bound, every prime $p\le 2^{ck^2-3}$ belongs to some block $P_{\ell}$ with $\ell\le k$. After the correction we lose only the finite exceptional set $E(c)$, so \[ A_{q,c}^{\circ}(x)\ge \pi(2^{ck^2-3})-|E(c)|. \] By the prime number theorem, \[ \pi(2^{ck^2-3})=2^{ck^2+o(k^2)}=x^{c+o(1)}, \] and subtracting the constant $|E(c)|$ does not change the asymptotic exponent.

For the upper bound, exactly the same block argument as in Round 2 gives \[ A_{q,c}^{\circ}(x)\le \pi(2^{c(k+2)^2-3})=x^{c+o(1)}. \] Hence \[ A_{q,c}^{\circ}(x)=x^{c+o(1)}. \] $\square$

5.5. Transfer of the Round-3 structure theorem

Lemma 5.5 (transfer principle).

After replacing each block $P_k$ by $P_k^{\circ}$, every Round 3 proof from the digit-definition stage onward applies verbatim to the corrected family. In particular:

1. the injectivity proof survives unchanged;
2. the exact repeated-sum window \[ (1-c)k_1^2<k_2^2<\frac{c}{1-c}k_1^2 \] survives unchanged;
3. the Round 3 bad-prime estimate \[ |B_k^{\circ}(c)|\le 2^{\gamma(c)k^2+o(k^2)}, \qquad \gamma(c)=\frac{3c-1}{1-c}, \] survives unchanged;
4. the Round 2 endpoint estimate at $c=\sqrt2-1$ survives unchanged.

Proof.

The only place where the original proof could fail was the attempted use of a discrete logarithm when $p=q_j$. By Definition 5.2 that case has been removed. All subsequent arguments in Rounds 2--3 are purely algebraic manipulations involving:

• carry-free addition in the mixed-radix digits;
• the congruences $g_j^{x_j(p)}\equiv p\pmod{q_j}$ for indexed primes;
• size comparisons using the block bounds $p\in P_k$ and the growth conditions on the $q_j$.

Those ingredients are now valid for every indexed prime in the corrected construction. Nothing else in the proofs depends on the discarded set $E(c)$. Therefore all those arguments transfer without change. $\square$

5.6. The corrected two-threshold theorem

Let \[ c_0:=\frac{3-\sqrt5}{2}, \qquad c_1:=\sqrt2-1. \]

Theorem 5.6 (corrected explicit family for the full Round-3 range).

For the corrected construction, the following hold.

1. If $0<c\le c_0$, then $A_{q,c}^{\circ}$ is a Sidon sequence and \[ A_{q,c}^{\circ}(x)=x^{c+o(1)}. \]
2. If $c_0<c<c_1$, then after the same largest-element deletion as in Round 2 one obtains a corrected Sidon sequence $A_{q,c}^{\circ,*}$ with \[ A_{q,c}^{\circ,*}(x)=x^{c+o(1)}. \]
3. If $c=c_1=\sqrt2-1$, then the corrected deleted sequence $A_{q,c_1}^{\circ,*}$ satisfies \[ A_{q,c_1}^{\circ,*}(x)=x^{\sqrt2-1+o(1)}. \]

Proof.

For part $1$, combine Lemma 5.5 with the Round 3 exact collision window. If a repeated sum existed, it would force \[ (1-c)k_1^2<k_2^2<\frac{c}{1-c}k_1^2. \] When $c\le c_0$, the interval on the right is empty because \[ \frac{c}{1-c}\le 1-c \quad\Longleftrightarrow\quad c\le \frac{3-\sqrt5}{2}. \] So no repeated sum exists, and $A_{q,c}^{\circ}$ is Sidon. The counting function is Proposition 5.4.

For part $2$, define the bad-prime sets $B_k^{\circ}(c)$ exactly as in Round 2 but inside the corrected family. By Lemma 5.5, the same largest-element deletion destroys all repeated sums, so the deleted corrected family is Sidon. Round 3 already identified the bad-prime exponent \[ |B_k^{\circ}(c)|\le 2^{\gamma(c)k^2+o(k^2)}, \qquad \gamma(c)=\frac{3c-1}{1-c}. \] If $c<c_1=\sqrt2-1$, then \[ \gamma(c)-c=\frac{c^2+2c-1}{1-c}<0. \] Hence $|B_k^{\circ}(c)|=o(2^{ck^2})$. Since Proposition 5.4 gives $|P_k^{\circ}|=2^{ck^2+o(k^2)}$, deleting the bad primes does not change the exponent, and \[ A_{q,c}^{\circ,*}(x)=x^{c+o(1)}. \]

For part $3$, when $c=c_1$, the exponent comparison becomes critical because $\gamma(c_1)=c_1$. Here one uses the sharper Round 2 constant-level estimate (transferred by Lemma 5.5), which shows that only a strict proportion of each block is removed. Therefore the deleted corrected family still has counting function \[ x^{\sqrt2-1+o(1)}. \] $\square$

5.7. The exact barrier inside the present method

Proposition 5.7 (sharp obstruction for the current deletion-counting route).

Inside the corrected mixed-radix/discrete-log framework, the Round 3 union-bound/divisor-counting method yields \[ |B_k^{\circ}(c)|\le 2^{\gamma(c)k^2+o(k^2)}, \qquad \gamma(c):=\frac{3c-1}{1-c}. \] Since \[ |P_k^{\circ}|=2^{ck^2+o(k^2)}, \] this method proves \[ |B_k^{\circ}(c)|=o(|P_k^{\circ}|) \] if and only if $c<\sqrt2-1$. At $c=\sqrt2-1$ the exponent comparison is critical, and for $c>\sqrt2-1$ the exponent comparison alone cannot yield $o(|P_k^{\circ}|)$.

Proof.

The upper bound \[ |B_k^{\circ}(c)|\le 2^{\gamma(c)k^2+o(k^2)} \] is exactly the Round 3 exponent calculation transferred by Lemma 5.5. Now \[ \gamma(c)-c = \frac{3c-1}{1-c}-c = \frac{c^2+2c-1}{1-c}. \] Because $1-c>0$, the sign of $\gamma(c)-c$ is the sign of $c^2+2c-1$. Its positive root is $c=\sqrt2-1$. Therefore:

• if $c<\sqrt2-1$, then $\gamma(c)<c$, so the bad-prime bound is power-saving and gives \[ |B_k^{\circ}(c)|=o(|P_k^{\circ}|); \]
• if $c=\sqrt2-1$, then $\gamma(c)=c$, so exponent comparison alone is inconclusive;
• if $c>\sqrt2-1$, then $\gamma(c)>c$, so this exponent comparison route by itself cannot prove that the bad set is negligible.

This is the exact obstruction claimed. $\square$

Interpretation.

To improve the explicit exponent beyond $\sqrt2-1$ inside this framework, one must introduce a new ingredient beyond the present union-bound/divisor-counting step. In other words, the barrier is not in the digit construction itself, but in the present way of estimating how many primes must be deleted.

6) ADVERSARIAL VERIFICATION

I tried to break the new work in the following ways.

• Could the exceptional set be infinite? No. Lemma 5.1 reduces the bad event $p=q_j$ with $j\le k(p)$ to the inequality \[ c(k-1)^2<2k+4, \] which can hold for only finitely many $k$.

• Did the correction change the exponent? No. The repair removes only the finite set $E(c)$, so Proposition 5.4 differs from the Round 2 growth proof only by subtracting a constant in the lower bound.

• Did I overstate the transfer principle? No. I checked where the old proof could fail: only at the discrete-log step. Once the illegal indices are removed, every later algebraic step in Round 2/3 is legitimate.

• Could the finite repair itself create a new Sidon obstruction? Deleting indices cannot create repeated sums, so all positive Sidon results remain valid after the correction.

• Could the repaired undeleted family actually stay Sidon for some $c>c_0$? Possibly; I did not settle that here. The correction removes only finitely many primes, so the most likely answer is "no", but I have not proved a persistence-of-collisions theorem after the repair. I therefore make no claim beyond the explicit range already justified by Theorem 5.6.

• Does Proposition 5.7 say the construction fails for $c>\sqrt2-1$? No. It says only that the current counting method is not strong enough there. A different argument could still conceivably control the bad primes beyond the critical exponent.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 4 found and repaired a genuine flaw in the Round 2/3 mixed-radix construction: the digits were not defined when $p=q_j$ with $j\le k(p)$. I proved that this obstruction is only finite, gave the minimal corrected statement, and showed that after the correction the full Round 3 explicit range survives: \[ 0<c\le \sqrt2-1 \quad\Longrightarrow\quad \text{there is an explicit corrected Sidon construction with }A(x)=x^{c+o(1)}. \] I also isolated a sharp obstruction: the present bad-prime counting route is power-saving exactly for $c<\sqrt2-1$, critical at $c=\sqrt2-1$, and therefore cannot by exponent comparison alone push the explicit construction past the Ruzsa--Cilleruelo exponent.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 83%

9) REFERENCES

References

• [Ruzsa1998] I. Z. Ruzsa, An infinite Sidon sequence, J. Number Theory 68 $1998$, no. 1, 63--71.
• [Cilleruelo2014] J. Cilleruelo, Infinite Sidon sequences, Adv. Math. 255 $2014$, 474--486.
• [ErdosProblems39] T. F. Bloom, Erdős Problem #39, problem page checked on 2026-03-07.