Erdős 699: cutoff 21, further complete cases, and the remaining twelve indices

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This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

P699 research · Erdős problem 699

Subsequent result. The cutoff-15 proof completely excludes $10,15,18,20$. The classification below records the preceding cutoff-21 stage.

There are no bad triples with $i\ge21$, or with $i\in\{11,13,16,17,19\}$. The threshold is the original $p\ge i$. Together with the earlier cases $i=1,2$, these are complete exclusions over every admissible row and column.

This improves the preceding cutoff $26$. Eight additional indices are completely excluded in this step:

$$11,13,16,19,21,22,24,25.$$

It does not finish every $i\ge4$. The exact remaining classification is:

Thus twelve indices at least four remain unresolved, of which nine still have an effectivity gap. The separate case $i=3$ is unchanged.

How the complete exclusions work

Three improvements fit together.

First, the asymmetric cofactor argument retains different exponents for the two endpoints in BFT's Padé proof. For example, it gives $u_5>x^{.29}$ or $u_7>x^{.20}$ above an explicit threshold. These alternatives reduce the numerical tail bounds at $i=11,13,16,21$ to $10^{914},10^{1046},10^{914},10^{1732}$, respectively. The exact weighted checker includes all collisions of maximal small-prime positions.

Second, on an already bounded height range, the finite prime-power method proves substantially stronger cofactor estimates. For fixed powers $P=p^a,Q=q^b$ and a bounded difference $d$, modular inversion supplies the unique minimal positive solution of $Pu-Qv=d$. Every other positive solution satisfies the required cube inequality automatically. Exact exponent-pair enumeration therefore proves

$$\max(u,v)^3\ge\min(Pu,Qv)$$

throughout the prescribed height range, apart from explicitly retained small exceptions. This finite calculation concerns prime-power exponents, not all integers below the stopping bound. It also works for pairs such as $7,11$ for which the retained general Padé estimates gave no saving.

Third, those estimates strengthen the local valuation bounds on entire row intervals. Their summed cofactor exponents often reduce the possible range again before the interval proof begins. The directed-rounding interval method then excludes every remaining interval. At two indices, a few small singleton rows need direct exact gcd checks.

Complete certificates

There are 1,903,249 certified intervals, all with strictly positive logarithmic margins, and 51 exact row/index cases. The latter have $n\le2060$ and require 11,510 admissible-column gcd checks in total. After removing the primes below $i$, every checked gcd still exceeds one. The interval partitions and singleton cases cover the whole bounded range; the analytic bounds cover the infinite tails.

Detailed proofs and records are retained in:

• Complete index-eleven proof, with certificate.
• Complete index-sixteen proof, with certificate.
• Further complete interval proofs, with four-index records, index-21 records, and index-13 records.

The preceding complete cases $17,23$ and $i\ge26$ remain inputs. Thus the new interval proofs give the stated cutoff $21$ and additional cases. These are computer-assisted mathematical proofs; they have not been formalized in Lean.

New effective cases: 15, 18 and 20

The new finite-dimensional interpolation argument supplies nonvanishing, not merely a polynomial that might vanish at the putative bad pair.

Prescribe the needed Hasse jets at the finite Kummer grid. If symmetric polynomials through degree $D$ map onto the symmetric jet algebra, let $T$ denote multiplication by $X+Y$ on that algebra. A rational right inverse produces integer polynomials of degree at most $D+1$ whose simultaneous vanishing at $(j,n-j)$ would make $n$ an eigenvalue of $T$. But all eigenvalues belong to $\{0,\ldots,i-1\}$, the grid's row labels. Thus at least one polynomial is nonzero throughout the physical region. The full lemma proves this directly.

Exact modular rank certificates and positive weighted gains give:

Both matrices have full row rank. The interpolation proof and constants give the whole-index bounds

$$\begin{aligned} i=15:&\quad n<\max\{X_{15}+15,10^{1\,100\,000\,000}\},\\ i=20:&\quad n<\max\{X_{20}+20,10^{4\,000\,000}\}. \end{aligned}$$

Here the $X_i$ are the specified, effectively computable BFT thresholds. The explicit powers of ten bound all polynomial coefficients and elementary losses. No exceptional algebraic curve remains. The same lemma first gave the numerical bound $10^{56000}$ at index 13; the subsequent simpler asymmetric argument and interval proof exclude that index completely.

At index $18$, two asymmetric alternatives improve the weighted polynomial $F_{12}$ enough to give gain $186/625$. The index-18 proof verifies all 65 cofactor vectors and 5707 collision cases, giving

$$n<\max\{X_{18},10^{2\,405\,782}\}+17.$$

Its algebraic term is only $10^{814}$; the displayed larger term includes the explicit asymmetric source thresholds. The remaining $X_{18}$ is effectively computable as specified in that proof.

What remains difficult below 15

At $i=4,5$, an elementary endpoint argument completely excludes additional congruence classes:

$$i=4\Longrightarrow4\mid n\ \text{or}\ 9\mid n, \qquad i=5\Longrightarrow8\mid n\ \text{or}\ 9\mid n$$

for any bad triple. The proof and exact finite certificate reduce the complementary classes to finitely many rational slopes and divisors of fixed constants.

The remaining cases include nearby integers $n=2^a u$, $n-1=3^b v$ with both cofactors unbounded. The present cofactor estimates and interpolation constructions do not give a positive uniform exponent in every such case. In particular, the results above do not establish the conjecture for all $i\ge4$.