Erdős 699: joint cofactor bounds with freely chosen Padé blocks

P699 research · Rational residual blocks · Numerical product thresholds

The rational-block argument applies to products of several nearby factors. It can also choose the block size according to the proposed cofactor bounds. This gives effective joint restrictions beyond the rectangular alternatives obtained by stopping at the first exhausted prime-power block.

The lemma below is conditional on a bounded product difference. It does not, by itself, settle an index of the binomial problem.

The product lemma

For distinct primes $p$, write

where $u_p$ are positive integers and $R\ge1$ is fixed. Partition the primes into two nonempty sets $I_0,I_1$ and assign positive multiplicities $b_p$ such that each product has degree $\nu$:

Assume $|X-Y|\le D$, where $D>0$ is fixed. Choose a valid BFT seed

and valid coefficient estimates for $s=c/d>1$ at every integer $m>m_0$. Let $\kappa_1,\kappa_2,\Omega_3,\Omega_4$ be the constants of BFT Theorem 2.4 for this seed and gap $D$. Suppose $\Omega_3,\Omega_4>1$ and put

Given proposed upper exponents $0\le e_p\le1$ and any real $t>0$, define

Lemma. If both margins $\sigma_1,\sigma_2$ are positive, then the inequalities $u_p\le h^{e_p}$ cannot all hold once

All constants in (1) are numerical when the coefficient-gcd estimate is numerical. Otherwise the resulting threshold is still effectively computable if the coefficient-gcd starting value $m_0$ is effective.

Proof. Set $m=\lfloor t\log h/c\rfloor>m_0$. The two residual products $U_m=X/P^{cm}$ and $V_m=Y/Q^{cm}$ are positive rationals. Their denominators divide

The integer Padé determinant argument, after multiplying by $Z_m$, gives exactly as in the two-prime proof

This argument uses that $P,Q$ have disjoint prime supports; it does not require them to be prime powers. The BFT determinant gives a nonzero rational linear combination, whose product with $Z_m$ is a nonzero integer.

If every $u_p\le h^{e_p}$, then $b_pk_p\log p\ge b_p(1-e_p)\log h$. Since $cm\le t\log h$,

Also $cm>t\log h-c$ and $Y\le(h+R)^\nu<2^\nu h^\nu$. Thus the right side of (2) is at most

contradicting (2). $\square$

The proof includes $X=Y$: the determinant supplies nonvanishing, and the term bounded by $D$ is then zero. Only the fixed Padé seed is required to have a nonzero difference.

The joint inequality and choice of block

For the balanced choice $t=\nu/J$, the two margins agree. In particular, for every fixed $\epsilon>0$ and sufficiently large $h$, necessarily

If a cofactor exceeds $h$, its exponent is already too large to belong to any of the upper boxes under consideration. Formula (3), interpreted with the actual exponents, also follows directly without the upper restriction $e_p\le1$.

One should not replace (3) with its rectangular endpoint alternatives when checking an entire region of possible exponents. Several positive terms can be relevant simultaneously. To exclude an upper box, the monotonicity in each $e_p$ makes it enough to evaluate its upper corner. A finite collection of such boxes may then cover a larger region.

The margins are continuous piecewise linear functions of $t$. Besides the balanced value, useful block sizes occur at the denominator breakpoints

The checker tries these finitely many choices. It approximates the selected choice by a positive rational and recomputes both margins with directed intervals. The proof uses only the verified margins for that rational; it does not depend on a claim that the search found the best block.

Numerical examples

For the AP $2r_3=r_2+r_7$ with distinct positions in $[0,13]$, the seed $28-27=1$, $s=4/3$, $L_1=1.4170$, $m_0=153$ gives

Here $h\le n-r_p\le h+13$ and the AP gap is at most $121$. This excludes an upper box which extends past one of the previous coordinatewise limiting targets from this seed.

For four distinct positions satisfying $r_3+r_5=r_2+r_{11}$, the seed $45-44=1$, $s=3/2$, and the explicit Dusart gcd bound at $m_0=10^8$ give

The final alternative is part of the displayed upper-box exclusion; it is not a claim that $u_{11}$ is bounded in the binomial problem. Other upper boxes can be passed to the same evaluator.

The checker and certificate record the positive margins and numerical thresholds. Product identities and nonzero gap bounds remain explicit hypotheses of the evaluator.

.venv/bin/python wiki/research/erdos_699/computations/erdos_699_rational_product_bounds.py

Two additional asymmetric prime-pair estimates

The same lemma with $\nu=1$ gives useful alternatives missed by optimizing only a symmetric maximum-cofactor exponent. For positive integer cofactors, a nonzero difference at most $196$, and input height $h$ equal to the smaller factor, the following implications hold:

For general inputs in this table, use $R=196$ in (1); the logarithmic $R$ term is smaller than the already certified threshold. In the binomial applications the factor difference is at most $13$ and the checker directly uses $R=13$.

The first seed is $49-3\cdot16=1$, with $s=7/4$, $L_1=1.6219$, $m_0=60$. The second is $13\cdot512-5\cdot1331=1$, with $s=9/4$, $L_1=1.7666$, $m_0=87$. Both coefficient-gcd bounds are retained entries of BFT Proposition 5.1. The new feature is the asymmetric use of these seeds, rather than a new estimate for their coefficient gcds.

The pair checker and certificates record exact seed factorizations, positive directed-interval margins, and the numerical stopping thresholds for these implications.