Erdős 699: off-diagonal Padé approximants and unequal prime-power blocks

P699 research · Joint rational-block estimates

The diagonal choice $A=C$ in BFT's application of Padé approximation is not necessary. The general polynomials in its Lemma 3.1 give two different prime-power block lengths. Retaining this difference improves some of the cofactor estimates. The conclusion below is effective; it does not assume an ineffective approximation theorem.

These are conditional source inequalities. Excluding one previously difficult cofactor vector does not establish an entire binomial index: every new alternative and every position assignment still needs coverage.

The exact identity and nonvanishing

Use the integral normalizations of BFT (3.1)--(3.4), so that

Their degrees are respectively $C,A,B$. Let $P',Q',E'$ refer to $(A+1,B-1,C+1)$, where $B\ge1$. The exponent $B+C+1$ in (1) is unchanged, and subtracting the two identities gives

The left side has degree at most $A+C+1$. Therefore

Indeed its coefficient of $z^{A+C+1}$ is $E(0)Q'(0)$, which is nonzero by the integral formulas, or by the binomial-coefficient expansions. This proves the required determinant statement for all rectangles, not only $A=C$. The integral normalization fixes the signs consistently; only absolute values are used below.

Choose a seed $aP_0-bQ_0=D_0>0$, and set $z=D_0/(aP_0)$. Let $G$ be the gcd of the coefficients of $Q_{A,B,C}$. The identity shows that $G$ also divides every coefficient of $P_{A,B,C}$ and $E_{A,B,C}$. Thus the integers

satisfy

Both block lengths $A+B+1$ and $B+C+1$ are preserved by the adjacent rectangle in (2). Hence the two choices still supply a nonzero integer linear combination after any residual denominators are cleared.

Using a common denominator power $\max(A,C)$ instead would lose this advantage. Formula (3) is the reason unequal rectangles help.

An effective coefficient-gcd lower bound

Fix positive integer directions $\lambda,\mu,\kappa$, and take

For a finite set of positive integers $t$ satisfying

put

This is an explicit positive rational number. Every summand comes from a disjoint prime interval in BFT Lemma 5.2. For any rational $\ell<L_T$,

holds above an effectively computable $m_0$, for both adjacent rectangles.

Here is why the bounded shifts in $A,B,C$ cause no difficulty. For an eligible $t$, the three fractional parts in (4) sum to $2$. None can be zero, since the other two are strictly less than $2$ in total. Consequently all the floors stabilize effectively under these shifts. The corresponding interval endpoints are $m$ times the displayed constants, up to bounded errors. The effective prime-number estimates already retained in the Dusart note give $\theta(x)/x\longrightarrow1$ with a computable rate. Apply them to the finitely many interval endpoints and absorb the strict difference $L_T-\ell$. This proves (5) with an effective threshold. No estimate for an infinite tail of (4) is needed.

The two exponential rates

Define

The two maxima are at unique points in $(0,1)$, determined by quadratic equations. They can therefore be bounded with directed intervals. The integral formulas and effective factorial estimates give the growth rates $q,e$ for $Q,E$, with at most a polynomial factor in $m$. Such a factor is bounded effectively by $e^{\epsilon m}$ for every fixed $\epsilon>0$.

After division by the coefficient gcd, the decay rates needed in the integer argument are

The block lengths per unit $m$ are $L_0=\lambda+\mu$ and $L_1=\mu+\kappa$.

The effective joint criterion

Write nearby factors as $f_p=p^{k_p}u_p\in[h,h+R]$, for distinct primes, and form products $X,Y$ of equal degree $d$ with positive multiplicities $b_p$. Assume $|X-Y|\le D$, with a fixed positive bound $D$. The primes occurring in each product must be precisely those in the corresponding seed side

For a proposed upper cofactor vector $e_p$ and a real $t>0$, set

where $s(p)$ is the seed side. If

then the simultaneous bounds $u_p\le h^{e_p}$ are impossible above an effectively computable height.

To prove this, use $m=\lfloor t\log h\rfloor$ and clear the residual denominators of $X/P_0^{L_0m}$ and $Y/Q_0^{L_1m}$. Their logarithmic cost is at most $\Delta(t)\log h$. Equations (2), (3), and (6) yield, for a computable constant $C_\epsilon$,

Since $Y\le2^d h^d$ for $h>R$, both terms tend effectively to zero when $\epsilon>0$ is chosen small enough relative to the strict margins in (7). This contradiction proves the criterion.

The actual product difference may be zero. Nonvanishing of the integer linear combination comes from the seed determinant (2), whose seed gap $D_0$ is positive; it does not come from $X-Y$. If $X=Y$, the term involving $X-Y$ simply vanishes. Thus the same joint constraints apply when two selected prime powers belong to the same numerator factor. Their valuation and factorization hypotheses must still be checked.

The integral-block alternative is an immediate special case. Provided $C_0>0$, every target strictly below

is a valid coordinate in a cofactor disjunction. The joint criterion (7) can be stronger than these coordinatewise limits.

The inequalities can also be replaced by rational linear alternatives for exact subsequent computations. Fix rational $t>0$, choose rational $g_p>tL_{s(p)}a_p\log p-b_p$, and choose a positive rational $R<\min(tC_0,tK_0-d)$. Above an effective threshold, necessarily

Otherwise both margins in (7) have a uniform positive lower bound. The last inequality is equivalent to the finite disjunction

The rational-clause checker rounds the intercepts upwards and the threshold downwards, then verifies every rounding and a strict source margin using directed intervals. Its example certificate contains only rational coefficients for the resulting linear alternatives. This separates the analytic source verification from exact polyhedral coverage calculations.

A four-prime example

For $P_0=847=7\cdot11^2$, $Q_0=845=5\cdot13^2$, $a=b=1$, use directions $(\lambda,\mu,\kappa)=(39,7,33)$. The finite sum (4) through $t=200$ exceeds $3.0421$, so take $\ell=3$. Directed evaluation gives

Thus, for four distinct positions satisfying $r_7+r_{11}=r_5+r_{13}$, the following alternative holds effectively:

The product gap is a fixed nonzero difference of the position products. This strengthens the diagonal source for the same seed. Applying it in a whole-index proof still requires covering every branch of the alternative.

The checker and certificate record the finite rational gcd sum, the two saddle points, directed bounds for both rates, and a strict margin for each listed exponent. The same checker exports the general growth calculation for other seeds and rectangles.

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