Erdős 699: boundary divisors forced when index 14 has its 13-adic maximum at zero

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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.

Suppose $(n,14,j)$ is bad for the original prime threshold, put $c=n-j$, and assume $13^2\mid n$. Thus the maximal 13-adic position in $n,n-1,\ldots,n-13$ is zero. The prime-index transfer shows that this is one of the possible position cases for a bad triple.

For an integer $a$, let $[a]_{\ge17}$ denote its part supported on primes at least 17, and define

$$D_j=\left[\gcd\!\left(j,\prod_{r=1}^{13}(n-r)\right)\right]_{\ge17}, \qquad D_c=\left[\gcd\!\left(c,\prod_{r=1}^{13}(n-r)\right)\right]_{\ge17}.$$

Then

$$D_j>1,\qquad D_c>1,\qquad \gcd(D_j,D_c)=1. \tag{1}$$

In particular, both $j$ and $c$ have a prime divisor at least 17, and $j<c$. The numerical index-13 certificate also gives, whenever $n-1\ge10^{1732}$,

$$\boxed{ D_j^{114}\ge 10^{-356876}(n-1)^{1287/200},\qquad D_c^{114}\ge 10^{-356876}(n-1)^{1287/200}. } \tag{2}$$

Thus each boundary gcd is at least

$$10^{-356876/114}(n-1)^{1287/22800},$$

where $1287/22800\approx0.05644736$. This is a conditional constraint on bad triples, not an effective exclusion of index 14.

The shifted binomial coefficients

Put $B=\binom nj$. The identities

$$\binom{n-1}{j-1}=\frac jn B, \qquad \binom{n-1}{j}=\frac cn B \tag{3}$$

relate the two shifted columns to $B$. Every prime $q\ge17$ in the numerator block $n-1,\ldots,n-13$ also divides $\binom n{14}$. Badness gives $v_q(B)=0$. If $q\mid n-r$ for $1\le r\le13$, then $q\nmid n$, so (3) gives the exact valuations

$$v_q\binom{n-1}{j-1}=v_q(j),\qquad v_q\binom{n-1}{j}=v_q(c). \tag{4}$$

Moreover $13$ does not divide $\binom{n-1}{13}$: its unique numerator multiple is $n-13$, whose 13-adic valuation is one, and $13!$ cancels that factor.

The complete index-13 theorem applies to both binomial pairs in (3). For the second pair, reflect its column if necessary. In both cases the smaller column exceeds 13: $j\ge15$ and $c\ge j$. The supplied common prime is therefore at least 17. Equation (4) proves the first two assertions of (1).

A prime $q\ge17$ dividing both $D_j$ and $D_c$ would divide $j+c=n$ and some $n-r$ with $1\le r\le13$, which is impossible. This proves coprimality. If $j=c$, the same contradiction already follows from any prime divisor of $D_j$, so the central column is excluded.

Keeping the extra divisors in the numerical proof

Apply the index-13 polynomial certificate to row $N=n-1$. Its row multiplicities, in the order $t=0,\ldots,12$, are

$$m=(114,111,111,103,99,99,89,78,70,63,57,51,46).$$

The certificate has total degree at most 641, and its exact exponent gain after accounting for primes $2,3,5,7,11$ is $1287/200$. These facts hold for every assignment of the small-prime maxima, including coinciding positions. Its source estimates and elementary error bounds apply for $N\ge10^{1732}$.

For the first shifted column define

$$Q_{r,j}=[\gcd(n-r,j)]_{\ge17}\qquad(1\le r\le13).$$

The exact localization lemma and (4) say that, at a prime $q\ge17$ dividing $n-r$, the missing valuation in the polynomial divisibility argument is exactly

$$m_{r-1}\min\{v_q(n-r),v_q(j)\}.$$

Consequently the index-13 argument remains valid on inserting the factor

$$\prod_{r=1}^{13}Q_{r,j}^{m_{r-1}}$$

on its right side. No new loss is required at 13: its numerator valuation is one, already covered by the certificate's factor $13^{114}$. All smaller-prime estimates, coefficient estimates, and conversion from $N-12$ to $N$ remain the same.

The complete numerical constant in that argument is less than $10^{356876}$: it combines the coefficient bound $10^{355721}$, $13^{114}<10^{128}$, the off-maximum factor $(12!)^{114}<10^{1026}$, and the remaining factor $1000/999<10$. Thus

$$N^{1287/200} \le 10^{356876}\prod_{r=1}^{13}Q_{r,j}^{m_{r-1}} \le 10^{356876}D_j^{114}. \tag{5}$$

For the last inequality, each prime at least 17 divides at most one of the thirteen consecutive factors, so $D_j=\prod_rQ_{r,j}$, and $m_{r-1}\le114$. The nonvanishing part of the certificate is applicable: both coordinates $(j-1,c)$ exceed 13. The same argument at $(j,c-1)$ gives the bound for $D_c$, proving (2).

The divisors contain whole prime powers

If $q\ge17$ divides $n-r$, put $A=v_q(n-r)$. Badness and Kummer give

$$j\bmod q^A\le r<q.$$

If $q\mid j$, this residue must be zero. Hence $q^A\mid j$. Likewise, $q\mid c$ implies $q^A\mid c$. Thus each prime contributing to $D_j$ or $D_c$ contributes its entire prime-power factor in the corresponding $n-r$. Its Kummer point is respectively $(0,r)$ or $(r,0)$ modulo that entire prime power.

What the bound adds

The usual row multiplicity argument uses only the total large-prime part of each $n-r$. The divisors in (2) locate a growing part of it at boundary Kummer points: a prime power dividing $j$ has first coordinate zero modulo that prime power, and a prime power dividing $c$ has second coordinate zero. A certificate that gives greater multiplicity at those boundary points can retain this information. Formula (2) alone leaves $j$ and $c$ free to be comparable to $n$, so it does not supply a stopping bound.