Erdős 699: a forced valuation gap at indices six and seven

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

P699 research · The corresponding row restriction at indices four and five

Proposition. If $(n,i,j)$ is bad with $i=6$ or $i=7$, then

In particular, every such row satisfies

This excludes entire row and column congruence classes, uniformly in their size. It does not bound the rows in the three alternatives of (1).

Fixed slope divisibilities

Put $j/n=a/q$ in lowest terms. Let $R_t$ be the part of $n-t$ supported on primes greater than five, and put $H=1$ at index six and $H=7$ at index seven. Badness and Lucas's theorem imply

For each prime-power factor $p^e$ in this divisor, the prime is active in $\binom ni$. Lucas gives $j\bmod p^e\le t$. Multiplying the relevant congruence $j\equiv r\pmod{p^e}$ by $q$, and using $qj=an$ and $n\equiv t\pmod{p^e}$, gives $p^e\mid at-qr$. At index seven, removing one factor of seven safely handles the possible cancellation in $7!$. When the numerator has seven-adic valuation at least two, Lucas gives the stronger congruence modulo the full power.

At $t=0$, Lucas itself gives $R_0/\gcd(R_0,H)\mid j$. Thus $q$ has no prime factors other than $2,3,5$ and possibly one factor of seven at index seven. If (1) fails, then

We prove the stronger assertion that no bad triple at index six has $q\mid60$, and no bad triple at index seven has $q\mid420$, without any independent restriction on the valuations of $n$.

Finite exact certificate

For each divisor $q\ge2$ of $60H$, each $1\le a\le q/2$ coprime to $q$, and each residue $s=n\bmod30$ compatible with $q\mid n$, choose $t\in\{1,\ldots,i-1\}$ satisfying

The checker verifies such a choice in every case. Since $a,q$ are coprime, $q\nmid t$ implies $F_t(a,q)\ne0$. Also $n-t$ is coprime to $30$, so (3) gives

Consequently each slope and residue class has finitely many candidates, obtained from the positive divisors of the fixed integer $H|F_t(a,q)|$. The checker uses the choice minimizing the number of divisors, followed by the constant and the shift.

Every one of the 73 residual candidates violates (3) at another shift. The record gives that shift, the required divisor, and the polynomial value. Thus no binomial factorization or search through a row interval is needed. This proves (4) impossible and establishes (1).

The integer checker uses only the Python standard library. Its certificate contains every slope and residue case and every rejected candidate.

python3 wiki/research/erdos_699/computations/erdos_699_low_crt_row_zero.py

Consequence for the selected positions

Condition (2) means that at least one of the primes $2,3,5$ has its unique maximal numerator valuation at the position $r=0$. Indeed, among $n,n-1,\ldots,n-6$, a multiple of eight, nine, or twenty-five cannot be accompanied by another multiple of that same power. The stronger statement (1) also requires that the column have a strictly smaller valuation at that prime.

This supplies a position restriction for the weighted interpolation argument. It does not exclude configurations in which a selected small-prime power at $n$ grows without bound.