Erdős 699: the five remaining indices

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details/: Proofs of the current restrictions and obstructions at indices three, four, five, six, and eight.

index_three: Rational abc implies cubic finiteness; unconditional residue restrictions and the nonexistence question remain open.

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Call $(n,i,j)$ bad if $1\le i<j\le n/2$ and no prime $p\ge i$ divides both $\binom ni$ and $\binom nj$. The completed arguments leave only $i\in\{3,4,5,6,8\}$.

For each of $4,5,6,8$, van Doorn and Rocca prove that there are only finitely many bad triples. No effective upper bound on all their rows has been established here. Their finiteness argument uses an ineffective $S$-part estimate; the available effective version saves too little in these cases. The index-three guide explains the different situation there: even global finiteness remains open.

Assuming rational $abc$, the cubic argument proves finiteness at $i=3$, and hence finiteness of all bad triples. Unconditional finiteness at $i=3$ remains open.

The results below restrict infinite families of possible counterexamples. They do not complete any of these four indices. An effective threshold means that the proof provides an algorithm to compute it; several such thresholds have not been evaluated numerically.

Indices four and five

The following necessary row conditions hold without a height restriction:

$$i=4:\quad 4\mid n\ \text{or}\ 9\mid n, \qquad i=5:\quad 8\mid n\ \text{or}\ 9\mid n.$$

The proof by fixed rational slopes excludes the complementary congruence classes completely. Its difficulty in the remaining classes is concrete: the denominator of $j/n$ can grow with the large powers of two or three dividing $n$.

Choose positions $r_2,r_3$ maximizing $v_2(n-r)$ and $v_3(n-r)$ among the numerator factors, and write

$$n-r_2=2^a u,\qquad n-r_3=3^b v.$$

At index four, sufficiently large bad triples must have distinct positions and satisfy

$$2^a u-3^b v=r_3-r_2\in\{-3,-2,-1,1,2,3\}, \qquad uv<4n^{5/6}.$$

The cofactors remain unbounded. The effective small-cofactor theorem does exclude every range $\min(u,v)\le(\log n)^A$, for each fixed $A$. It gives a computable lower bound of the form $\exp(c\log n/\log\log n)$ for both cofactors. The intermediate range is still too large for the current argument.

At index five, the confinement is stronger. Beyond an effective threshold,

$$\{r_2,r_3\}=\{0,1\},\qquad (n-4)^{1/100}<u,v<(n-4)^{57/100}, \qquad uv<\frac{240}{4^{6/5}}n^{3/5}.$$

The quintic cofactor bounds include the proof and link the position-confinement argument. They leave an explicit range of intermediate cofactor sizes. Merely allowing more polynomial factors does not close it: the index-five obstruction proves that every polynomial with row multiplicities $m_r$ satisfies

$$\deg F\ge .13m_0+.34m_1+m_2+m_3+m_4.$$

Thus the strict degree saving needed by the current divisor argument cannot hold at that profile. This is an all-degree theorem, not a failed search through a small catalog of polynomials.

The tested column refinements do not yet exclude this profile. The higher-digit note writes the complete Lucas condition as exact inequalities in the slope $j/n=A/q$. On a candidate branch, including odd columns with $j\equiv2\pmod3$ when $8\mid n$, the two- and three-adic overlaps equal the selected powers up to a factor two. A one-prime lemma identifies why higher digits do not automatically improve polynomial valuations; it does not establish compatibility across primes for an actual column. On that branch $q>n^{43/100}$ and $A$ is bounded below by a positive power of $n$. No uniform effective upper bound for these parameters is obtained, and the branch is not proved to occur.

Indices six and eight

Here it helps to distinguish row information from information about the actual column. Put $H=n-i+1$, select maximal positions $r_p$, and write $P_p=p^{v_p(n-r_p)}$, $u_p=(n-r_p)/P_p$ for each $p<i$.

At six, consider $(r_2,r_3,r_5)=(2,1,0)$. If

$$H^{.31}\le u_2\le H^{.4},\qquad H^{.36}\le u_3\le H^{.4},\qquad u_5\ge H^{.31},$$

then a bad triple with $H\ge10^{425}$ must satisfy

$$j\bmod P_2>2,\qquad j\bmod P_3>1,\qquad j\bmod P_5>0.$$

The carry restriction at six uses a degree-twelve product and an exact valuation identity. It also forces the five-part of $\binom nj$ above $\sqrt H$, while the five-parts of $j$ and $n-j$ lie below $\sqrt H$. The remaining branches, where all three displayed residue inequalities hold, have not been excluded.

At eight, consider $(r_2,r_3,r_5,r_7)=(1,3,0,2)$ with $u_2,u_3,u_5\ge H^{.3}$. For $H\ge10^{850001}$, the carry restriction at eight forces $j\bmod P_7>2$. It also forces the seven-part of $\binom nj$ above $\sqrt H$ and the seven-parts of $(j)_3$ and $(n-j)_3$ below $\sqrt H$. Again, the remaining branch is open.

These are difficult configurations, not a claim that every possible bad triple lies in the displayed cofactor boxes. Two exact geometric obstructions explain why further row-only interpolation is insufficient:

Both bounds hold in every degree and permit arbitrary curve factors. They do not show that the profiles occur arithmetically. Stronger cofactor estimates or restrictions on the column could exclude them.

What a new argument still has to supply

For these four indices, numerical improvements to existing source bounds are not enough by themselves: some permitted configurations have no strict asymptotic saving. The useful remaining targets are stronger cofactor inequalities, additional restrictions from the actual column, or a new relation confining bad pairs to manageable algebraic curves.

The effective Runge criterion would handle many such fixed curves. At four and five, three suitable finite fibers of the map $(n,j)\mapsto n$, together with infinity, suffice. The missing global step is to put all remaining bad pairs on finitely many curves satisfying that criterion. No such confinement has been proved.