Erdős 699: simultaneous carry restrictions in the hard index-six region

P699 research · Column restrictions at index eight · Actual common-prime valuations

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The hard index-six assignment is

$$(r_2,r_3,r_5)=(2,1,0).$$

The row cofactor profile alone does not settle this assignment. The actual column can nevertheless be restricted without another interpolation calculation.

Proposition. Let $(n,6,j)$ be bad, set $c=n-j$, $h=n-5$, and choose the maximal small-prime positions as above. Put

$$P_p=p^{v_p(n-r_p)},\qquad u_p=(n-r_p)/P_p.$$

Assume

$$h^{31/100}\le u_2\le h^{2/5},\qquad h^{9/25}\le u_3\le h^{2/5},\qquad u_5\ge h^{31/100}. \tag{1}$$

For $h\ge10^{425}$, necessarily

$$\boxed{ j\bmod P_2>2,\qquad j\bmod P_3>1,\qquad j\bmod P_5>0. } \tag{2}$$

In particular, each selected prime produces a carry across its selected prime-power boundary. There are also quantitative restrictions at 5:

$$\left[\binom nj\right]_5>\sqrt h, \qquad u_5<\sqrt h, \qquad [j]_5,[c]_5<\sqrt h, \tag{3}$$

where $[a]_p=p^{v_p(a)}$. In fact, (3) needs only the two lower bounds on $u_2$ and $u_3$ in (1), without either upper bound or a lower bound on $u_5$.

These are conditional restrictions on bad triples. The branches in (2) are not excluded, so this proposition does not establish effectivity for the entire index.

A column identity allowing row labels at least the prime

The valuation identity from the index-eight argument extends to a row label $r$ that may be at least $p$. Let

$$n-r=P u,\qquad P=p^A>r,\qquad p\nmid u,\qquad r<j\le n.$$

Then

$$\boxed{ \left[\binom nj\right]_p[(j)_{r+1}]_p \le[r!]_p(n-r). } \tag{4}$$

Indeed,

$$(j)_{r+1}\binom nj=(n)_{r+1}\binom{n-r-1}{j-r-1}.$$

Because $P>r$, the valuation of $(n)_{r+1}$ is $A+v_p(r!)$. The bottom $A$ digits of $n-r-1=p^Au-1$ are all $p-1$, so

$$v_p\binom{n-r-1}{j-r-1} =v_p\binom{u-1}{\lfloor(j-r-1)/P\rfloor} \le\lfloor\log_pu\rfloor.$$

This proves (4). It also applies at the complementary column $c$.

If $b=j\bmod P\le r$, the $p$-valuation of the falling factorial is at least $A+v_p(b!)+v_p((r-b)!)$. Thus, writing

$$E_p=p^{\min\{A,v_p\binom nj\}},$$

we obtain

$$E_p\le\min\!\left(P,\left[\binom rb\right]_p u\right). \tag{5}$$

For the three positions under consideration this gives

$$\begin{array}{c|c} \text{low-residue branch}&\text{overlap bound}\\ \hline j\bmod P_2\le2&E_2\le2u_2\\ j\bmod P_3\le1&E_3\le u_3\\ j\bmod P_5=0&E_5\le\min(P_5,u_5)\le\sqrt n. \end{array} \tag{6}$$

The factor 2 in the first row is required: $\binom21=2$. At the heights in the proposition, the upper bounds in (1) guarantee $P_2,P_3\ge h^{3/5}>2$, so all uses of $P>r$ are justified.

One explicit polynomial

Use

$$F(X,Y)=(XY)^3((X-1)(Y-1))^2(X-2)(Y-2).$$

Its degree is 12, and at every grid point $x+y=t<6$ it has multiplicity at least $6-t$. It is positive at every admissible $(j,c)$. The exact localization inequality therefore gives

$$\prod_{t=0}^5(n-t)^{6-t} \le (5!)^6 F(j,c)E_2^4E_3^5E_5^6. \tag{7}$$

Each unselected small-prime contribution is covered by $(5!)^6$. The large-prime hypothesis is precisely the original threshold $p\ge6$.

Since $jc\le n^2/4$ and $n\le2h$,

$$F(j,c)\le(jc)^6\le h^{12}.$$

The multiplicities in (7) sum to 21, giving

$$\boxed{E_2^4E_3^5E_5^6\ge h^9/120^6.} \tag{8}$$

The cofactor lower bounds imply

$$P_2\le2h^{69/100},\quad P_3\le2h^{16/25},\quad P_5\le2h^{69/100}.$$

Using $E_p\le P_p$ for the other two primes in turn yields

$$\begin{aligned} E_2^4&\ge\frac{h^{83/50}}{120^6 2^{11}},\\ E_3^5&\ge\frac{h^{21/10}}{120^6 2^{10}},\\ E_5^6&\ge\frac{h^{76/25}}{120^6 2^9}. \end{aligned} \tag{9}$$

Each row only needs the cofactor lower bounds at the other primes. Their limiting overlap exponents are respectively $83/200$, $21/50$, and $38/75$.

Now substitute the corresponding low-residue bound from (6) into (9). The three conclusions would be

$$h^{3/50}\le120^6 2^{15},\qquad h^{1/10}\le120^6 2^{10},\qquad h^{1/25}\le120^6 2^{12},$$

respectively. But

$$120^6 2^{15}<10^{17},\qquad 425/25=17.$$

None is possible for $h\ge10^{425}$. This proves (2), without a matrix rank calculation or an asymptotic analytic estimate.

To verify (3), put $C_5=120^6 2^9$. The third inequality in (9) gives $E_5\ge C_5^{-1/6}h^{38/75}$. Both $u_5=n/P_5$ and the quantities $[j]_5,[c]_5$ from (4) are at most

$$2C_5^{1/6}h^{37/75}.$$

The exact integer comparison

$$\bigl(2^6C_5\bigr)^{25}<10^{425}$$

shows these are below $\sqrt h$ at the stated height. The same comparison also gives $E_5>\sqrt h$, proving all of (3).

The checker and record verify the multiplicities, rational exponent arithmetic, and numerical thresholds. The proof uses only the explicit product $F$ and the column valuation identity; it does not repeat an interpolation search.