Erdős 699: effective power bounds for both quintic cofactors

P699 research · The maximal-position confinement · Unequal-block Padé approximants

---

Proposition. Beyond an effectively computable threshold, a bad triple at index five satisfies

$$n-r_2=2^a u,\qquad n-r_3=3^b v,\qquad \{r_2,r_3\}=\{0,1\},\qquad \boxed{(n-4)^{1/100}<u,v<(n-4)^{57/100}.} \tag{1}$$

The cofactors are those obtained by removing the maximal powers of two and three from their respective numerator factors. The threshold is not numerically evaluated. This excludes every family in which either cofactor is at most a fixed hundredth power of the row, with the harmless shift in (1). It does not make the whole index effective.

Unlike the earlier consequence of Yu's theorem, the lower exponent here is a fixed positive constant. The additional input is the quintic polynomial argument, which first gives an effective upper bound for each cofactor.

An effective upper bound at either position

Put $h=n-4$. The previous confinement theorem leaves only the two maximal positions zero and one. Suppose that the cofactor at one of those positions is at least $h^{57/100}$.

If that position is zero, use

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

It has degree 12 and row multiplicities $(6,5,4,3,2)$ on the quintic Kummer grid. The gain in the weighted divisor inequality is

$$6\cdot\frac{57}{100}+4+3+2-12=\frac{21}{50}>0.$$

The product is nonzero for $X=j>5$ and $Y=n-j>5$. Thus this case has an effective row bound.

For a cofactor of size at least $h^{57/100}$ at position one, use the following symmetric interpolation data:

$$\begin{aligned} (h_0,h_1,h_2,h_3,h_4)&=(101,76,27,0,0),\\ (m_0,m_1,m_2,m_3,m_4)&=(0,223,190,142,114),\\ D&=164. \end{aligned}$$

Here $\prod_{k=0}^4((X-k)(Y-k))^{h_k}$ is the initial positive product. The residual multiplicity at $(x,y)$ is $\max(0,m_{x+y}-h_x-h_y)$. The symmetric jet matrix of degree at most 164 has size $6817\times6889$ and full row rank modulo 251. The surjective-jet nonvanishing lemma therefore gives a finite residual family of degree at most 165, with a nonzero member at each physical pair. After restoring the initial product, its total degree is

$$2(101+76+27)+165=573.$$

The gain is

$$\frac{57}{100}\,223+190+142+114-573=\frac{11}{100}>0.$$

All coefficients can be computed rationally from the finite matrix, so the corresponding row bound is effective. Replacing $h$ by $n$ in the exponent comparisons introduces only bounded factors of $n/(n-4)$. Consequently, sufficiently large bad triples have both $u,v<h^{57/100}$.

Turning the upper bound into a lower bound

Two further applications of the seed $9-8=1$ give the effective alternatives

$$\begin{aligned} u>h^{1/100}&\quad\text{or}\quad v>h^{57/100},\\ u>h^{23/40}&\quad\text{or}\quad v>h^{1/100}. \end{aligned}\tag{2}$$

Their Padé directions and coefficient-gcd rates are

In each case the exact prime-interval sum truncated at $t=1000$ exceeds the chosen gcd rate. Directed interval arithmetic gives a margin greater than $10^{-4}$ for both targets. The general unequal-block construction makes the starting indices effective.

The first alternative in (2), together with $v<h^{57/100}$, forces $u>h^{1/100}$. The second, together with $u<h^{57/100}<h^{23/40}$, forces $v>h^{1/100}$. This proves (1).

Certificate and remaining restriction

The checker verifies the source inequalities, the elementary product, and the full modular rank. Its record contains the exact data. The --arithmetic-only option rechecks the inequalities without repeating the matrix rank.

The earlier upper product bound remains

$$uv<\frac{240}{4^{6/5}}n^{3/5},$$

and all four cofactor alternatives in the maximal-position note still hold. Intermediate cofactor sizes satisfy these necessary conditions, so a complete quintic stopping bound has not been obtained.