Erdős 699: a curve obstruction for one remaining index-five profile

P699 research · Weighted polynomial certificates

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At index five, consider selected small-prime positions $0,1$ and cofactor exponents $e_0=13/100$, $e_1=17/50$. The row capacities are

$$\alpha=(13/100,17/50,1,1,1).$$

The current construction combines fixed curve factors with a residual polynomial obtained by counting linear conditions. For this profile, adding more fixed plane curves cannot lower the normalized residual quadratic cost below

$$\boxed{\frac{310000}{270691}=1.14521724032\ldots>1.}\tag{1}$$

This is an exact statement in every degree, proved by nine lines. It explains why searches for additional conics, cubics, or higher curves cannot complete this particular construction. It does not exclude this arithmetic profile from bad triples, and it does not obstruct stronger cofactor estimates, additional column congruences, or estimates using the particular value of a polynomial at a column.

Exact point weights

Let $S=\{(a,b)\in\mathbf Z_{\ge0}^2:a+b\le4\}$. Assign symmetric weights $w_{a,b}=w_{b,a}$ using the following numerators, all divided by $3100$.

Direct addition gives

$$\sum_{a+b=r}w_{a,b}=\alpha_r,\qquad \sum_{P\in S}w_P^2=\frac{270691}{310000}<1.\tag{2}$$

These weights obey, for every nonzero polynomial $F\in\mathbf C[X,Y]$,

$$\boxed{\sum_{P\in S}w_P\operatorname{ord}_P F\le\deg F.}\tag{3}$$

In particular, a polynomial having multiplicity at least $m_r$ at every point of row $r$ necessarily satisfies

\deg F\ge\frac{13}{100}m_0+rac{17}{50}m_1+m_2+m_3+m_4.\tag{4}

Thus a strict comparison of weighted row multiplicities against total degree cannot succeed at this profile. This statement concerns that specific exponent comparison; equality cases and stronger arithmetic information are not addressed by it.

A nine-line proof

Give the following lines the indicated positive coefficients, now with common denominator $15500$.

Write these coefficients as $\lambda_L$. At every $P\in S$,

$$w_P\le\sum_{L\ni P}\lambda_L,\qquad \sum_L\lambda_L=\frac{7103}{7750}<1.\tag{5}$$

The only strict pointwise excesses are $737/15500$ at $(0,0)$ and $1107/15500$ at each of $(0,4),(4,0)$.

If an irreducible curve $C$ is different from these nine lines, Bézout's theorem gives

$$\sum_{P\in S\cap L}\operatorname{mult}_P C\le\deg C$$

for each line $L$: the local intersection multiplicity with a line is at least the curve multiplicity, and the total projective intersection number is $\deg C$. Therefore

$$\sum_{P\in S}w_P\operatorname{mult}_P C \le\frac{7103}{7750}\deg C<\deg C.\tag{6}$$

Each of the nine lines itself has weighted incidence at most one. Seven have incidence exactly one; $X=2$ and $Y=2$ each have incidence $2757/3100$. Factoring a nonzero polynomial into irreducible factors now proves (3), since multiplicities and degrees are additive.

Consequence for every collection of fixed curve factors

Let $C_\ell$ be arbitrary curves, with degrees $d_\ell$ and point multiplicities $\nu_{\ell,P}$. Suppose nonnegative real factor weights $t_\ell$, row multiplicities $m_r$, and residual point multiplicities $z_P$ satisfy

$$m_{a+b}\le z_{(a,b)}+\sum_\ell t_\ell\nu_{\ell,(a,b)} \quad((a,b)\in S).$$

Their gain before constructing the residual polynomial is

$$G=\sum_r\alpha_r m_r-\sum_\ell t_\ell d_\ell.$$

Multiplication by the point weights, followed by (2) and (3), gives

$$G\le\sum_P w_Pz_P.$$

For the normalization $G\ge1$, Cauchy's inequality therefore forces

$$1\le\left(\sum_Pw_P^2\right)\left(\sum_Pz_P^2\right),\qquad \sum_Pz_P^2\ge\frac{310000}{270691}.$$

This proves (1) for every finite collection of curve factors, with no degree restriction and no positivity restriction on the curves. The usual residual dimension count needs a quadratic cost strictly below one, so no enlargement of the fixed-curve catalog can meet that criterion at these capacities.

The lower bound is attained using just the factors $X,Y,X-1,Y-1$. Put $q=310000/270691$, take row multiplicities

$$\frac q{3100}(1181,1107,1420,1065,852),$$

and give $X,Y$ factor multiplicity $389q/3100$ each and $X-1,Y-1$ factor multiplicity $191q/3100$ each. The resulting residual at $P$ is exactly $q w_P$, the gain is exactly one, and its quadratic cost is exactly $q$. Thus (1) is the minimum, not just a lower bound.

The integer/rational checker and its output verify the weights, line covering, incidence bounds, row sums, and exact quadratic cost. Their proof input is the elementary Bézout argument above; no optimization solver or finite degree search is needed.