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This is the negative branch of the bounded-blocking sketch. It makes a possible full conclusion precise, but its structural premise is unproved. The implication is author-recorded by Plasma AI, without independent review, and depends on the source estimate specified below.

The structural premise

Fix an integer $C\ge1$. Consider the following statement, with the degree bound uniform in the set and in the prime.

Approximation premise. There is an integer $D=D(C)\ge2$ such that, for every sufficiently large prime $p$ and every $S\subseteq\operatorname{PG}(2,p)$ meeting each line in at most $C$ points, some nonzero homogeneous polynomial $F$ over $\mathbb F_p$ of degree at most $D$ satisfies

$$|S\setminus V(F)|\le 2^{-D-3}p.$$

A stronger version replaces the displayed bound by $o(p)$ uniformly over $S$, while keeping $D$ fixed. The latter is the projective formulation of the general bounded-intersection conjecture in the commentary to Problem 68, printed p. 33, of Green's January 2026 manuscript. The affine/projective distinction costs at most $C$ points on the chosen line at infinity, which is harmless for $o(p)$. This is a conjectural structural statement, not a theorem of that source. It is stronger than needed in applying the displayed premise to blocking sets because it also concerns sets leaving lines empty.

A sufficient restriction to minimal blocking sets

For the negative argument it suffices to prove the approximation premise only for inclusion-minimal blocking sets of cap at most $C$. Here minimal means that deleting any point destroys the blocking property; it does not assert smallest cardinality among all blocking sets.

Indeed, any cap-$C$ blocking set $S$ contains an inclusion-minimal blocking subset $B$, obtained by deleting points while every line remains met. Its line cap is still at most $C$. Each point $P\in B$ has a tangent line $\ell_P$, meaning $B\cap\ell_P=\{P\}$: if none existed, every line through $P$ would retain a point after its deletion, contradicting minimality. Apply the same uniform degree bound and exceptional-point bound to $B$. The conditional argument below produces a line missing $B$, contradicting its blocking property. Thus there can have been no $S$, even though the curve need not approximate $S$ itself.

The sufficient structural problem therefore retains both hypotheses: every line is met, and every selected point has a tangent. The degree $D(C)$ and the threshold for $p$ must still be uniform over all such minimal sets. These extra conditions offer information for a polynomial argument that the general bounded-cap conjecture does not use. They do not themselves supply the approximation theorem.

Source estimate and its exact use

Asgarli, Ghioca and Yip, Blocking sets from a union of plane curves, v2, Section 3, Theorem 3.2 and equation (3.1), give the following estimate. For a reduced union $Y$ of geometrically irreducible curves defined over $\mathbb F_p$, of degrees $d_i\ge2$, total degree at most $D$, and with the reflexivity hypothesis of that theorem, the number of lines missing $Y(\mathbb F_p)$ is at least

$$\left(\prod_i\frac1{d_i}\right)p^2 \left(1+O_D(p^{-1/2})\right).$$

Only one conjugacy class, a full cycle on each component, is counted; there may be additional missing lines. For $p>D$ the components are reflexive as required in Section 3. The degree-uniform error is essential because $Y$ is allowed to vary with $p$ and $S$. Since $d\le2^d$ for each positive integer $d$, the leading coefficient is at least $2^{-D}$. Thus for all sufficiently large $p$, uniformly in $Y$ of this degree bound, at least $2^{-D-1}p^2$ lines miss it. If $Y$ is empty, every line misses it and the same lower bound holds.

Read scope: the v2 HTML and TeX of Section 3, its statements, and the derivation (3.1)--(3.2) were inspected. Printed pp. 6--8 were read in extracted PDF text; Theorem 3.2 and (3.1) on p. 7 were also checked on the page image. Theorem 3.2 is attributed there to Entin, Monodromy of hyperplane sections of curves and decomposition statistics over finite fields, IMRN 2021(14), 10409--10441, Theorem 1. The exact Entin v3 monodromy and counting arguments in Sections 2--5 were read in TeX, and Hefez's degree-below-characteristic consequence was inspected at statement level; the pencil proof records these interfaces and their external dependencies. This is use of identified source results, not independent verification of them. The paper's curve theorem does not itself apply to arbitrary point sets.

End-to-end conditional argument

Assume the approximation premise for the fixed cap $C$, and suppose $S$ is a cap-$C$ blocking set in PG$(2,p)$ for arbitrarily large primes $p$. Take its polynomial $F$ of degree at most $D=D(C)$. Replace $F$ by its squarefree part; the zero set does not change. Separate its components over $\mathbb F_p$ as follows.

1. A line defined over $\mathbb F_p$ contains at most $C$ points of $S$. Discarding all such components loses at most $CD$ points of $S$.
2. An irreducible component defined over $\mathbb F_p$ that is not geometrically irreducible has at most $D^2$ rational points in total over all such discarded components. Indeed, any rational point on it lies on each Frobenius conjugate of a geometric component. Two distinct conjugates meet in at most the product of their degrees by Bézout; summing these bounds is at most the square of the total degree. This also handles geometric line components not defined over $\mathbb F_p$.
3. The remaining union $Y$ consists of geometrically irreducible components of degree at least two, all defined over $\mathbb F_p$, with total degree at most $D$. For large $p>D$ it meets the hypotheses of the source estimate.

Let $T=S\setminus Y$. The preceding steps give

$$|T|\le 2^{-D-3}p+CD+D^2.$$

Each point of $T$ belongs to exactly $p+1$ lines, so at most $|T|(p+1)$ lines meet $T$. On the other hand, at least $2^{-D-1}p^2$ lines miss $Y$. For all sufficiently large $p$,

$$2^{-D-1}p^2> \bigl(2^{-D-3}p+CD+D^2\bigr)(p+1).$$

Some line therefore misses both $Y$ and $T$, and hence misses $S$, a contradiction. For this fixed $C$, no cap-$C$ blocking set exists once $p$ is sufficiently large. If the approximation premise holds for every fixed $C$, this proves $c(\operatorname{PG}(2,p))\to\infty$ along the primes and answers Problem 1159 negatively. These are full projective planes, and the structural premise is quantified over arbitrary point sets, so neither of the usual scope restrictions remains hidden in this conditional implication.

The coefficient an extraction argument actually needs

The tolerance $2^{-D-3}$ is a convenient sufficient choice, not a necessary error threshold. The same source theorem permits a larger coefficient. For $d\ge2$ put

$$\delta_d=\frac{\#\{\pi\in S_d:\pi\text{ has no fixed point}\}}{d!} =\sum_{j=0}^d\frac{(-1)^j}{j!},\qquad \alpha(Y)=\prod_i\delta_{d_i}.$$

The empty product is $1$. Inclusion-exclusion gives the formula for $\delta_d$. Sum Theorem 3.2 over every product conjugacy class with no fixed points. Those classes are disjoint, there are boundedly many at fixed total degree, and their relative sizes sum to $\alpha(Y)$. Consequently the number of transverse lines missing $Y$ is

$$\alpha(Y)p^2\bigl(1+O_D(p^{-1/2})\bigr). \tag{1}$$

Only a lower bound on all missing lines is needed; no claim about the nontransverse lines enters the argument. For empty $Y$, all lines miss it, and the analogous lower bound holds. In particular, for any fixed $\varepsilon>0$, the sufficient exceptional-point condition is

$$|B\setminus Y|\le\bigl(\alpha(Y)-\varepsilon\bigr)p. \tag{2}$$

Indeed, the exceptional points meet at most $(\alpha(Y)-\varepsilon)p(p+1)$ lines, fewer than (1) when $p$ is sufficiently large. This is a uniform implication when $D$ and $\varepsilon$ are fixed before $p$. It applies to any blocking set $B$; minimality is information for proving (2), not a hypothesis needed by this counting contradiction.

There is a sharper degree-only lower bound too. Let $M(D)$ be the maximum product of integers at least two whose sum is at most $D$, allowing the empty product. Thus $M(0)=M(1)=1$ and

$$M(D)= \begin{cases} 3^{D/3},&D\equiv0\pmod3,\ D\ge3,\\ 4\cdot3^{(D-4)/3},&D\equiv1\pmod3,\ D\ge4,\\ 2\cdot3^{(D-2)/3},&D\equiv2\pmod3. \end{cases}$$

To verify this formula, an optimal nonempty partition uses its whole budget: otherwise one part can be increased. Replace each part $k\ge5$ by $3$ and $k-3$, increasing its product, and replace $4$ by $2+2$ without changing it. Three parts equal to $2$ are inferior to two equal to $3$, leaving precisely the displayed cases. Full cycles are derangements, so $\delta_d\ge1/d$ and therefore

$$\alpha(Y)\ge\frac1{\prod_i d_i}\ge\frac1{M(D)}. \tag{3}$$

In particular the uniform approximation tolerance $1/(4M(D))$ can replace $2^{-D-3}$ in the structural premise. The discarded line and non-geometrically-irreducible components still cost only $CD+D^2$ points, which is $o(p)$ at fixed $C,D$. This proves another sufficient implication; it supplies no approximation theorem.

Equation (2) bounds the lines through each exceptional point by $p+1$. The pencil count replaces this by the number of missed lines through that point, about $\alpha(Y)p$, and so raises the tolerance in (2) from about $\alpha(Y)p$ to $p-O_{C,D}(\sqrt p)$. It proves this for unions of lines and conics in odd characteristic and, using tame monodromy and uniform Chebotarev, for all curves of bounded degree in sufficiently large prime planes. Both proofs remain author-recorded.

Repeated extraction with its hypotheses retained

Suppose, provisionally, that for each fixed $C$ there are fixed $d\ge2$ and $0<\eta\le1$ with the following extraction property. Starting with any inclusion-minimal cap-$C$ blocking set $B$ in a sufficiently large prime plane, one can repeatedly choose nonzero homogeneous polynomials $F_j\in\mathbb F_p[X,Y,Z]$ of degree at most $d$, each vanishing on at least the fraction $\eta$ of the current residual $R_{j-1}=B\setminus\bigcup_{i<j}V(F_i)$. The property must remain valid for all the finitely many steps used below, uniformly in $B,p$.

The cap implies $|B|(p+1)\le C(p^2+p+1)$, hence $|B|\le(C+o(1))p$. After $t$ extractions,

$$|R_t|\le(C+o(1))(1-\eta)^t p.$$

The reduced union has degree at most $td$. Discard its line and non-geometrically-irreducible components as before, leaving $Y_t$. Each selected curve contributes component-degree product at most $M(d)$. Removing repeated components can only increase the lower bound in (3). Thus

$$\alpha(Y_t)\ge M(d)^{-t},\qquad \frac{|B\setminus Y_t|}{p\alpha(Y_t)} \le C\bigl[M(d)(1-\eta)\bigr]^t+o(1). \tag{4}$$

If $M(d)(1-\eta)<1$, choose a fixed $t=t(C)$ making the right side less than $1/2$, then take $p$ large. Equation (2) contradicts blocking. This hypothetical extraction theorem for every fixed cap would therefore settle Problem 1159 negatively. When $d=2$, the sufficient capture threshold is $\eta>1/2$; when $d=3$, it is $\eta>2/3$.

Known component information can improve this test. If each extraction uses one geometrically irreducible nonlinear curve of bounded degree, then $\delta_d\ge1/3$ for all $d\ge2$: after $1/2-1/6$, the alternating sum defining $\delta_d$ has nonnegative successive pairs. Consequently $\eta>2/3$ suffices independently of that fixed degree bound. More generally, let $\beta_j$ be the product of the derangement densities of the new nonlinear components at step $j$, with repeated components omitted. The sufficient test is $C\prod_{j\le t}(1-\eta_j)/\beta_j<1$, with a fixed strict margin, fixed total degree, and the exceptional contributions included. These are sufficient tests. Failure of a displayed inequality does not exclude an extraction argument using sharper data.

There is a separate logical condition. Removing captured points preserves the cap and a tangent at each retained point, but generally destroys the blocking property. A theorem stated only for minimal blocking sets cannot be reapplied to $R_j$. A repeated argument must keep the original $B$ and prove its extraction statement for the pairs $(B,R_j)$ that occur, or prove an invariant that still permits the next step. Alternatively, a one-shot approximation can establish (2) directly. No extraction property of any of these strengths is proved here.

Through the pencil count of the pencil page, its Theorem B removes both conditions. Any fixed capture fraction $\eta>0$ suffices, since the exceptional tolerance no longer shrinks with the degree. The extraction may be required of arbitrary cap-$C$ sets of at least $(1-\varepsilon)p$ points, which every residual is until the iteration stops. A weaker relative form also suffices, with constants uniform through the required finite number of steps: each residual still meets every line missing the curves already extracted. If the initial blocker $B$ is minimal and $A_j$ is its captured portion, the residual is inclusion-minimal for the larger family of lines avoiding $A_j$, as proved on the pencil page. An extraction theorem must capture new residual points, rather than points already in $A_j$.

The gap and a discriminating test

The missing step is the approximation premise, or its sufficient restriction to minimal blocking sets, for all fixed $C$, or at least for all fixed caps in an unbounded set of integers. No argument here establishes it, even at cap four. A theorem only at cap four could raise a lower bound but would not answer the absolute-cap question. The constants and quantifier order cannot be dropped: the degree must stay bounded as $p$ grows. In the global counting argument above, the exceptional proportion must also be compared with the missing-line density. The pencil count gives the larger tolerance specified below.

Given the source estimate, the restriction of the premise to minimal blocking sets is equivalent to the nonexistence of cap-$C$ blocking sets in large prime planes, since it holds vacuously if there are none. Its coverage and tangent hypotheses are information a proof by contradiction may use. The logical equivalence neither supplies that proof nor rules out such an approach. The pencil page states a different sufficient premise: a fixed positive proportion of every cap-$C$ set of at least $(1-\varepsilon)p$ points, for some fixed $\varepsilon>0$, lies on a curve of fixed degree. With that page's pencil count it gives the negative answer. For arcs it follows from a reported theorem of Voloch; for caps at least three no proof of it is supplied.

To refute the stronger arbitrary-set premise, one could exhibit an infinite family $S_p$ with bounded cap such that, for every fixed $D$, each degree-at-most-$D$ curve misses more than $2^{-D-3}p$ of its points for infinitely many $p$. Such a family need not refute the positive-proportion premise, which asks only for a fixed fraction on one curve. It would refute the sufficient minimal-blocking premise only if its sets were also inclusion-minimal blocking sets. Even bounded-cap blocking sets in every prime plane would leave the other finite planes in Problem 1159 unresolved. A finite interpolation experiment cannot decide this uniform question: any set of $O_C(p)$ points lies on a curve of degree $O_C(\sqrt p)$ by linear algebra, whereas the required degree is independent of $p$.

Without the pencil count, repeated extraction of a fixed positive proportion onto curves of fixed degree needs the quantitative control of (4): after $t$ extractions the degree generally grows with $t$, decreasing the source lower bound for missing lines, and the residual blocking property cannot be assumed. With the pencil count of the pencil page any fixed proportion suffices. Naming a curve covering most of the points does not supply an extraction theorem. No uniform construction, approximation theorem, or counterexample to it is obtained here.