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Statement

Setting (pp. 131--133). A 2-design on $v$ points, a near pencil and breaking up a line are defined on the pages for Theorem 1 and Theorem 3.

Theorem 4 (p. 133, quoted). "Let $v=p^2+p+1$ and $\mathbf A=\{A_1,\ldots,A_b\}$ a 2-design which is neither a projective plane nor a near pencil nor is obtained from a projective plane by "breaking up" one of its lines. Then $b>p^2+(2+c)p$ where $c$ can be taken as 0.147899."

The constant comes from the proof (p. 140): $c$ is chosen with $0\ge c^4+6c^3+11c^2+5c-1$, and $c=0.147899$ to within six decimal places. The trivial design whose only line is the whole point set is not excluded by the printed hypotheses; the proof's Lemma 1 (p. 135) sets it aside together with the near pencil.

Consequence (p. 133; abstract, p. 131). Let $v=p^2+p+1$ with $p=q^2+q$. A line of a projective plane of order $p$ has $p+1=q^2+q+1$ points, and by the de Bruijn--Erdős theorem and Theorem 2 a 2-design with more than one line on those points has $q^2+q+1$ lines or at least $q^2+2q+1$. So breaking up that line gives $b=(p^2+p+1)+p$ or $b\ge(p^2+p+1)+p+q$. By Theorem 4 the latter bound also holds, when $b>v$ and $v>v_0$, for the 2-designs on $v$ points not obtained by breaking up a line of a projective plane. So for $v>v_0$ the interval $[v+p+1,v+p+q-1]$ is disjoint from $M_v$.

Problem 1 (p. 141). The authors say Theorem 4 is not best possible and conjecture that it holds with $b\ge p^2+3p+O(1)$.

Proof pointer

Pp. 138--141, combinatorial. Assume $b\le p^2+(2+c)p$. Counting ordered pairs of distinct points on a common line gives at least $p^2+1$ lines of length $p+1$ when the longest shorter line has at most $\sqrt{1/(2+c)}\,p$ points, and a degree count around a line of length $p+1$ gives the same when it is longer, provided $c$ satisfies the quartic above. Vanstone's theorem (the paper's reference [9]) then embeds the lines of length $p+1$ in a projective plane of order $p$, and counting the short lines needed to cover the pairs inside the $p-t+1$ missing lines gives $b\ge p^2+2p+1$, with equality only when exactly one line was broken up, and $b\ge p^2+3p-1$ otherwise.

Read depth

Claims checked: Theorem 4, the choice of $c$, the consequence on p. 133 and Problem 1 were read clause by clause on the page images of the print, and the combinatorial proof was followed for structure. Nothing here is independently reviewed.

Dependencies

Lemma 1 (p. 135) and Theorem 2 for the consequence. External inputs named by the paper: the de Bruijn--Erdős theorem and Vanstone's embedding theorem.

Source. P. Erdős, J. C. Fowler, V. T. Sós and R. M. Wilson, On 2-designs, J. Combin. Theory Ser. A 38 (1985), no. 2, 131--142; the edition read is named on the source card.

Bears on

• Problem 903: the paper's combinatorial proof of Theorem 2, the problem's assertion, goes through Theorem 4.