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Released under the CC BY-ND license (International 4.0).} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If needed, include a line break (\\) at an appropriate place in the title. \title{A New Lower Bound\\ for the Size of an Affine Blocking Set} % Input author, affiliation, address and support information as follows; % The address should include the country, but does not have to include % the street address. Give at least one email address. \author{Maarten De Boeck\\%\thanks{Supported by NASA grant ABC123.} \small Vakgroep Wiskunde\\[-0.8ex] \small Universiteit Gent\\[-0.8ex] \small Flanders, Belgium\\ \small\tt maarten.deboeck@ugent.be\\ \and Geertrui Van de Voorde\\ \small School of Mathematics and Statistics\\[-0.8ex] \small University of Canterbury\\[-0.8ex] \small Christchurch, New Zealand\\ \small\tt geertrui.vandevoorde@canterbury.ac.nz} \begin{document} \maketitle % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract may be distributed without the rest of the % paper so it must be entirely self-contained. Try to include all words % and phrases that someone might search for when looking for your paper. \begin{abstract} A blocking set in an affine plane is a set of points $B$ such that every line contains at least one point of $B$. The best known lower bound for blocking sets in non-desarguesian affine planes was derived in the 1980's by Bruen and Silverman \cite{bsil}. In this note, we improve on this result by showing that a blocking set of an affine plane of order $q$, $q\geq 25$, contains at least $q+\lfloor\sqrt{q}\rfloor+3$ points.. \end{abstract} \section{Introduction and preliminaries} \begin{definition} A blocking set in an affine plane $\A$ (or an {\em affine blocking set}) is a point set $B$ such that every line of $\A$ contains at least one point of $B$. \end{definition} In general, little is known about the smallest affine blocking sets, especially when compared to the knowledge about small blocking sets in projective planes (see amongst others \cite{bier,bruen,drake,cyrus}). For the classical (desarguesian) affine plane $\AG(2,q)$ we have the following result, obtained independently by Jamison and by Brouwer and Schrijver. \begin{theorem}[\cite{bs} and \cite{jam}]\label{jamison} A blocking set of $\AG(2,q)$ has size at least $2q-1$. \end{theorem} If an affine blocking set $B$ in a non-desarguesian plane of order $q$ contains a line, then it is easy to see that $B$ contains at least $2q-1$ points. However, without the assumption that the blocking set contains a line, %Note that this bound is also valid for affine blocking sets (so, in arbitrary affine planes) containing a line. T the bound from Theorem \ref{jamison} cannot be generalised to non-desarguesian affine planes. This was first noticed in \cite{bdr} (see also \cite{bier2}) where it was shown that each of the non-desarguesian affine planes of order 9 admits a blocking set of order 16. More generally, it is shown in \cite{db} that for $q$ square, $q\geq 9$, there exist an affine plane of order $q$ admitting a blocking set of size at most $\frac{4}{3}q+\frac{5}{3}\sqrt{q}$. %The smallest non-desarguesian affine planes have order 9. Each of these planes admits a blocking set of order 16 (see \cite{bier2,bdr}). The following result by Bruen and Thas gives a lower bound on the size of an affine blocking set. \begin{theorem}[{\cite[Cor. 2]{bt}}] If $\mathcal{S}$ is a blocking set of an affine plane of order $q$, then \[ |\mathcal{S}|\geq q+1+\frac{\sqrt{4q^{3}-4q+1}-1}{2q}\;. \] If $q$ is a square, this implies $|\mathcal{S}|\geq q+\sqrt{q}+1$. \end{theorem} This result was improved by Bruen and Silverman. \begin{theorem}[{\cite[Theorem 3.1]{bsil}}]\label{silverman} If $\mathcal{S}$ is a blocking set of an affine plane of order $q>3$, % and $\mathcal{S}$ does not contain a line, then \[ |\mathcal{S}|>q+\sqrt{q}+1\;. \] If $q\geq4$ is a square, this implies $|\mathcal{S}|\geq q+\sqrt{q}+2$. \end{theorem} We can embed an affine plane $\mathcal{A}$ in a projective plane $\mathcal{P}$ by adding the parallel classes as points and a unique line $\ell_{\infty}$ containing precisely these points (the line `at infinity'). This process is called \emph{completion}. By adding one point on $\ell_{\infty}$ to a blocking set of $\mathcal{A}$ we get a blocking set of $\mathcal{P}$. Using this correspondence, we can see that the non-square case of the Theorem \ref{silverman} also follows from the main result of \cite{bier}. In this paper, we will prove the following: \begin{maintheorem} A blocking set of an affine plane of order $q$, $q\geq 25$, contains at least $q+\left\lfloor\sqrt{q}\right\rfloor+3$ points. \end{maintheorem} Note that all known blocking sets of non-desarguesian affine planes of order $q<25$ also have size at least $q+\left\lfloor\sqrt{q}\right\rfloor+3$. We mentioned the affine planes of order 9 above. In \cite{db} the construction of a blocking set in an affine plane derived from the Hall plane of order 16 is described: it has size at least 24 and at most 28. \section{The proof of the main theorem} The following lemma will be used in the proof of our main theorem. \begin{lemma}\label{afschatting} Let $b>1$ be an integer and let $k,m\in\N$ be such that $k\equiv m\pmod{b-1}$ and $m b \end{cases}\\ a_{i} &\text{else} \end{cases}\;. \end{align*} We find that $\sum_{i=1}^{b}(i-1)a'_{i}=k$ and that $\sum_{i=1}^{b}(i-1)^{2}a_{i}< \sum_{i=1}^{b}(i-1)^{2}a'_{i}$. If $a_{1},\dots,a_{b}\in\N$ satisfy $\sum_{i=1}^{b}(i-1)a_{i}=k$ and we can find $a_{j},a_{j'}\geq1$ for $2\leq j\neq j'\leq b-1$, then set \begin{align*} a'_{i}=\begin{cases} a_{i}-1 &\text{if }i\in\{j,j'\}\\ a_{i}+1 &\text{if }\begin{cases} i=j+j'-1 &\text{if }j+j'\leq b\\ i\in\{j+j'-b,b\}&\text{if }j+j'> b \end{cases}\\ a_{i} &\text{else} \end{cases}\;. \end{align*} We find that $\sum_{i=1}^{b}(i-1)a'_{i}=k$ and that $\sum_{i=1}^{b}(i-1)^{2}a_{i}< \sum_{i=1}^{b}(i-1)^{2}a'_{i}$. From these arguments the claim follows. Now, we find immediately that \begin{align*} \sum_{i=1}^{b}(b-i)(i-1)a_{i}&=(b-1)\sum_{i=1}^{b}(i-1)a_{i}-\sum_{i=1}^{b}(i-1)^{2}a_{i}\\ &\geq(b-1)k-\left(\ell(b-1)^{2}+m^{2}\right)=m(b-1-m)\;. &\qedhere \end{align*} \end{proof} We will prove our main theorem in the dual setting. A line $\ell$ of a plane $\Pi$ is said to {\em cover} a point $P$ if $P$ lies on $\ell$. It is now easy to see that the following lemma holds. \begin{lemma} The dual of an affine blocking set (considered as a subset of the projective plane $\Pi$ obtained through completion), is a set of lines in the dual of $\Pi$, covering all the points but one. \end{lemma} \begin{definition} Let $\L$ be a set of lines of a projective plane $\mathcal{P}$. A {\em $k$-knot} is a point of $\mathcal{P}$ that lies on exactly $k$ lines of $\L$. \end{definition} \begin{lemma}\label{Beq} Let $P$ be a point of an axiomatic projective plane $\mathcal{P}$ of order $q$, and let $\mathcal{L}$ be a set of lines in $\mathcal{P}$ such that all points but $P$ are on a line of $\mathcal{L}$, and $P$ is not. Let $k$ be such that $\mathcal{L}$ has no $k'$-knot for $k'>k$. Let $x_i$ denote the number of $i$-knots in $\mathcal{P}$. Then, for all $\overline{k}$: \begin{align*} \sum_{i=1}^{k}(i-1)(\overline{k}-i)x_{i}%=-\sum_{i=1}^{k}i(i-1)x_{i}+\overline{k}\sum_{i=1}^{k}ix_{i}-\overline{k}\sum_{i=1}^{k}x_{i}\nonumber\\ =-|\mathcal{L}|(|\mathcal{L}|-1)+\overline{k}|\mathcal{L}|(q+1)-\overline{k}(q^{2}+q)\;.%\nonumber\\ \end{align*} \end{lemma} \begin{proof} \par We denote the number of $i$-knots (points on exactly $i$ lines of $\mathcal{L}$) by $x_{i}$, $i=1,\dots,k$. We do the standard countings: first we count the number of points, then the number of pairs $\{(P,\ell)\mid P\ \mbox{is a point on the line }\ell\in\L\}$ and finally the number of triples $\{(P,\ell,m)\mid \ell,m\in\L,\ell\neq m,\ P\mbox{ a point on }\ell\mbox{ and }m\}$. We find \[ \sum_{i=1}^{k}x_{i}=q^{2}+q\;,\quad\sum_{i=1}^{k}ix_{i}=|\mathcal{L}|(q+1)\;,\quad\sum_{i=1}^{k}i(i-1)x_{i}=|\mathcal{L}|(|\mathcal{L}|-1)\;. \] Now we can execute the following calculation: \begin{align*} \sum_{i=1}^{k}(i-1)(\overline{k}-i)x_{i}&=-\sum_{i=1}^{k}i(i-1)x_{i}+\overline{k}\sum_{i=1}^{k}ix_{i}-\overline{k}\sum_{i=1}^{k}x_{i}\nonumber\\ &=-|\mathcal{L}|(|\mathcal{L}|-1)+\overline{k}|\mathcal{L}|(q+1)-\overline{k}(q^{2}+q)\;.%\nonumber\\ &\qedhere \end{align*} \end{proof} \begin{lemma}\label{eerste} Let $P$ be a point of an axiomatic projective plane $\mathcal{P}$ of order $q$, $q\geq9$, and let $\mathcal{L}$ be a set of less than $2q-1$ lines in $\mathcal{P}$ such that all points but $P$ are on a line of $\mathcal{L}$, and $P$ is not. Let $k$ be such that $\mathcal{L}$ admits a $k$-knot but no $k'$-knot for $k'>k$. Then $k\left\lfloor\sqrt{q}\right\rfloor$.%$k\leq\left\lfloor\sqrt{q}\right\rfloor$ \end{lemma} \begin{proof} It is immediate that $\mathcal{L}$ does not admit a $(q+1)$-knot as the $q+1$ lines through some fixed point cover all points. If $\mathcal{L}$ admits a $q$-knot $R$, then $\mathcal{L}$ has to contain at least $q-1$ lines not through $R$, in order to cover the points of $PR$ different from $P$ and $R$. So, $|\mathcal{L}|\geq 2q-1$, a contradiction. \par Let $k$ be such that $\mathcal{L}$ admits a $k$-knot, but no $k'$-knot for $k'>k$, and let $K$ be a $k$-knot. We may assume that $k\leq q-1$, and so we can find a line $\ell$ through $K$ not in $\mathcal{L}$ and not through $P$. Since all points on $\ell$ different from $K$ are contained in a line of $\mathcal{L}$ and these lines are necessarily different, we have $|\mathcal{L}|\geq q+k$. \par If $k\leq\left\lfloor\sqrt{q}\right\rfloor$, then it follows from Lemma \ref{Beq} that \begin{align*} 0&\leq\sum_{i=1}^{k}(i-1)(k-i)x_{i}=-|\mathcal{L}|(|\mathcal{L}|-1)+k(q+1)\left(|\mathcal{L}|-q\right)\\&\leq-|\mathcal{L}|(|\mathcal{L}|-1)+\left\lfloor\sqrt{q}\right\rfloor(q+1)\left(|\mathcal{L}|-q\right) \end{align*} where we used that $|\mathcal{L}|-q\geq k>0$. We set $|\mathcal{L}|=q+\left\lfloor\sqrt{q}\right\rfloor+m$ and we find that \begin{align*} 0&\leq-(q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2})-m^{2}-2q\left\lfloor\sqrt{q}\right\rfloor-2mq-m\left\lfloor\sqrt{q}\right\rfloor+q+\left\lfloor\sqrt{q}\right\rfloor+m+mq\left\lfloor\sqrt{q}\right\rfloor\\ &\leq(m-2)(q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-m-1)-3q-\left\lfloor\sqrt{q}\right\rfloor-2\\ &=-(m-2)^{2}+(m-2)(q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-3)-3q-\left\lfloor\sqrt{q}\right\rfloor-2\\ &\leq (m-2)(q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-3)-3q-\left\lfloor\sqrt{q}\right\rfloor-2\;. \end{align*} We know that $\left\lfloor\sqrt{q}\right\rfloor>\sqrt{q}-1$ and hence $q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-3> q\sqrt{q}-3q-\sqrt{q}-2$. We can see that $q\sqrt{q}-3q-\sqrt{q}-2>0$ for $q\geq12$ and hence that $q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-3>0$ for $q\geq9$. Since $q\left\lfloor\sqrt{q}\right\rfloor-2q-\left\lfloor\sqrt{q}\right\rfloor-3>0$ for $q\geq9$ and $-3q-\left\lfloor\sqrt{q}\right\rfloor-2<0$, the previous inequality gives a contradiction for $m\leq 2$. \end{proof} \begin{lemma}\label{plus1} Let $P$ be a point of an axiomatic projective plane $\mathcal{P}$ of order $q$, $q\geq 25$, and let $\mathcal{L}$ be a set of at most $q+\left\lfloor \sqrt{q}\right\rfloor+3$ lines in $\mathcal{P}$ such that all points but $P$ are on a line of $\mathcal{L}$, and $P$ is not. Let $k$ be such that $\mathcal{L}$ admits a $k$-knot but no $k'$-knot for $k'>k$. Then $k>\left\lfloor\sqrt{q}\right\rfloor+1$. \end{lemma} \begin{proof} Note that $2q-1>q+\sqrt{q}+3\geq q+\left\lfloor\sqrt{q}\right\rfloor+3$ since $q\geq7$. From Lemma \ref{eerste}, we know that $k>\left\lfloor\sqrt{q}\right\rfloor$. If $k=\left\lfloor\sqrt{q}\right\rfloor+1$, then we have $|\mathcal{L}|\geq q+\left\lfloor\sqrt{q}\right\rfloor+1$. Setting $|\mathcal{L}|=q+\left\lfloor\sqrt{q}\right\rfloor+1+m$ it follows from Lemma \ref{Beq} that \begin{align}\label{Beq0} &\quad\sum_{i=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(i-1)(\left\lfloor\sqrt{q}\right\rfloor+1-i)x_{i}\nonumber\\&=-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+\left\lfloor\sqrt{q}\right\rfloor+1+m\left((q-1)\left\lfloor\sqrt{q}\right\rfloor-q-m\right)\;. \end{align} \par We need to show that for $m=0$ and $m=1$, we find a contradiction. \paragraph*{Case 1: $m=0$.}\mbox{}\\ \indent If $m=0$ the total contribution of the $i$-knots, $i>0$, in the left hand side of \eqref{Beq0} equals $\left\lfloor\sqrt{q}\right\rfloor+1-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)$. We know that 1-knots and $(\left\lfloor\sqrt{q}\right\rfloor+1)$-knots contribute zero to this sum, 2-knots and $\left\lfloor\sqrt{q}\right\rfloor$-knots contribute $\sqrt{q}-1$ to this sum, and all other knots contribute at least $2\left\lfloor\sqrt{q}\right\rfloor-4$ to this sum. Since $\left\lfloor\sqrt{q}\right\rfloor\geq5$ we have $0<\left\lfloor\sqrt{q}\right\rfloor-1<2\left\lfloor\sqrt{q}\right\rfloor-4<2\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)$. We look at the different cases. \begin{itemize} \item If $\left\lfloor\sqrt{q}\right\rfloor+1-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)=0$, then either $\left\lfloor\sqrt{q}\right\rfloor+1=0$ or $\left\lfloor\sqrt{q}\right\rfloor+1\geq q$, which both lead to a contradiction. \item If $\left\lfloor\sqrt{q}\right\rfloor+1-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)=\left\lfloor\sqrt{q}\right\rfloor-1$, then $q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)=2$, which also leads to contradiction since $q\geq3$. \item If $\left\lfloor\sqrt{q}\right\rfloor+1-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)\geq2\left\lfloor\sqrt{q}\right\rfloor-4$, then $q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+\left(\left\lfloor\sqrt{q}\right\rfloor-5\right)\leq 0$. Since both terms are non-negative, it follows that $q=25$ and there is equality in the assumption. So, in this case we find that there is exactly one 3-knot or exactly one 4-knot, and all other points (different from $P$) are 1-knots or 6-knots. Say there is a 3-knot $P'$ (the argument for a 4-knot is analogous). We count the tuples $(Q,\ell)$ with $\ell\in\mathcal{L}$ and $Q$ the intersection point of $\ell$ and $PP'$. Reducing modulo 5 we find $1\equiv|\mathcal{L}|\equiv(25-1)\cdot1+3\equiv 2\pmod{5}$, a contradiction.% (Note that any line through $P'$ could take the role of $PP'$.) \end{itemize} \paragraph*{Case 2: $m=1$.}\mbox{}\\ \indent If $m=1$ the total contribution of the $i$-knots, $i>0$, in the left hand side of \eqref{Beq0} equals $q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)$. We know that $\L$ admits a $(\left\lfloor\sqrt{q}\right\rfloor+1)$-knot, say $K$. Let $\ell$ be a line through $K$ and denote the number of $i$-knots on $\ell$ by $a_{i}$. \par If $\ell\notin\mathcal{L}$, then we have $\sum_{i=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}ia_{i}=q+\left\lfloor\sqrt{q}\right\rfloor+2$ by counting the tuples $(Q,n)$ with $Q$ on $n$ and $n\in\mathcal{L}$. We also have $\sum_{i=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}a_{i}=q+1-\delta$ where $\delta=0$ if $P\notin\ell$ and $\delta=1$ if $P\in\ell$. We find $\sum_{i=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(i-1)a_{i}=\left\lfloor\sqrt{q}\right\rfloor+1+\delta$. Using Lemma \ref{afschatting} with $b=\left\lfloor\sqrt{q}\right\rfloor+1$, $k=\left\lfloor\sqrt{q}\right\rfloor+\delta+1$, $m=\delta+1$, we find that the points on $\ell$ contribute at least $\left\lfloor\sqrt{q}\right\rfloor-1$ to the sum in \eqref{Beq0} if $P\notin\ell$ (so if $\delta=0$) and that the points on the line $KP$ (so, for $\delta=1$) contribute at least $2(\left\lfloor\sqrt{q}\right\rfloor-2)$. \par If $\ell\in\mathcal{L}$ we have $\sum_{i=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(i-1)a_{i}=q+\left\lfloor\sqrt{q}\right\rfloor+1$ by counting the tuples $(Q,n)$ with $Q$ on $n$ and $n\in\mathcal{L}\setminus\{\ell\}$. We now distinguish between two cases. \begin{itemize} \item[(i)] If $q+\left\lfloor\sqrt{q}\right\rfloor+1\not\equiv0\pmod{\left\lfloor\sqrt{q}\right\rfloor}$, then we find that not all points on $\ell\in\mathcal{L}$ can be 1-knots and $\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)$-knots, hence the points on $\ell$ contribute at least $\left\lfloor\sqrt{q}\right\rfloor-1$ to the sum in \eqref{Beq0}. It follows that \begin{align*} q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)&=\sum_{i=1}^{\sqrt{q}+1}(i-1)(\left\lfloor\sqrt{q}\right\rfloor+1-i)x_{i}\\&\geq q(\left\lfloor\sqrt{q}\right\rfloor-1)+2(\left\lfloor\sqrt{q}\right\rfloor-2)\;, \end{align*} hence \[ q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+2\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)\leq 0\;, \] a contradiction since $q>4$. Note that the contribution of $K$ is necessarily 0 so it does not matter that we counted this point $q+1$ times. \item[(ii)] If $q+\left\lfloor\sqrt{q}\right\rfloor+1\equiv0\pmod{\left\lfloor\sqrt{q}\right\rfloor}$, then $q=\left\lfloor\sqrt{q}\right\rfloor\left(\left\lfloor\sqrt{q}\right\rfloor+\varepsilon\right)-1$ for some $\varepsilon\in\{1,2\}$ since $\left\lfloor\sqrt{q}\right\rfloor\left(\left\lfloor\sqrt{q}\right\rfloor+0\right)-1\leq\sqrt{q}\sqrt{q}-1(\sqrt{q}-1)(\sqrt{q}+2)-1\geq q$ for $q\geq9$. Using the previous argument, only looking at lines through $K$ not in $ \L$, we then find \begin{align*} q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)&=\sum_{i=1}^{\sqrt{q}+1}(i-1)(\left\lfloor\sqrt{q}\right\rfloor+1-i)x_{i}\\ &\geq (q-\left\lfloor\sqrt{q}\right\rfloor-1)(\left\lfloor\sqrt{q}\right\rfloor-1)+2(\left\lfloor\sqrt{q}\right\rfloor-2)\;, \end{align*} hence \begin{align*} &q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)-\left(\left\lfloor\sqrt{q}\right\rfloor^{2}-2\left\lfloor\sqrt{q}\right\rfloor+3\right)\leq 0\;,\\ \Leftrightarrow\quad&\varepsilon\left\lfloor\sqrt{q}\right\rfloor^{3}+(\varepsilon^{2}-2)\left\lfloor\sqrt{q}\right\rfloor^{2}-2(\varepsilon-1)\left\lfloor\sqrt{q}\right\rfloor-2\leq 0 \end{align*} a contradiction for $\varepsilon\in\{1,2\}$ since $q\geq4$.\qedhere \end{itemize} \end{proof} \begin{lemma}\label{5superknots} Let $P$ be a point of an axiomatic projective plane $\mathcal{P}$ of order $q$, $q\geq 25$, and let $\mathcal{L}$ be a set of lines in $\mathcal{P}$ such that all points but $P$ are on a line of $\mathcal{L}$, and $P$ is not. Let $k$ be such that $\mathcal{L}$ admits a $k$-knot but no $k'$-knot for $k'>k$. Suppose that $k=\lfloor \sqrt{q}\rfloor+2$ and that $\mathcal{L}$ has size $q+\lfloor \sqrt{q}\rfloor+2$, then the number of $(\lfloor \sqrt{q}\rfloor+2)$-knots is at most $5$. \end{lemma} \begin{proof} We will first prove that there can be at most three $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots on a line. Assume that $K_{1},K_{2},K_{3},K_{4}$ are four $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots on a line; necessarily this line is in $\mathcal{L}$. The point $P$ is not on $\ell$. On the line $PK_{1}$ there is precisely on 2-knot $T$; the remaining $q-1$ points are 1-knots. Consider the lines $TK_{i}$, $i=2,3,4$. They do not contain the point $P$, and hence, if $TK_i$ is not a line of $\L$, all $q-1$ points, different from $T$ and $K_i$ have to be covered by one of the remaining $q+\left\lfloor\sqrt{q}\right\rfloor+2-(\left\lfloor\sqrt{q}\right\rfloor+2)-2=q-2$ lines of $\L$ not through $T$ and $K_i$ which is impossible. We conclude that the lines $TK_i$, $i=2,3,4$ are lines of $\L$. But this is a contradiction as $T$ is on exactly two lines of $\mathcal{L}$. \par We now prove that $d\leq 5$. Let $K_{1},\dots,K_{6}$ be six $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots. For any $K_{i}$, $i=1,\dots,6$, we know that there is exactly one 2-knot $T_{i}$ on the line $PK_{i}$. If $i\neq j$, then $T_{i}K_{j}$ is a line not containing $P$, hence, using the same argument as above, $T_iK_j$ is a line of $\mathcal{L}$. As there are only two lines of $\mathcal{L}$ through $T_{i}$, $i=1,\dots,6$, one of them contains at least three $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots. So, we know that three of the points $K_{1},\dots,K_{6}$ are collinear, say $K_{1}$, $K_{2}$ and $K_{3}$, and we denote the line they are on by $\ell$. As there are only two lines of $\mathcal{L}$ through a 2-knot we have immediately that $T_{i}\in\ell$ for $i=4,5,6$. Moreover, the other line of $\mathcal{L}$ through $T_{i}$ contains $K_{j}$ and $K_{j'}$ for $\{i,j,j'\}=\{4,5,6\}$. Now, the two lines of $\mathcal{L}$ through $T_{1}$ must be $T_{1}K_{2}$ and $T_{1}K_{3}$. Each of the $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots $K_{4}$, $K_{5}$ and $K_{6}$ must be on one of these two lines, so one of these two lines contains at least two of them. Say, without loss of generality that $K_{4}$ and $K_{5}$ are on $T_{1}K_{2}$. However, then $T_{6}$ is also on this line, which implies $T_{6}=K_{2}$ since both points are also on $\ell$, a contradiction. \end{proof} \begin{theorem} Let $P$ be a point of an axiomatic projective plane $\mathcal{P}$ of order $q$, and let $\mathcal{L}$ be a set of lines in $\mathcal{P}$ such that all points but $P$ are on a line of $\mathcal{L}$, and $P$ is not. If $q\geq25$, then $|\mathcal{L}|\geq q+\left\lfloor\sqrt{q}\right\rfloor+3$. \end{theorem} \begin{proof} By Lemma \ref{eerste}, we have that $|\mathcal{L}|\geq q+\left\lfloor\sqrt{q}\right\rfloor+3$ if $k\geq\left\lfloor\sqrt{q}\right\rfloor+3$. By Lemma \ref{plus1}, $k>\left\lfloor\sqrt{q}\right\rfloor+1$. So it follows that we only need to rule out the case $k=\left\lfloor\sqrt{q}\right\rfloor+2$ in order to obtain that $|\mathcal{L}|\geq q+\left\lfloor\sqrt{q}\right\rfloor+3$. \par If $k=\left\lfloor\sqrt{q}\right\rfloor+2$, then we know $|\mathcal{L}|\geq q+\left\lfloor\sqrt{q}\right\rfloor+2$ and we may assume that $|\mathcal{L}|=q+\left\lfloor\sqrt{q}\right\rfloor+2$. We set $d=x_{\left\lfloor\sqrt{q}\right\rfloor+2}$ for ease of notation. It then follows from Lemma \ref{Beq} that \begin{align}\label{Beq1} \sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(j-1)(\left\lfloor\sqrt{q}\right\rfloor+1-j)x_{j}=q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+d(\left\lfloor\sqrt{q}\right\rfloor+1)\;. \end{align} Now, let $K$ be an $i$-knot with $4\leq i\leq\left\lfloor\sqrt{q}\right\rfloor$, let $\ell\notin\mathcal{L}$ be a line through $K$ and denote the number of $j$-knots on $\ell$ different from $K$ and $P$ by $a_{j}$. Then we have $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}ja_{j}=q+\left\lfloor\sqrt{q}\right\rfloor+2-i$ by counting the tuples $(Q,n)$ with $\{Q\}=\ell\cap n$, $Q\neq K$ and $n\in\mathcal{L}$. We also have $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}a_{j}=q-\delta$ where $\delta=0$ if $P\notin\ell$ and $\delta=1$ if $P\in\ell$. We find $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}(j-1)a_{j}=\left\lfloor\sqrt{q}\right\rfloor+2-i+\delta$. Moreover, as $\left\lfloor\sqrt{q}\right\rfloor+2-i+\delta\leq\left\lfloor\sqrt{q}\right\rfloor-1$, we have $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(j-1)a_{j}=\left\lfloor\sqrt{q}\right\rfloor+2-i+\delta$. So we find that the points on $\ell\neq KP$ different from $K$ contribute at least $\left(\left\lfloor\sqrt{q}\right\rfloor+2-i\right)(i-2)$ to the sum in \eqref{Beq1} by Lemma \ref{afschatting}; the points on $KP$ different from $K$ contribute at least $\left(\left\lfloor\sqrt{q}\right\rfloor+3-i\right)(i-3)$ to the sum in \eqref{Beq1}. None of these contributions includes the point $K$, but $K$ itself contributes $\left(\left\lfloor\sqrt{q}\right\rfloor+1-i\right)(i-1)$ to the sum in \eqref{Beq1}. We find that \begin{align}\label{Beq2pro} q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+d(\left\lfloor\sqrt{q}\right\rfloor+1)&=\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(j-1)\left(\left\lfloor\sqrt{q}\right\rfloor+1-j\right)x_{j}\nonumber\\ &\geq (q-i)\left(\left\lfloor\sqrt{q}\right\rfloor+2-i\right)(i-2)\nonumber\\&\ \quad+\left(\left\lfloor\sqrt{q}\right\rfloor+3-i\right)(i-3)\nonumber\\&\ \quad+\left(\left\lfloor\sqrt{q}\right\rfloor+1-i\right)(i-1)\;. \end{align} Hence, \begin{align}\label{Beq2} 0&\geq i^{3}-(q+\left\lfloor\sqrt{q}\right\rfloor+6)i^{2}+(q\left\lfloor\sqrt{q}\right\rfloor+4q+4\left\lfloor\sqrt{q}\right\rfloor+12)i-d(\left\lfloor\sqrt{q}\right\rfloor+1)\nonumber\\&\qquad+q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2}-3q\left\lfloor\sqrt{q}\right\rfloor-3q-4\left\lfloor\sqrt{q}\right\rfloor-10\;. \end{align} For $i=4$ we find, using that $d\leq 5$ from Lemma \ref{5superknots}, \[ 0\geq q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2}+q\left\lfloor\sqrt{q}\right\rfloor-3q -4\left\lfloor\sqrt{q}\right\rfloor+6-d\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)\geq q\left\lfloor\sqrt{q}\right\rfloor-3q -9\left\lfloor\sqrt{q}\right\rfloor+1\;, \] which is a contradiction for $q\geq25$. For $i=\left\lfloor\sqrt{q}\right\rfloor$ we find, again using that $d\leq 5$ from Lemma \ref{5superknots}, \begin{align*} 0&\geq q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2}+q\left\lfloor\sqrt{q}\right\rfloor-3q-2\left\lfloor\sqrt{q}\right\rfloor^{2}+8\left\lfloor\sqrt{q}\right\rfloor-10-d\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)\\ &\geq q\left\lfloor\sqrt{q}\right\rfloor-3q-2\left\lfloor\sqrt{q}\right\rfloor^{2}+3\left\lfloor\sqrt{q}\right\rfloor-15\;, \end{align*} which is a contradiction for $q>25$. Since the right hand side of \eqref{Beq2} is function of degree 3 in $i$ it can easily be checked that it is first increasing and then decreasing on the interval $[4,\left\lfloor\sqrt{q}\right\rfloor]$ for $q\geq25$: looking at its derivative we see that the function is decreasing on $\left]\frac{q+\left\lfloor q\right\rfloor-\sqrt{q^{2}-q\left\lfloor q\right\rfloor+\left\lfloor q\right\rfloor^{2}}}{3}+2,\frac{q+\left\lfloor q\right\rfloor+\sqrt{q^{2}-q\left\lfloor q\right\rfloor+\left\lfloor q\right\rfloor^{2}}}{3}+2\right[$ and \[ 4\leq\frac{q+\left\lfloor q\right\rfloor-\sqrt{q^{2}-q\left\lfloor q\right\rfloor+\left\lfloor q\right\rfloor^{2}}}{3}+2\leq\left\lfloor q\right\rfloor\leq\frac{q+\left\lfloor q\right\rfloor+\sqrt{q^{2}-q\left\lfloor q\right\rfloor+\left\lfloor q\right\rfloor^{2}}}{3}+2\;. \] So, the minimum of this function on this interval is to be found at one of its endpoints. So, as we have a contradiction for $i=4$ and $i=\left\lfloor\sqrt{q}\right\rfloor$, we have a contradiction for all $i$ with $4\leq i\leq\left\lfloor\sqrt{q}\right\rfloor$. \par For $q=25$ we did not yet exclude $i=5$. We find a contradiction in \eqref{Beq2} if $0\leq d\leq 4$, and we have equality in \eqref{Beq2} if $d=5$. In the latter case we also have equality in \eqref{Beq2pro} and hence the points on the lines of $\mathcal{L}$ through the $5$-knot $K$ should all contribute zero to the sum in \eqref{Beq1}. If $\ell\in\mathcal{L}$ through $K$ we have \[ \sum_{j=1}^{7}(j-1)a_{j}=q+\left\lfloor\sqrt{q}\right\rfloor+2-i=27 \] by counting the tuples $(Q,n)$ with $\{Q\}=\ell\cap n$, $Q\neq K$ and $n\in\mathcal{L}\setminus\{\ell\}$. Hence, not all points on $\ell$ different from $K$ can be 1-knots or 6-knots; more precisely, there should be at least two 7-knots on $\ell$. It follows that the total number of 7-knots is at least 10, a contradiction. \par So far we have concluded that $x_{i}=0$ for all $i$ with $4\leq i\leq\left\lfloor\sqrt{q}\right\rfloor$. Now, let $K'$ be a 3-knot, let $\ell\notin\mathcal{L}$ be a line through $K'$ different from $K'P$ and denote the number of $j$-knots on $\ell$ different from $K'$ by $a_{j}$. Counting the tuples $(Q,n)$ with $\{Q\}=\ell\cap n$, $Q\neq K'$ and $n\in\mathcal{L}$ and using $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}a_{j}=q$ we find $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}(j-1)a_{j}=\left\lfloor\sqrt{q}\right\rfloor-1$, but as $a_{j}=0$ for all $j=4,\dots,\left\lfloor\sqrt{q}\right\rfloor$, we find that \[ a_{2}+2a_{3}+\left\lfloor\sqrt{q}\right\rfloor a_{\left\lfloor\sqrt{q}\right\rfloor+1}+\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)a_{\left\lfloor\sqrt{q}\right\rfloor+2}=\left\lfloor\sqrt{q}\right\rfloor-1 \] and hence that $a_{\left\lfloor\sqrt{q}\right\rfloor+1}=a_{\left\lfloor\sqrt{q}\right\rfloor+2}=0$. So, the points on $\ell$ different from $K$ contribute \begin{align*} a_{2}\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)+a_{3}\left(2\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)\right)&=a_{2}\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)+\frac{\left\lfloor\sqrt{q}\right\rfloor-1-a_{2}}{2}\left(2\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)\right)\\ &=a_{2}+\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)\left(\left\lfloor\sqrt{q}\right\rfloor-2\right) \end{align*} hence at least $\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)$ to the sum in \eqref{Beq1}. The point $K$ itself also contributes $2\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)$ to this sum. We find \begin{align*} q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-q\left(q-\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+d\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)&=\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+1}(j-1)\left(\left\lfloor\sqrt{q}\right\rfloor+1-j\right)x_{j}\\ &\geq (q-3)\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)\\&\qquad+2\left(\left\lfloor\sqrt{q}\right\rfloor-2\right)\\ &=q\left\lfloor\sqrt{q}\right\rfloor^{2}-3q\left\lfloor\sqrt{q}\right\rfloor+2q-3\left\lfloor\sqrt{q}\right\rfloor^{2}\\&\qquad+11\left\lfloor\sqrt{q}\right\rfloor-10\;. \end{align*} So, we have, using Lemma \ref{5superknots}, that \[ 0\geq q^{2}-4q\left\lfloor\sqrt{q}\right\rfloor+3q-3\left\lfloor\sqrt{q}\right\rfloor^{2}+11\left\lfloor\sqrt{q}\right\rfloor-10-d\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)\geq q^{2}-4q\left\lfloor\sqrt{q}\right\rfloor+6\left\lfloor\sqrt{q}\right\rfloor-15\;, \] which is a contradiction for $q\geq16$. \par So far, we know that $x_{i}=0$ for all $i\notin\{1,2,\left\lfloor\sqrt{q}\right\rfloor+1,\left\lfloor\sqrt{q}\right\rfloor+2\}$. We call the $\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)$-knots and the $\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)$-knots {\em big knots}. It follows from the standard countings that \begin{align*} &\left\{\begin{aligned} x_{1}+x_{2}+x_{\left\lfloor\sqrt{q}\right\rfloor+1}&=q^{2}+q-d\\ x_{1}+2x_{2}+\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)x_{\left\lfloor\sqrt{q}\right\rfloor+1}&=(q+1)\left(q+\left\lfloor\sqrt{q}\right\rfloor+2\right)-\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)d\\ 2x_{2}+\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)\left\lfloor\sqrt{q}\right\rfloor x_{\left\lfloor\sqrt{q}\right\rfloor+1}&=\left(q+\left\lfloor\sqrt{q}\right\rfloor+2\right)\left(q+\left\lfloor\sqrt{q}\right\rfloor+1\right)\\&\qquad-\left(\left\lfloor\sqrt{q}\right\rfloor+2\right)\left(\left\lfloor\sqrt{q}\right\rfloor+1\right)d \end{aligned}\right.\\ \Leftrightarrow\quad&\left\{\begin{aligned} x_{1}&=q^{2}-q\left\lfloor\sqrt{q}\right\rfloor-q-d-1+\frac{q^{2}-q-2}{\left\lfloor\sqrt{q}\right\rfloor}\\ x_{2}&=q\left\lfloor\sqrt{q}\right\rfloor+2q+d-\frac{q^{2}-q-2d}{\left\lfloor\sqrt{q}\right\rfloor-1}\\ x_{\left\lfloor\sqrt{q}\right\rfloor+1}&=1-d+\frac{q^{2}-q-2(d-1)\left\lfloor\sqrt{q}\right\rfloor-2}{\left\lfloor\sqrt{q}\right\rfloor\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)} \end{aligned}\right. \end{align*} Since $x_{1}$ is an integer, there must be an integer $a$ such that $q^{2}-q-2=a\left\lfloor\sqrt{q}\right\rfloor$. Since $x_{2}$ is an integer too, $q^2-q-2d=a\left\lfloor\sqrt{q}\right\rfloor-2(d-1)\equiv0\pmod{\left\lfloor\sqrt{q}\right\rfloor-1}$. So, there must be an integer $b$ such that $a=2(d-1)+b(\left\lfloor\sqrt{q}\right\rfloor-1)$. We find that \begin{align*} \left\{\begin{aligned} x_{1}&=q^{2}-q\left\lfloor\sqrt{q}\right\rfloor-q+b\left\lfloor\sqrt{q}\right\rfloor+d-3-b\\ x_{2}&=q\left\lfloor\sqrt{q}\right\rfloor+2q-d-b\left\lfloor\sqrt{q}\right\rfloor+2\\ x_{\left\lfloor\sqrt{q}\right\rfloor+1}&=1-d+b \end{aligned}\right. \end{align*} By counting the tuples $(Q,n)$ with $\{Q\}=\ell\cap n$, $Q\neq P$ and $n\in\mathcal{L}$ and using $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}a_{j}=q$ we find that for a line $\ell\ni P$ (necessarily not in $\mathcal{L}$) we have that $\sum_{j=1}^{\left\lfloor\sqrt{q}\right\rfloor+2}(j-1)a_{j}=\left\lfloor\sqrt{q}\right\rfloor+2$ with $a_{j}$ the number of $j$-knots on $\ell$. We see that there can be at most 1 big knot on a line through $P$. Hence, $d+x_{\left\lfloor\sqrt{q}\right\rfloor+1}$, the number of big knots, is at most $q+1$. It follows that $b\leq q$. Consequently, \[ q^{2}-q-2=\left\lfloor\sqrt{q}\right\rfloor\left(2(d-1)+b(\left\lfloor\sqrt{q}\right\rfloor-1)\right)\leq q\left\lfloor\sqrt{q}\right\rfloor^{2}-q\left\lfloor\sqrt{q}\right\rfloor+2(d-1)\left\lfloor\sqrt{q}\right\rfloor\;. \] However, then, using that $d\leq 5$ from Lemma \ref{5superknots}, \begin{align*} 0&\geq q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2}+q\left\lfloor\sqrt{q}\right\rfloor-q-2(d-1)\left\lfloor\sqrt{q}\right\rfloor-2\\&\geq \left(q^{2}-q\left\lfloor\sqrt{q}\right\rfloor^{2}\right)+q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-8\left\lfloor\sqrt{q}\right\rfloor-2\\ &\geq q\left(\left\lfloor\sqrt{q}\right\rfloor-1\right)-8\left\lfloor\sqrt{q}\right\rfloor-2\;, \end{align*} which is a contradiction for $q\geq10$. \end{proof} \begin{corollary} A blocking set of an affine plane of order $q$ contains at least $q+\left\lfloor\sqrt{q}\right\rfloor+3$ points if $q\geq25$. \end{corollary} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection*{Acknowledgements} This research was performed when the first author was visiting the School of Mathematics and Statistics at the University of Canterbury. 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