[proofplan] We prove both directions directly from the incidence axioms. For a projective plane of order $q$, we count points by fixing a line and partitioning the remaining points according to the lines through one point not on that line; the same argument counts the lines. For the converse, the $2$-design axiom gives a unique block through two points, and symmetry forces any two distinct blocks to meet in exactly one point. Finally, we verify the nondegeneracy axiom by explicitly constructing four points with no three collinear. [/proofplan]

[step:Count the points and lines in a projective plane of order $q$]Let $\Pi = (P,\mathcal L)$ be a finite [projective plane](/page/Projective%20Plane) of order $q$. Thus every line contains exactly $q+1$ points, any two distinct points lie on a unique line, any two distinct lines meet in a unique point, and there exist four points no three of which are collinear. First fix a point $p \in P$ and a line $M \in \mathcal L$ with $p \notin M$. Such a line exists by the nondegeneracy axiom. Indeed, choose four points $a,b,c,d \in P$ no three of which are collinear. If $p$ is one of these four points, take $M$ to be the line through two of the other three points. If $p$ is not one of them, then at least one of the lines through two of $a,b,c,d$ does not contain $p$; otherwise the lines through $a,b$ and through $a,c$ would both contain the two distinct points $a$ and $p$, forcing them to be equal and making $a,b,c$ collinear. Define $\mathcal L(p) := \{L \in \mathcal L : p \in L\}$, the set of lines through $p$. Define the map $\Phi_p : \mathcal L(p) \to M$ as follows: for each line $L \in \mathcal L(p)$, let $\Phi_p(L)$ be the unique point in $L \cap M$. This map is well-defined because every line through $p$ meets $M$ in a unique point. It is injective because if $L_1,L_2 \in \mathcal L(p)$ satisfy $L_1 \cap M = L_2 \cap M = \{m\}$, then both $L_1$ and $L_2$ contain the two distinct points $p$ and $m$, so uniqueness of the line through two points gives $L_1=L_2$. It is surjective because for each $m \in M$, the unique line through $p$ and $m$ lies in $\mathcal L(p)$ and meets $M$ at $m$. Hence $|\mathcal L(p)|=|M|=q+1$. Now fix a line $B \in \mathcal L$ and a point $p \notin B$. The $q+1$ lines through $p$ meet $B$ in the $q+1$ distinct points of $B$. Every point $z \in P \setminus \{p\}$ lies on exactly one of these lines, namely the unique line through $p$ and $z$. Each such line contains $q$ points other than $p$, and these $q+1$ sets of points are pairwise disjoint away from $p$, since two distinct lines through $p$ meet only at $p$. Therefore \begin{align*} |P| = 1 + (q+1)q = q^2+q+1. \end{align*} Finally, since every point lies on exactly $q+1$ lines and every line contains exactly $q+1$ points, counting incident pairs $(p,L) \in P \times \mathcal L$ with $p \in L$ gives \begin{align*} |P|(q+1)=|\mathcal L|(q+1). \end{align*} Because $q+1>0$, we obtain \begin{align*} |\mathcal L|=|P|=q^2+q+1. \end{align*}[/step]

[guided]We need the design parameters, so we first extract the numerical consequences of the projective-plane axioms. The key auxiliary count is the number of lines through a fixed point. Let $p \in P$ be a point. We choose a line $M \in \mathcal L$ with $p \notin M$. This is possible because the projective plane contains four points $a,b,c,d \in P$ no three of which are collinear. If $p$ is one of these four points, then the line through two of the remaining three cannot contain $p$. If $p$ is not one of them, at least one line through two of $a,b,c,d$ must avoid $p$; otherwise the lines through $a,b$ and through $a,c$ would both contain $a$ and $p$, so uniqueness of the line through two distinct points would force those two lines to coincide, contradicting that $a,b,c$ are not collinear. Define $\mathcal L(p) := \{L \in \mathcal L : p \in L\}$, the set of all lines through $p$. We compare $\mathcal L(p)$ with the points of $M$ by defining the map $\Phi_p : \mathcal L(p) \to M$ as follows: for each line $L \in \mathcal L(p)$, let $\Phi_p(L)$ be the unique point in $L \cap M$. The map is well-defined because the projective-plane axiom says any two distinct lines meet in exactly one point. It is injective: if two lines through $p$ meet $M$ at the same point $m$, then both lines contain $p$ and $m$, and the uniqueness of the line through two distinct points forces the two lines to be equal. It is surjective: given $m \in M$, the unique line through $p$ and $m$ belongs to $\mathcal L(p)$ and maps to $m$. Thus $\Phi_p$ is a bijection, so \begin{align*} |\mathcal L(p)|=|M|=q+1. \end{align*} Now fix a line $B \in \mathcal L$ and a point $p \notin B$. There are exactly $q+1$ lines through $p$. Each one meets $B$ in exactly one point, and the bijection above shows that these intersection points run through all points of $B$. Each line through $p$ contains $q+1$ points total, hence $q$ points other than $p$. Two distinct lines through $p$ share no point except $p$, because if they shared another point then two distinct points would determine two different lines. Every point $z \in P \setminus \{p\}$ lies on one of these lines through $p$, because the unique-line axiom supplies a unique line containing $p$ and $z$. Therefore the point set $P$ is partitioned into the point $p$ together with $q+1$ disjoint groups of $q$ further points. Hence \begin{align*} |P| = 1 + (q+1)q = q^2+q+1. \end{align*} To count the lines, count the same incidence set in two ways: \begin{align*} I := \{(p,L) \in P \times \mathcal L : p \in L\}. \end{align*} Counting by points gives $|I|=|P|(q+1)$ because each point lies on $q+1$ lines. Counting by lines gives $|I|=|\mathcal L|(q+1)$ because each line contains $q+1$ points. Therefore \begin{align*} |P|(q+1)=|\mathcal L|(q+1). \end{align*} Since $q+1>0$, cancellation gives \begin{align*} |\mathcal L|=|P|=q^2+q+1. \end{align*}[/guided]

[step:Read the projective plane as a symmetric design] Take the points of the design to be the elements of $P$ and the blocks to be the lines $L \in \mathcal L$, regarded as subsets of $P$. From the previous step, \begin{align*} |P|=|\mathcal L|=q^2+q+1. \end{align*} Every block has size $q+1$ by the definition of order $q$. Finally, any two distinct points of $P$ lie on a unique line, so every two-element subset of $P$ is contained in exactly one block. Hence the design parameter is $\lambda=1$. Thus $(P,\mathcal L)$ is a [$2$-design](/page/Block%20Design) with parameters $2$-$(q^2+q+1,q+1,1)$, and it is [symmetric](/page/Symmetric%20Design) because the number of blocks equals the number of points. [/step]

[step:Show that two blocks in the symmetric design meet in one point]Conversely, let $(X,\mathcal B)$ be a [symmetric design](/page/Symmetric%20Design) with parameters $2$-$(q^2+q+1,q+1,1)$. Thus \begin{align*} |X|=|\mathcal B|=q^2+q+1, \end{align*} each block $B \in \mathcal B$ has $|B|=q+1$, and every two distinct points of $X$ lie in exactly one block. We first prove that any two distinct blocks meet in exactly one point. Fix a point $x \in X$, and let $r_x$ denote the number of blocks containing $x$. Count the pairs \begin{align*} T_x := \{(y,B) \in (X \setminus \{x\}) \times \mathcal B : x \in B \text{ and } y \in B\}. \end{align*} Counting by points $y \ne x$, the $2$-design axiom with $\lambda=1$ gives exactly one block containing both $x$ and $y$, so \begin{align*} |T_x|=|X|-1=q^2+q. \end{align*} Counting by blocks containing $x$, each such block has $q$ points other than $x$, so \begin{align*} |T_x|=r_x q. \end{align*} Since $q \ge 2$, division by $q$ gives \begin{align*} r_x=q+1. \end{align*} Thus every point lies in exactly $q+1$ blocks. Fix a block $B_0 \in \mathcal B$. Count the pairs \begin{align*} S := \{(x,B) \in B_0 \times (\mathcal B \setminus \{B_0\}) : x \in B\}. \end{align*} Counting by points $x \in B_0$, each such $x$ lies in $r_x-1=q$ blocks other than $B_0$, so \begin{align*} |S|=(q+1)q=q^2+q. \end{align*} But $|\mathcal B \setminus \{B_0\}|=q^2+q$, so the average value of $|B \cap B_0|$ over $B \ne B_0$ is $1$. For every $B \ne B_0$, the intersection $B \cap B_0$ is nonempty. Indeed, if $B \cap B_0=\varnothing$, then the $q+1$ points of $B$ and the $q+1$ points of $B_0$ determine $(q+1)^2$ distinct blocks joining one point of $B$ to one point of $B_0$; together with $B$ and $B_0$, this gives at least \begin{align*} (q+1)^2+2=q^2+2q+3 \end{align*} blocks, which is larger than $|\mathcal B|=q^2+q+1$ for $q \ge 2$, a contradiction. Therefore each of the $q^2+q$ blocks other than $B_0$ contributes at least one point to $S$, and since the total contribution is exactly $q^2+q$, each contributes exactly one point. Hence every block distinct from $B_0$ meets $B_0$ in exactly one point. Since $B_0$ was arbitrary, any two distinct blocks meet in exactly one point.[/step]

[guided]We need the line-intersection axiom for the incidence structure obtained from the design. In a projective plane, two distinct lines must meet in exactly one point, so here we must prove that two distinct blocks of the symmetric design meet in exactly one point. First compute the replication number directly from the $2$-design axiom. Fix a point $x \in X$, and let $r_x$ denote the number of blocks containing $x$. We count the set \begin{align*} T_x := \{(y,B) \in (X \setminus \{x\}) \times \mathcal B : x \in B \text{ and } y \in B\}. \end{align*} Counting by points $y \in X \setminus \{x\}$ gives $|T_x|=|X|-1=q^2+q$, because the design axiom with $\lambda=1$ says that each pair of distinct points $x,y$ lies in exactly one block. Counting by blocks containing $x$ gives $|T_x|=r_x q$, because each block containing $x$ has $q+1$ points total and therefore $q$ points other than $x$. Hence \begin{align*} r_x q=q^2+q. \end{align*} Since $q \ge 2$, division by $q$ gives \begin{align*} r_x=q+1. \end{align*} The point $x$ was arbitrary, so every point lies in exactly $q+1$ blocks. Now fix one block $B_0 \in \mathcal B$. We count how all other blocks meet $B_0$. Define \begin{align*} S := \{(x,B) \in B_0 \times (\mathcal B \setminus \{B_0\}) : x \in B\}. \end{align*} For each point $x \in B_0$, the computation above gives $r_x=q+1$ blocks containing $x$, one of which is $B_0$ itself. Hence there are $q$ blocks other than $B_0$ containing $x$. Since $|B_0|=q+1$, counting by points gives \begin{align*} |S|=(q+1)q=q^2+q. \end{align*} There are also \begin{align*} |\mathcal B \setminus \{B_0\}|=(q^2+q+1)-1=q^2+q \end{align*} blocks other than $B_0$. Thus the average intersection size $|B \cap B_0|$ for $B \ne B_0$ is $1$. An average of $1$ alone does not rule out some block missing $B_0$ and another block meeting $B_0$ twice. So we prove that no block can miss $B_0$. Suppose, for contradiction, that $B \in \mathcal B$ satisfies $B \ne B_0$ and $B \cap B_0=\varnothing$. For each ordered pair $(x,y) \in B \times B_0$, the $2$-design axiom gives a unique block containing $x$ and $y$. These blocks are all distinct: if the same block contained both pairs $(x,y)$ and $(x',y')$ with either $x \ne x'$ or $y \ne y'$, then it would contain two distinct points of $B$ or two distinct points of $B_0$, forcing it to be $B$ or $B_0$ by uniqueness of the block through two points; this is impossible because the block also contains a point from the other disjoint block. Hence the pairs produce $(q+1)^2$ distinct blocks, all different from $B$ and $B_0$. Therefore the design would contain at least \begin{align*} (q+1)^2+2=q^2+2q+3 \end{align*} blocks. Since \begin{align*} q^2+2q+3 > q^2+q+1 \end{align*} for $q \ge 2$, this contradicts $|\mathcal B|=q^2+q+1$. Thus every block other than $B_0$ meets $B_0$ at least once. Since the sum of all intersection sizes with $B_0$ is exactly the number of other blocks, each such intersection has size exactly $1$. Because $B_0$ was arbitrary, any two distinct blocks meet in exactly one point.[/guided]

[step:Verify the projective plane axioms for the design] Regard the elements of $X$ as points and the blocks $B \in \mathcal B$ as lines. The $2$-design axiom with $\lambda=1$ says that any two distinct points of $X$ lie on exactly one block, so there is a unique line through any two distinct points. The previous step says that any two distinct blocks meet in exactly one point, so any two distinct lines meet in exactly one point. Each line contains exactly $q+1$ points by the block-size parameter. It remains to verify the nondegeneracy axiom. Choose a block $B \in \mathcal B$. Since $q \ge 2$, the block $B$ has $q+1 \ge 3$ points; choose distinct points $x,y \in B$. Since \begin{align*} |X|-(q+1)=q^2>0, \end{align*} there exists a point $u \in X \setminus B$. Let $L_{xu}$ denote the unique block containing $x$ and $u$, and let $L_{yu}$ denote the unique block containing $y$ and $u$. Since $u \notin B$, neither $L_{xu}$ nor $L_{yu}$ equals $B$. By the previous step, $L_{xu} \cap B=\{x\}$ and $L_{yu} \cap B=\{y\}$. Outside $B$, the block $L_{xu}$ contains $u$ and $q-1$ further points, and the block $L_{yu}$ contains $u$ and $q-1$ further points. Hence the union \begin{align*} (L_{xu} \cup L_{yu}) \setminus B \end{align*} has at most \begin{align*} 1+2(q-1)=2q-1 \end{align*} points. Since $X \setminus B$ has $q^2$ points and \begin{align*} q^2-(2q-1)=(q-1)^2>0, \end{align*} there exists a point \begin{align*} v \in (X \setminus B)\setminus (L_{xu} \cup L_{yu}). \end{align*} We claim that $x,y,u,v$ are four points no three of which are collinear. The line through $x$ and $y$ is $B$, and neither $u$ nor $v$ lies on $B$. The point $v$ was chosen not to lie on $L_{xu}$ or $L_{yu}$, so neither $\{x,u,v\}$ nor $\{y,u,v\}$ is collinear. Finally, if $x$ lay on the line through $u$ and $v$, then that line would be the unique line through $x$ and $u$, namely $L_{xu}$, forcing $v \in L_{xu}$, contrary to the choice of $v$; the same argument with $y$ and $L_{yu}$ rules out $y$ lying on the line through $u$ and $v$. Thus no three of $x,y,u,v$ are collinear. Therefore the incidence structure is a projective plane of order $q$. [/step]