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\begin{document}

\title{\bf Block Designs with SDP Parameters}
\author{Harold N. Ward \\
\small Department of Mathematics\\[-0.8ex]
\small University of Virginia\\[-0.8ex]
\small Charlottesville, VA 22904\\[-0.8ex]
\small United States of America\\
\small \texttt{hnw@virginia.edu}\\[1ex]
\textsc{For Virginia Warfield}\\[-0.4ex]
\textsc{on her retirement from the}\\[-0.4ex]
\textsc{University of Washington}}
\date{\dateline{Apr 19, 2012}{Jul 27, 2012}{Aug 9, 2012}}

\maketitle

\begin{abstract}
It is traditional to call a quasi-symmetric design with certain parameters
an SDP design if the symmetric difference of two different blocks is either
a block or a block complement. In this note, we delete the requirements on
the parameters and demand just that the symmetric difference of two blocks
be a block, a block complement, or either the empty set or the whole point
set. We obtain the parameters of such designs and use the result to prove
Kantor's theorem on the parameters of a symmetric SDP design. A spin-off of
an exponential Diophantine equation considered by Ramanujan is at the core.
\end{abstract}

\section{GP designs\label{SectGP}}

For a $(v,k,\lambda )$ design $\mathcal{D}$, let $\mathcal{P}$ be the point
set, of size $v$, and let $\mathcal{B}$ be the block set, of size $b=\lambda
v(v-1)/(k(k-1))$. Each block has size $k$ and each point is in $r=\lambda
(v-1)/(k-1)$ blocks. If $\mathcal{D}$ is a symmetric design, $\mathcal{D}$
is said to be an SDP (Symmetric Difference Property) design when the
symmetric difference of any three blocks is either a block or a block
complement. We identify subsets of $\mathcal{P}$ with their binary
characteristic vectors, making the symmetric difference of subsets their
(binary) sum. The complement of an SDP symmetric design is also one.

W. M. Kantor proved that any symmetric SDP design (apart from the trivial $%
(2,1,0)$ design) has parameters%
\begin{equation}
v=q,\quad k=\frac{q}{2}+\delta \frac{\sqrt{q}}{2},\quad \lambda =\frac{q}{4}%
+\delta \frac{\sqrt{q}}{2},  \label{SDP}
\end{equation}%
where $q$ is an even power of 2 and $\delta =\pm 1$ \cite[Theorem 3]{K}.
Both the derived and residual designs, with respect to a block, have the
property that the sum of any two different blocks is either a block or a
block complement (Lemma \ref{LemmaDer Res}, ahead). The derived and residual
designs have parameters%
\begin{eqnarray}
\text{derived}\text{: } &&v=\frac{q}{2}+\delta \frac{\sqrt{q}}{2},\quad k=%
\frac{q}{4}+\delta \frac{\sqrt{q}}{2},\quad \lambda =\frac{q}{4}+\delta 
\frac{\sqrt{q}}{2}-1  \label{GP} \\
\text{residual}\text{: } &&v=\frac{q}{2}-\delta \frac{\sqrt{q}}{2},\quad k=%
\frac{q}{4},\quad \lambda =\frac{q}{4}+\delta \frac{\sqrt{q}}{2},  \notag
\end{eqnarray}%
again with $q$ an even power of 2. The designs of all of these parameter
sets have orders $r-\lambda =q/4$. The complements of these designs also
have the block-sum property. This collection of parameters also contains
those of the complements (recall that the complement of a derived [residual]
design is the residual [derived] design of the complement of the original).
The block-sum property can be rephrased to say that the set $\left\{
B,B^{\prime }|B\in \mathcal{B}\right\} \cup \left\{ \varnothing ,\mathcal{P}%
\right\} $, where $B^{\prime }$ is the complement $\mathcal{P}+B$ of $B$, is
a group (an elementary Abelian 2-group). We make the following definition:

\begin{definition}
A design is called a GP (group property) design if the set%
\begin{equation*}
\mathcal{G=}\left\{ B,B^{\prime }|B\in \mathcal{B}\right\} \cup \left\{
\varnothing ,\mathcal{P}\right\}
\end{equation*}
is a group under symmetric difference.
\end{definition}

\noindent The complement of a GP design is also a GP design. It is
traditional to call a non-symmetric design an SDP design if it is a GP
design and it (or its complement) has the parameters in (\ref{GP}) (see the
summary by V. D. Tonchev in \cite[VII.1.9]{CD}). The purpose of this note is
to show that any nonsymmetric GP design (or its complement) that is not a
Hadamard 3-design does have the parameters in (\ref{GP}). From that we can
infer Kantor's theorem for symmetric SDP designs. Part of the point of doing
this is that the proof of our result is an elementary number-theoretic
argument, and obtaining Kantor's theorem as a corollary avoids some of the
complexities of that theorem's proof. Moreover, the customary attendant
specification of the parameters for GP designs can almost be omitted.

Let $\mathcal{D}$ be a GP design. If $B_{1}+B_{2}=B_{3}$, with $B_{i}\in 
\mathcal{B}$, then as $\left\vert B_{1}+B_{2}\right\vert =2(k-\left\vert
B_{1}\cap B_{2}\right\vert )$, $\left\vert B_{1}\cap B_{2}\right\vert =k/2$.
If $B_{1}+B_{2}=B_{3}^{\prime }$ instead, then $\left\vert B_{1}\cap
B_{2}\right\vert =(3k-v)/2$. Suppose first that these two intersection
numbers are the same, which means that $v=2k$. Then $\mathcal{D}$ cannot be
symmetric, for if so, the second design equation 
\begin{equation}
r(k-1)=\lambda (v-1)  \label{DE}
\end{equation}%
would entail the impossibility $k(k-1)=\lambda (2k-1)$. The only other
conceivable intersection number is 0, so $\mathcal{D}$ must be a
quasi-symmetric design with intersection numbers $x=0$ and $y=k/2$ (thus the
complement of some block is also a block; that fact would imply that $v=2k$%
). Then \cite[Proposition 3.17]{SS} applies:%
\begin{equation}
(r-1)(y-1)=(k-1)(\lambda -1).  \label{SS}
\end{equation}%
Solving this along with (\ref{DE}) gives $r=2k-1$ and $\lambda =k-1$. As now 
$b=2v-2$, Theorem 5.8 of \cite{CvL} implies that $\mathcal{D}$ is either a
Hadamard 3-design or the unique $(6,3,2)$ design \cite[II Example 1.18]{CD}.
But this latter citation shows that the $(6,3,2)$ design does not qualify as
a GP design. The Hadamard matrices involved in the 3-designs must actually
be of Sylvester type because of the group property. (A Sylvester type
Hadamard matrix is one equivalent to a Kronecker power of $\left[ 
\begin{array}{cc}
1 & 1 \\ 
1 & -1%
\end{array}%
\right] $, or (equivalently!) to the character table of an elementary
Abelian 2-group; see \cite[Example 1.31]{CvL}, for instance.)

However, we can see all this identification rather more directly: if $B$ is
any block of $\mathcal{D}$, then on counting the blocks other than $B$ that
meet $B$ (a standard maneuver), we get $k(r-1)/y=4k-4=b-2$. The missing two
blocks can only be $B$ and $B^{\prime }$; so the complement of \emph{any}
block is also a block. If the group $\mathcal{G}$ for the GP property has
size $2q$, $q$ a power of 2, then the design parameters are%
\begin{equation}
v=q,\quad b=2q-2,\quad r=q-1,\quad k=q/2,\quad \lambda =q/2-1.  \label{H3D}
\end{equation}%
Moreover, if $B_{1}$ and $B_{2}$ are different blocks with $B_{2}\neq
B_{1}^{\prime }$, then $B_{1}+B_{2}$ must be a block. From this we infer
that $\mathcal{D}$ is a Hadamard 3-design based on a Sylvester matrix.
Another way to present the design is as the set of supports of the words of
weight $q/2$ in the first order Reed-Muller code $\mathcal{R}(1,m)$, where $%
q=2^{m}$ (see \cite{AK} for these codes). Thus we have

\begin{proposition}
\label{Propv2k}If $\mathcal{D}$ is a GP design for which $v=2k$, then $%
\mathcal{D}$ is a Hadamard 3-design corresponding to a Sylvester type
Hadamard matrix.
\end{proposition}

Now take $v\neq 2k$. Then no block complement is also a block, and the two
possible intersection numbers $k/2$ and $(3k-v)/2$ are different. Could $%
\mathcal{D}$ be symmetric? If so, then $\lambda =k/2$ or $(3k-v)/2$. Say
that $\lambda =k/2$. Then $k(k-1)=k/2\times (v-1)$ gives $v=2k-1$. Hence $%
\mathcal{D}$ is a Hadamard 2-design, and the group property again implies
that the corresponding Hadamard matrix is of Sylvester type. If $\lambda
=(3k-v)/2$, then $v=2k+1$ and $\lambda =(k-1)/2$. This is also a Hadamard
2-design, with Sylvester matrix. Thus

\begin{proposition}
\label{PropSymmSD2}If $\mathcal{D}$ is a symmetric GP design, then $\mathcal{%
D}$ is a Hadamard 2-design, and again the corresponding Hadamard matrix is
of Sylvester type.
\end{proposition}

\noindent The design with $v=2k-1$ can also be realized as the set of
supports of the words of weight $q/2$ in the punctured first-order
Reed-Muller code $\mathcal{R}(1,m)^{\ast }$, $q=2^{m}$. Incidentally, the
sum of two different blocks is a block; but for $k>1$, there are three
blocks whose sum is $\varnothing $. Thus the design is not an SDP design--as
it better well not be!

There is one more special case: it could be that the intersection number $%
(3k-v)/2$ is 0, that is, that $v=3k$. Now when we solve (\ref{SS}) and the
design equations we get $k=3-8/(r+3)$. The possibilities are $r=1$ and $5$,
giving the trivial $(3,1,0)$ design and the $(6,2,1)$ design whose blocks
are the 2-subsets of a 6-set. The parameters are those of (\ref{GP}) with $%
q=4$ and $\delta =-1$ in the residual set and with $q=16$ and $\delta =-1$
in the derived set.

Thus finally we may assume that $\mathcal{D}$ is a quasi-symmetric design
with intersection numbers $x=(3k-v)/2$ and $y=k/2$, $x$ and $y$ different ($%
\mathcal{D}$ is \emph{proper}) and both positive. We may also take $k<v/2$
by replacing $\mathcal{D}$ with its complement, if necessary. Let the order
of $\mathcal{G}$ be $2q$, $q$ a power of 2 as before, so that now $b=q-1$,
since no block complement is a block. The equation generalizing (\ref{SS}) is%
\begin{equation*}
k(r-1)(x+y-1)+xy(1-b)=k(k-1)(\lambda -1)
\end{equation*}%
\cite[Lemma 3.23(i)]{SS}. This becomes (after a cancellation of $k$)%
\begin{equation*}
(r-1)(2k-\frac{v}{2}-1)-(q-2)\frac{3k-v}{4}=(k-1)(\lambda -1).
\end{equation*}%
Solving it with the design equations gives%
\begin{equation*}
r=\frac{k(v-1)}{v-(v-2k)^{2}},\quad \lambda =\frac{k(k-1)}{v-(v-2k)^{2}}%
,\quad q=\frac{4k(v-k)}{v-(v-2k)^{2}}.
\end{equation*}%
Substituting $v-(v-2k)^{2}=4k(v-k)/q$ from the third equation into the other
two, we get 
\begin{equation*}
\lambda =\frac{q(k-1)}{4(v-k)},\quad r=\frac{q(v-1)}{4(v-k)}.
\end{equation*}%
Now by \cite[Corollary 3.9]{SS}, $y-x$ divides both $k-x$ and $r-\lambda $.
Here%
\begin{equation*}
y-x=\frac{v-2k}{2},\quad k-x=\frac{v-k}{2},\quad r-\lambda =\frac{q}{4}.
\end{equation*}%
Thus $(v-2k)$ divides $q/2$, so that $v-2k=q_{0}$, $q_{0}$ also a power of
2, with $q_{0}<q$ (we have assumed that $k<v/2$). Then $v=q_{0}+2k$ and the
equation for $q$ is 
\begin{equation*}
q=\frac{4k(q_{0}+k)}{q_{0}+2k-q_{0}^{2}}.
\end{equation*}%
Thus%
\begin{equation*}
4k^{2}+(4q_{0}-2q)k+qq_{0}^{2}-qq_{0}=0,
\end{equation*}%
making%
\begin{equation}
k=\frac{q}{4}-\frac{q_{0}}{2}+\delta \frac{q_{0}}{2}\sqrt{\left( \frac{q}{%
2q_{0}}\right) ^{2}-q+1},  \label{keq}
\end{equation}%
$\delta =\pm 1$. So it must be that%
\begin{equation}
\left( \frac{q}{2q_{0}}\right) ^{2}-q+1=z^{2}  \label{discr}
\end{equation}%
for some integer $z$.

Equation (\ref{discr}) is a particular case of the exponential Diophantine
equation%
\begin{equation*}
2^{N}\pm 2^{M}\pm 2^{L}=z^{2},
\end{equation*}%
$N,M,L$, and $z$ being integers (in the notation of \cite{S}, which gives
some of the equation's background, along with references). The prototype is $%
2^{N}-2^{3}+1=z^{2}$, conjectured by S. Ramanujan in 1913 to have solutions
just for $N=3,4,5,7,15$. This was proved to be true by T. Nagell in 1948.
For $2^{N}-2^{M}+1=z^{2}$, one has the trivial solutions $N=2t,M=0,z=2^{t}$, 
$t\geq 0$; the cases $N=5,7,15$ in Ramanujan's equation ($M=3$); and two
parameterized families%
\begin{eqnarray*}
N &=&2t,\quad M=t+1,\quad z=2^{t}-1,\quad t\geq 2 \\
N &=&M=t,\quad z=1,\quad t\geq 1
\end{eqnarray*}%
\cite[Theorem 2]{S}. We shall give a direct proof for the case relevant to
the designs in Lemma \ref{LemmaDioph}.

For $\mathcal{D}$, $2^{N}=(q/2q_{0})^{2}$ and $2^{M}=q$. We want $q>1$, of
course, so the possibilities are%
\begin{eqnarray*}
\frac{q}{2q_{0}} &=&\frac{q}{2}=2^{t},\quad \text{with }z=\frac{q}{2}-1, \\
\frac{q^{2}}{4q_{0}^{2}} &=&q,\quad \text{with }z=1
\end{eqnarray*}%
(the Ramanujan solutions are out because $N$ must be even). The first makes $%
q_{0}=1$. Then $k=q/4-1/2+\delta (q/2-1)/2$. Only $\delta =1$ works, giving $%
k=q/2-1$ and $v=q-1$. But then $\mathcal{D}$ is a symmetric design and
already dealt with (also excluded here by $x\neq y$). The second has $q_{0}=%
\sqrt{q}/2$, $q$ now being an even power of 2. Then equation (\ref{keq})
becomes 
\begin{equation*}
k=\frac{q}{4}-\frac{\sqrt{q}}{4}+\delta \frac{\sqrt{q}}{4}=\frac{q}{4}\text{
or\ }\frac{q}{4}-\frac{\sqrt{q}}{2}.
\end{equation*}%
This, with complements, produces the parameters of (\ref{GP}).

\begin{lemma}
\label{LemmaDioph}Suppose that $X$ and $Y$ are two powers of 2 for which $%
X^{2}-Y+1=Z^{2}$, with $Z>0$. Then one of the following holds:%
\begin{eqnarray*}
Y &=&1\text{ and }Z=X \\
Y &=&2X\text{ and }Z=X-1 \\
Y &=&X^{2}\text{ and }Z=1.
\end{eqnarray*}
\end{lemma}

\begin{proof}
Rewrite the equation as $(X+Z)(X-Z)=Y-1$. Certainly $Z\leq X$, so write $%
Z=X-A$, with $0\leq A<X$. Then we need $A(2X-A)=Y-1$. If $A=0$, we have $Y=1$
and $Z=X$. If $A=1$, then $Y=2X$ and $Z=X-1$. In addition, $A\leq Y-1$; but
if $A=Y-1$, then $Y=2X$ again, but now $Z=1-X\leq 0$. Thus we may assume
that $1<A<Y-1$ and $2X-A<Y-1$, wherewith $X<Y-1$. Then $2X\leq Y$, as $X$
and $Y$ are powers of 2. Since $Y=2X$ has been covered, we take $2X<Y$. Let $%
B=2X-A$. As $A>1$, so $X\geq 2$; both $Z$ and $A$ are odd.

Now let $A=2A^{\prime }-\Delta $, with $\Delta =\pm 1$ chosen to make $%
A^{\prime }$ odd. Then 
\begin{equation*}
(A+\Delta )(B+\Delta )=AB+\Delta (A+B)+1
\end{equation*}%
implies that%
\begin{eqnarray*}
2A^{\prime }(B+\Delta ) &=&Y+2\Delta X \\
&=&2X(\frac{Y}{2X}+\Delta ).
\end{eqnarray*}%
The second factor here is odd since $2X<Y$. Thus $X$ is the exact power of 2
dividing $B+\Delta $, and $B=XB^{\prime }-\Delta $, with $B^{\prime }$ odd.
Then%
\begin{equation*}
2X=A+B=2A^{\prime }+XB^{\prime }-2\Delta ,
\end{equation*}%
and 
\begin{equation*}
\Delta =A^{\prime }+\frac{X}{2}(B^{\prime }-2).
\end{equation*}%
Therefore $X/2(B^{\prime }-2)=\Delta -A^{\prime }\leq 0$, and it can only be
that $B^{\prime }=1$. Thus $B=X-\Delta $, $A=X+\Delta $, and $Y-1=AB=X^{2}-1$%
, giving the third possibility, $Y=X^{2}$.
\end{proof}

The grand summary is then

\begin{proposition}
\label{PropGP3}If $\mathcal{D}$ is a $(v,k,\lambda )$ GP design with $v\neq
2k$ that is not symmetric, then $(v,k,\lambda )$ is one of the triples given
in (\ref{GP}).
\end{proposition}

\section{SDP designs}

In this section, we present a proof of Kantor's theorem giving the
parameters of (symmetric) SDP designs. First we verify that the derived and
residual designs of an SDP design are GP designs.

\begin{lemma}
\label{LemmaDer Res}If $\mathcal{D=(P},\mathcal{B})$ is an SDP design and $D$
is a block of $\mathcal{D}$, then both the derived and residual designs of $%
\mathcal{D}$ with respect to $D$ are GP designs.
\end{lemma}

\begin{proof}
For $X,Y\subseteq \mathcal{P}$, the characteristic function of $X\cap Y$ is
the component-wise product $XY$. If $B_{1},B_{2}\in \mathcal{B}$, then $%
B_{1}+B_{2}+D$ is either a block $B$ or its complement $B^{\prime }$. That
is, $(B_{1}+B_{2}+D)D=BD$ or $B^{\prime }D$. Hence $B_{1}D+B_{2}D=BD+D$ or $%
BD$. Thus the sum of two blocks of the derived design, if not $\varnothing $%
, is either a block or its complement. So the derived design is a GP design.
The proof for the residual is similar.
\end{proof}

Now let $\mathcal{D=(P},\mathcal{B)}$ be a $(v,k,\lambda )$ SDP design, so
that the parameters of a residual design $\mathcal{D}^{\prime }$ are%
\begin{equation}
v^{\prime }=v-k,\quad b^{\prime }=v-1,\quad r^{\prime }=k,\quad k^{\prime
}=k-\lambda ,\quad \lambda ^{\prime }=\lambda  \label{Res}
\end{equation}%
(we can safely assume that $v>2$). By Lemma \ref{LemmaDer Res}, these must
be the parameters of a GP design. We run through the possibilities presented
in the propositions in Section \ref{SectGP}. The only way $\mathcal{D}%
^{\prime }$ can be a symmetric design (Proposition \ref{PropSymmSD2}) is
that $k=1$. Then SDP requires $\mathcal{D}$ to be the excluded trivial $%
(2,1,0)$ design or the equally trivial $(3,1,0)$ design we encountered
before. Suppose then that $v^{\prime }=2k^{\prime }$ (Proposition \ref%
{Propv2k}), so that $\mathcal{D}^{\prime }$ is a Hadamard 3-design with
parameters given by (\ref{H3D}) for some $q$ (this will be true then for 
\emph{any} residual design). Then $v=2q-1$, $k=q-1$, $\lambda =q/2-1$, and $%
\mathcal{D}$ is a Hadamard 2-design. The key point here is that if $D\in 
\mathcal{B}$, then in the residual with respect to $D$, each block
complement is also a block. Let $B_{1},B_{2}\in \mathcal{B}$ be such that
the residual blocks $B_{1}D^{\prime }$ and $B_{2}D^{\prime }$ are
complements. Since $\left\vert B_{1}B_{2}\right\vert $, $\left\vert
B_{1}D\right\vert $, and $\left\vert B_{2}D\right\vert $ are all $q/2-1$,
and $B_{1}D^{\prime }$ and $B_{2}D^{\prime }$ are disjoint, it can only be
that $B_{1}D=B_{2}D$. That makes 
\begin{eqnarray*}
B_{1}+B_{2}+D &=&(B_{1}+B_{2}+D)D^{\prime }+(B_{1}+B_{2}+D)D \\
&=&B_{1}D^{\prime }+B_{2}D^{\prime }+DD^{\prime }+B_{1}D+B_{2}D+DD \\
&=&D^{\prime }+D=\mathcal{P}.
\end{eqnarray*}%
But this gainsays SDP.

Thus we are left with $\mathcal{D}^{\prime }$ being described by Proposition %
\ref{PropGP3}. Only the parameters in the residual list of (\ref{GP}) are
quasi-residual ($\lambda v=k(k+\lambda -1)$), so the values in (\ref{Res})
would match a triple in that list. That is,%
\begin{equation*}
v-k=v^{\prime }=\frac{q}{2}-\delta \frac{\sqrt{q}}{2},\quad k-\lambda
=k^{\prime }=\frac{q}{4},\quad \lambda =\lambda ^{\prime }=\frac{q}{4}%
+\delta \frac{\sqrt{q}}{2}.
\end{equation*}%
So%
\begin{equation*}
v=q,\quad k=\frac{q}{2}+\delta \frac{\sqrt{q}}{2},\quad \lambda =\frac{q}{4}%
+\delta \frac{\sqrt{q}}{2},
\end{equation*}%
a parameter triple from (\ref{SDP}).

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UK, 1992.

\bibitem[CvL]{CvL} P. J. Cameron and J. H. van Lint, Designs, Graphs, Codes
and their Links,\ London Math. Soc. Student Texts \textbf{22}, Cambridge
University Press, Cambridge, UK, 1991.

\bibitem[CD]{CD} Handbook of Combinatorial Designs,\ second edition, C. J.
Colbourn and J. H. Dinitz, editors, Chapman and Hall/CRC, Boca Raton, FL,
2007.

\bibitem[K]{K} W. M. Kantor, Symplectic groups, symmetric designs, and line
ovals, \emph{J. Algebra }\textbf{33} (1975), 43--58.

\bibitem[SS]{SS} M. S. Shrikhande and S. S. Sane, Quasi-Symmetric Designs,\
London Math. Soc. Lecture Note Series \textbf{164}, Cambridge University
Press, Cambridge, UK, 1991.

\bibitem[S]{S} L. Szalay, The equations $2^{N}\pm 2^{M}\pm 2^{L}=z^{2}$, 
\emph{Indag. Math. (N.S.)} \textbf{13}(1) (2002), 131--142.
\end{thebibliography}

\end{document}