# Combinatorial design difference set proof

Suppose $D$ is the $(4t-1,2t-1,t-1)$-difference set obtained from the following theorem:

Suppose $v$ is a prime power congruent to $3\bmod 4$. Then there is a $(4t-1,2t-1,t-1)$-difference set (also called a Paley difference set).

For some prime power $4t-1$: Prove that $D \cap \{0\}$ is a $(4t-1,2t,t)$-difference set.

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How does $v$ relate to $t$? – Thomas Andrews Feb 29 '12 at 19:55
To answer your question: v = 4t - 1 – Sarah Martin Mar 1 '12 at 16:08

I think you mean $D\cup\{{0\}}$, not $D\cap\{{0\}}$.
Let's look at an example, $t=2$. Then if I understand your notation, we are working modulo $4t-1=7$, and we have a set with $2t-1=3$ elements, and every non-zero number comes up $t-1=1$ time as a difference. An example would be the set $D=\{{1,2,4\}}$, where we get the differences $1=2-1$, $2=4-2$, $3=4-1$, $4=1-4$, $5=2-4$, and $6=1-2$.
Now let's take the union with zero, $\{{0,1,2,4\}}$. The size of the set is now $2t=4$, and we get the new differences $1=1-0$, $2=2-0$, $3=0-4$, $4=4-0$, $5=0-2$, and $6=0-1$, so now every nonzero number comes up $t=2$ times as a difference.
So the thing you have to understand is why when you look at the numbers $d-0$ and $0-d$ for $d$ in $D$ you get each nonzero number exactly once. That has something to do with the special nature of prime powers congruent to $3$ modulo $4$, and the special nature of the Paley difference sets. So, do you understand how the Paley sets are constructed?