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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 $t$?2012-02-29
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    To answer your question: v = 4t - 12012-03-01

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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?

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    Yes it was suppose to be the union. I understand what you are showing me bit I am still unclear about what exactly a Paley set is constructing.2012-03-01
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    See Proposition 5 of http://www4.ncsu.edu/~singer/437/proj12.pdf. It doesn't use the name Paley, but I'm pretty sure that's what it's about. If you don't know about quadratic residues, the pages leading up to Proposition 5 should help. Come back if you have more questions after reading.2012-03-02
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    I really would like to understand this as well; specifically, why when you look at the numbers $d-0$ and $0-d$ for $d$ in $D$ do you get each nonzero number exactly once?2018-03-18
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    @Dan, I refer you to the last two sentences of my answer.2018-03-18
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    Well if $p^n\equiv 3$ (mod 4), then -1 is nonquadratic. Also, I think that every nonzero nonquadratic element is the negative of a quadratic element. So does this imply that $d\neq -d$ (by starting with $1-0\neq 0-1$ and multiplying by each nonzero number)? And since $D\cup \{0\}$ gives rise to $4t-2$ new differences (each of which must be unique from above), it follows that $d-0$ and $0-d$ for $d\in D$ produces each nonzero number exactly once. Is that how it goes? Is there something big that I'm missing?2018-03-18
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    You got it. For each nonzero $d$, exactly one of $d$ and $-d$ is a quadratic residue.2018-03-19
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    Okay, here's what I've come up with: Since $v$ is a prime power congruent to 3 (mod 4), $-1$ is nonquadratic. Hence, given some $d\in D$, we have $0-d=(-1)d=-d$ is nonquadratic if $d$ is quadratic; i.e. $d$ is quadratic iff $-d$ is nonquadratic. Hence, $d\neq-d \implies d-0\neq 0-d$ for all $d\in D$. Now, appending $\{0\}$ to $D$ gives precisely $2k=2(2t-1)=4t-2$ new differences. Since $d\neq -d$ for all $d\in D$, it follows that the $4t-2$ must produce precisely one of each nonzero number of $\mathbb{Z}_v$. Since $\lambda=t-1$ for $D$, $\lambda = (t-1)+1=t$ for $D\cup\{0\}$.2018-03-19
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    Oh I didn't see your confirmation above. Thanks a lot Gerry!2018-03-19