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Construct a (13,26,6,3,1) BIBD.

I think there may be some proposition I should use involving prime numbers but I am having a hard time finding it.

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    You either spend a few nights thinking about it, or you do a quick internet search, I suppose. See my answer.2012-01-31

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Wikipedia is always a good start for a search.

You will find out that the $2-(13,3,1)$ design you are looking for a is Steiner triple system, denoted by S(2,3,13) = STS(13). According to wikipedia there are two non-isomorphic such designs and both can be found in the database of t-designs.

According to Mathworld Wolfram one of these designs is a finite hyperbolic plane, but I wouldn't know about that.

(edit) Actually, here is a book, describing constructions for these Steiner triples --- and some discussion showing it it not embeddable in a (Desarguesian) projective plane.

(edit2) As you, yourself, point out in the comments below, this problem can actually be solved by considering difference sets. This explains the construction from the book, by noting that $\{1,2,5\}$ and $\{1,6,8\}$ is a difference family for $\mathbb Z/13\mathbb Z$ with $s=2$. I'm guessing that's what you should be looking for, after noticing that a difference set won't do the trick, because the parameters don't match, i.e. they don't satisfy $k^2-k = \lambda (v-1)$.

$ \begin{array}{c|ccc} - &1&2&5\\ \hline 1&0&1&4\\2&12&0&3\\5&9&10&0\end{array} \text{ and } \begin{array}{c|ccc} - &1&6&8\\ \hline 1&0&5&7\\6&8&0&2\\8&6&11&0\end{array} $

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    @Jackson I have expanded my answer a bit. But if this is not what you are looking for, then you should edit your question and try to be (a lot) more specific.2012-01-31