Building a 3D matrix of positive integers I'm trying to build a 3D matrix made up of positive integers that has very specific properties.
The matrix dimensions are $N \times N \times (N+1)$ where $N$ is a positive integer. The matrix has two properties:


*

*Every one of the $(N+1)$ "slices" of size $N \times N$  of the matrix contains each of the numbers $1$ through $N^2$ exactly once.

*Assuming we look at each such slice as made up of rows and columns, then if we pick any two rows belonging to different slices, they have to have exactly one number in common (due to property 1, it might be sufficient to impose that the two rows have at most one number in common and still get property 2).


Any ideas on whether this is easy or hard to achieve, and in the former case, what would be a way to achieve it?
 A: We start from a simple counting remark which will show us the way.  The properties of the matrix imply that any pair of different numbers can belong to at most one of the slice rows. Since there are ${N^2\choose 2}$ such pairs, each of $N(N+1)$ slice rows contains ${N\choose 2}$ such pairs, and ${N^2\choose 2}= N(N+1) {N\choose 2}$, we see that the set of slice rows constitutes the set of blocks for a Steiner system $S(2,N, N^2)$. Conversely, given such a Steiner system, we can construct from it the required matrix, provided we manage to group the blocks into slices such that each slice consists of $N$ blocks and 
contains each of $N^2$ numbers exactly once. We use an explicit construction of a Steiner system $S(2,N, N^2)$ as a finite affine plane.
Let $F$ be a finite field of $N$ elements. As  elements of the Steiner system we take elements of $F^2=F\times F$. We index blocks by triples $(a,b,c)$ from $F^3$ such that at least one of $a$ and $b$ is non-zero. Namely, a triple $(a,b,c)$ indexes a block $B(a,b,c)=\{(x,y)\in F^2: ax+by+c=0\}$. Remark, that triples $(a,b,c)$ and $(a’,b’,c’)$ index the same block iff there exists a non-zero $\lambda\in F$ such that $(a,b,c)=\lambda (a’,b’,c’)$. Using that $F$ is a field, we can easily check that each pair of elements of $F^2$ belongs to exactly one of the blocks, that is we indeed constructed a Steiner system $S(2,N, N^2)$. It remains to group the blocks to slices $S_i$, which we can do 
putting $S_t=\{B(t,1,c):c\in F\}$ for each $t\in F$ and $S_q=\{B(1,0,c):c\in F\}$.
The problem is that a finite field of $N$ elements exists iff $N$ is a power of a prime number (see, for instance [vdW, $\S$ 43]). So I have asked whether there exists a Steiner system $S(2, N, N^2)$ for other $N$, and it turned out to be a notorious unsolved problem in combinatorics. But it is already known that the answer is negative for $N=6$ and $N=10$. So I guess that I have to finish my answer at this point. 
References
[vdW] B. L. van der Waerden, Algebra (Russian edition).
