Riddle: Assigning Students into Groups Suppose you had a classroom with 25 students. You want to assign 6 homework assignments over the course of the term and for each of these assignments students will work in groups of 5. But you want to do it so that no two students work in the same group for two different assignments. Is this possible, and if so how?
I worked it out for the case of 25 students into groups of 5 and (I believe) $m^2$ students grouped into groups of $m$ if $m$ is a prime power.
But these aren't all the possibilities. The conditions for situations that work are clearly if you have $n$ people put into groups of size $k$ you should have that $k$ divides n. But if you want it to work out so that you can have everyone work with everyone else exactly once with constant group sizes it should be the case $k-1$ divides $n-1$ since each student works with $k-1$ new students each time and they have a total of $n-1$ students they need to eventually work with. It turns out that the numbers n that satisfy this are $k + s k(k-1)$ for any nonnegative integer $s$. So, $s = 0$ is trivial and $s = 1$ corresponds to the squares.
So my question is: is it necessarily possible to solve this when $s > 1$. That is:

Is it possible to take a class with $k+s k(k-1)$ students, group them into groups of size $k$ for a series of assignments so that everyone works with everyone else exactly one time?
Also, what about cases where there isn't a prime power number of students?

 A: The case $s=1$ with $n=k^2$ students given $k+1$ homework assigments in groups of $k$ is directly related to the problem of finding $k-1$ mutually orthogonal Latin squares of order $k$. From $m$ homework assignments to $k$ groups of $k$ students each, we can construct $m-2$ mutually orthogonal Latin squares of order $k$. Let $a_{ij}$ be the number of the group in which student $i$ works on the $j$-th assignment. Associate each student $j$ with the entry $(a_{i1},a_{i2})$ of the Latin squares to be constructed, that is, the student's group in the first assignment determines the row and the group in the second assignment determines the column. Since no two students work in the same group twice, this establishes a bijection between the students and the entries. Now use all remaining assignments to construct one Latin square each, by setting the entry $(a_{i1},a_{i2})$ to $a_{ij}$ (for $j=3,\ldots,n$). Since the students in one group for assignment $j$ are in different groups for assignment $1$, each number is only written once per row, and since they're in different groups for assignment $2$, each number is only written once per column, so these are indeed Latin squares. And since they're in different groups for any assignment $j'\neq j$, two entries can't coincide both in $j$ and in $j'$, so the Latin squares are pairwise orthogonal.
The thirty-six officers problem proposed by Euler in $1782$ asks for a pair of orthogonal Latin squares of order six. Tarry ($1901$) proved through extensive casework that no such pair exists, and Stinson ($1984$) and Burger, Kidd, and van Vuuren ($2011$) have since offered shorter proofs; publicly accessible links to all three proofs are provided in answers to Orthogonal Latin Square 6*6.
The non-existence of a pair of orthogonal Latin squares of order $6$ implies that not only can you not assign $36$ students to $6$ groups of $6$ for $7$ homework assignments in the desired manner; you cannot even do it for $4$ homework assignments.
This is the only result that I could find that gives a negative answer to your question. For all other cases, i.e., $s=1$ with $k\gt6$ and $s\gt1$ with any $k$, the published results are compatible with the existence of such assignments. The Handbook of Combinatorial Designs contains a table with lower bounds on the number of mutually orthogonal Latin squares of given order. In most cases, the lower bound is quite far from the upper bound $n-1$ that you need for your assignments, i.e. there may be such assignments but so far none have been found. OEIS sequence A001438 gives known maximal numbers of mutually orthogonal Latin squares, but they are only known up to $n=9$, so there's no new information there since all numbers up to $9$ except for $6$ are prime powers and thus allow $n-1$ squares. Harvey had a results page (now archived) for the social golfer problem, but the results seem rather weak – the lower bounds on the diagonal don't take into account the prime-power construction; the upper bounds below the diagonal (corresponding to your $s\gt1$) are all just the trivial upper bounds $\left\lfloor\frac{n-1}{k-1}\right\rfloor$, and the lower bounds are mostly below the upper bounds. The only case for $s\gt1$ and $k\gt2$ for which the lower bound is equal to the upper bound (so that a solution to your problem is known to exist) is $s=2$, $k=3$, $n=15$, for which $7$ assignments can be given to $5$ groups of $3$. Thus, little seems to be known about the existence of solutions for $s\gt1$.
