## Overlapping latin subsquares and full products

We derive necessary and sufficient conditions for there to exist a latin square
of order n containing two subsquares of order a and b that
intersect in a subsquare of order c. We also solve the case
of two disjoint subsquares. We use these results to show that:
- A latin square of order n cannot have more than
n{n choose h} / m{m choose h} subsquares of order m, where
h≥(m+1)/2. Indeed, the number of subsquares of order m
is bounded by a polynomial of degree at most √(2m)+2 in n.
- For all n≥5 there exists a loop of order n in which every
element can be obtained as a product of all n elements in some order
and with some bracketing.

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