Let’s say that a set *contains an singular matrix* if there exists an matrix with determinant zero whose entries are distinct elements of . For example, the set of prime numbers does not contain any singular matrices; however, for every it contains infinitely many singular matrices.

I don’t know of an elementary proof of the latter fact. By a 1939 theorem of van der Corput, the set of primes contains infinitely many progressions of length 3. (Much more recently, Green and Tao replaced 3 with an arbitrary .) If every row of a matrix begins with a 3-term arithmetic progression, the matrix is singular.

In the opposite direction, one may want to see an example of an infinite set that contains **no** singular matrices of any size. Here is one:

Let : these numbers are 1 less than Fermat’s should-have-been-primes. Suppose that an matrix is filled with distinct elements of . Then

where each product over is a power of 2. Moreover, these powers of 2 are distinct by the uniqueness of binary representation of integers. It follows that the sum over is nonzero — again, by the uniqueness of binary representation.

This set is very sparse, because the sequence grows super-exponentially. Is there such an example with exponential growth? or even with polynomial growth?

**Disclaimer**: This being a weekend blog post (and not a research proposal), I did not study the literature.

**Credit**: Once again, to Colin Carroll for “Prime Matrices I,II”.