# Integer sets without singular matrices

Let’s say that a set $A\subset \mathbb N$ contains an $n\times n$ singular matrix if there exists an $n\times n$ matrix with determinant zero whose entries are distinct elements of $A$. For example, the set of prime numbers does not contain any $2\times 2$ singular matrices; however, for every $n\ge 3$ it contains infinitely many $n\times n$ 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 $k$.) 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 $A\subset \mathbb N$ that contains no singular matrices of any size. Here is one:
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