[[UPDATE: This work has now been published at SIAM J Discrete Math.: Formulae for the Alon–Tarsi Conjecture.]]

By equating two formulae (one congruence by Glynn (1) (which has just appeared) and one unpublished formula) for the number of even Latin squares minus the number of odd Latin squares, we find the following result.

For odd primes $p$ we have \[\sum_{A \in B} (-1)^{\sigma_0(A)} \equiv 1 \pmod p\] where $B$ is the set of $(p-1) \times (p-1)$ $\\,(0,1)$-matrices whose determinant is indivisible by $p$ and $\sigma_0(A)$ is the number of zeroes in $A$. It happens to be true for $p=2$ also (but it does not follow from Glynn's result).

Is this result already known? If so, it would provide an alternate proof of Glynn's result.

To illustrate, consider when $p=3$. The (0,1)-matrices whose determinants are indivisible by $p$ are

```
01 10 01 10 11 11
10 01 and 11 11 01 10
```

So the sum becomes $+2-4=-2 \equiv 1 \pmod 3$.

It is equivalent to the congruence \[\sum_{A \in C} (-1)^{\sigma_0(A)} \det(A)^{p-1} \equiv 1 \pmod p\] where $C$ is the set of all $(p-1) \times (p-1)$ $\\,(0,1)$-matrices (via Fermat's Little Theorem).

(1) Glynn, D., 2010. The conjectures of Alon-Tarsi and Rota in dimension prime minus one. SIAM J. Discrete Math., 24 (2010), 394-399.