Bruck–Ryser–Chowla theorem

Remarks

Proof idea: the incidence matrix $N$ of the design is a square matrix with $N N^T = (k - \lambda) I + \lambda J$, where $J$ is the all-ones matrix. For $v$ even, taking determinants gives $\det(N)^2 = k^2 (k - \lambda)^{v - 1}$, and since $v - 1$ is odd, $k - \lambda$ must be a square. For $v$ odd, the identity says that two quadratic forms are equivalent over $\Q$; Lagrange's four-square theorem and a variable-by-variable reduction turn this into the stated equation. For a finite projective plane of order $n$, take $v = n^2 + n + 1$, $k = n + 1$ and $\lambda = 1$. If $n \equiv 1$ or $2 \pmod 4$, then $(v-1)/2 = n(n+1)/2$ is odd, and the condition becomes: $n$ is a sum of two squares. Example: $6 \equiv 2 \pmod 4$ and $6$ is not a sum of two squares, so there is no projective plane of order $6$. The theorem gives only a necessary condition: $10 = 1^2 + 3^2$ passes the test, yet there is no plane of order $10$.

Going deeper

For a student

The theorem is the main known non-existence result for symmetric designs and projective planes. It refines Fisher's inequality, whose equality case defines symmetric designs, by looking at the incidence matrix over the rationals instead of only its rank. The open question of whether every projective plane has prime power order lies beyond its reach.