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Statement

Setting (p. 88). A projective plane geometry is a system of points and of sets of points called lines (the print says "at least two in number"), such that two distinct points lie on a unique common line, two distinct lines have a unique common point, and every line has at least three points. It is finite when it has finitely many points. A finite plane has a positive integer $N$ such that every line has exactly $N+1$ points and every point lies on exactly $N+1$ lines; it then has $N^2+N+1$ points and $N^2+N+1$ lines (the paper cites this, its references [3], [6], [13]). The corpus calls this $N$ the order of the plane; the paper speaks of $N+1$ points on a line.

Theorem 1 (p. 88, quoted). "If $N\equiv1$ or $2$ mod $4$ and if the square free part of $N$ contains at least one prime factor of the form $4k+3$, then there does not exist a finite projective plane geometry with $N+1$ points on a line."

The squarefree part of $N$ is the product of the primes that divide $N$ to an odd power. By the two-squares theorem, its having a prime factor $\equiv3\pmod4$ is the same as $N$ not being a sum of two integer squares, so the theorem says: a plane of order $N\equiv1,2\pmod4$ exists only if $N=x^2+y^2$ for some integers $x,y$. That reformulation is the corpus's; the paper states only the squarefree-part form.

Consequences stated in the paper (p. 88). In particular no plane exists for $N=2p$ with $p$ a prime of the form $4k+3$. Since a plane with $N+1$ points on a line can be built from a complete set of mutually orthogonal Latin squares of order $N\ge3$ (the paper cites its references [1], [8]), for every $N$ covered by Theorem 1 there is no complete set of mutually orthogonal Latin squares of order $N$.

Postscript (b) (pp. 92--93). The authors note that Euler's 1782 conjecture, that no pair of orthogonal Latin squares of order $N$ exists when $N$ has the form $4k+2$, would, if true, give the nonexistence of planes with $N\equiv2\pmod4$, and so imply and improve one half of Theorem 1; they cite MacNeish's claimed proof of the conjecture and record that its correctness has been questioned.

Proof pointer

Section 4 (pp. 91--92), resting on Theorems 2 and 3 (section 2) and on the theory of rational congruence of quadratic forms recalled in section 3 (pp. 89--91: the Hilbert norm-residue symbol, the invariant $c_p$, and the Minkowski--Hasse theorem, Theorem 6, which the paper cites and does not prove). Let $B$ be the matrix of order $n=N^2+N+1$ with $N+1$ on the diagonal and $1$ elsewhere. The paper computes, for every odd prime $p$, $c_p(B)=(-1,N)_p^{N(N+1)/2}$, its equation (E). If a plane exists, Theorem 2 gives $B=A^TA$ with $A$ rational and nonsingular, so $B$ is rationally congruent to the identity and $c_p(B)=c_p(I)=+1$ for every odd $p$. When $N\equiv1,2\pmod4$ the exponent $N(N+1)/2$ is odd, and a prime $p\equiv3 \pmod4$ dividing the squarefree part of $N$ gives $(-1,N)_p=-1$, a contradiction. Postscript (a) (p. 92) records Marshall Hall's remark that $B$ is rationally congruent to the diagonal matrix $(1,N,\ldots,N)$, which gives a simpler route to (E).

Read depth. Claims checked: the setting, Theorem 1, the consequences on p. 88 and the postscript were read clause by clause on the page images of the print, and the proof in section 4 was followed in outline. The norm-residue facts (Theorems 4 and 5) and the Minkowski--Hasse theorem (Theorem 6) are cited by the paper and were not checked. Nothing here is independently reviewed.

Dependencies

Theorem 2 (a plane gives an incidence matrix satisfying (M)). External inputs named by the paper: Hilbert's norm-residue symbol (its reference [5]) and the Minkowski--Hasse theorem (its references [4], [9]).

Source. R. H. Bruck and H. J. Ryser, The nonexistence of certain finite projective planes, Canad. J. Math. 1 (1949), 88--93, doi:10.4153/CJM-1949-009-2; the edition read is named on the source card.

Bears on

• Problem 723: the problem asks whether every finite projective plane has prime-power order. Theorem 1 excludes every order $N\equiv1,2\pmod4$ whose squarefree part has a prime factor $\equiv3\pmod4$, among them $6$, $14$, $21$ and $22$ (the values are checked here; the paper names only the family $N=2p$). No prime power is among the excluded orders. The theorem excludes no order $N\equiv0,3 \pmod4$, such as $12$, and no order that is a sum of two squares, such as $10$, so it does not settle the problem.