The fewest and the most toroidal conflicts of an n-queens solution

A solution of the n-queens problem places n non-attacking queens on an n × n board. Viewed on the torus, two of its queens may share a wrapped diagonal; Engelhardt calls such a pair a conflict. Pólya's theorem says that a solution with no conflicts exists exactly when gcd(n,6)=1, and Engelhardt asked whether a solution can have exactly one conflict. We show that it cannot, and that the fewest conflicts of a solution is exactly 0, 2 or 3, according as neither, exactly one, or both of 2 and 3 divide n. The same counting rules out further levels: no solution has exactly 2 conflicts when n ≥ 5 is prime, and none has exactly 3 when n=2q with q ≥ 5 prime. Finally, every solution with the fewest conflicts is balanced modulo every prime K ≥ 5 dividing n: each residue class of i+p(i) and of i-p(i) modulo K receives exactly n/K queens, as for a toroidal solution. An example on the 75 × 75 board shows that this can fail for composite K. At composite n prime to 6, by contrast, solutions with exactly 2 conflicts exist; we give examples for n=25,35,49,55,65. We also answer Engelhardt's second question: no solution has a cycle of conflicts, and no cycle of conflicts can even be placed on the board with all its conflicts broken by the edges. The inequality behind this, combined with a weighting of the rows and columns, shows that a solution has at most (√5-1)n/2 ≈ 0.618n conflicts, and the floor of this bound is the exact maximum for every n from 6 to 50 except 10,13,18, at nineteen larger n up to 233 and, through products with toroidal solutions, at 221 further n below 10^5, but not at n=89, where the same argument gives one less; the same products show that the largest number of conflicts reaches 144n/233 for infinitely many n.