Vamshi Jandhyala
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AI for mathematics

Seven problems from Erich Friedman's *Math Magic*

Three armies of five bishops fit on a 5 × 5 board and three of eight on 6 × 6, and no more: Friedman's two values.

Open problem settled, verified by computation

Read the paper Code
Three armies of bishops

Erich Friedman’s Math Magic poses a problem each month and keeps a list of unsolved problems. As of 9 October 2026 the list shows problems 15, 21 and 24 as open, and the March 2005 page on armies of bishops asks for two exact values. The paper settles these four, and makes progress on problems 6, 28 and 33, which the list also shows as open:

  • Armies of bishops. B(3,5) = 5 and B(3,6) = 8: three armies of five bishops fit on a 5 × 5 board and three of eight on a 6 × 6 board, and no more. A placement is valid exactly when every diagonal carries one army, which turns the question into an exhaustive count over labellings of the diagonals.
  • Touch cycles (unsolved problem 24). No finite arrangement of three colours of king has touch counts (1,3,6), (1,6,3) or (2,3,4), nor (1,4,5). An extreme king reduces any finite arrangement to a finite formula, refuted with a DRUP certificate. Each listed triple does have a periodic arrangement of the whole plane, so finiteness is essential.
  • Capture digraphs (unsolved problem 21). The 2-regular digraphs q, r and s on six points are not the capture graph of any chess position on any board.
  • Magic polyomino squares (unsolved problem 15). The first listed heptomino can never be arranged in a square with equal row and column counts. The proof is an explicit weighting of rows and columns.
  • Slab cubes (unsolved problem 6). A cube tiled by one k-slab (a k × ik × jk box) for each k = 1, …, n has side at most n(n+1)/2, with equality only for n = 1, 3; no such cube exists for n = 4, 5, 6. The bound is proved; the non-existence is certified by two programs, one with a constraint solver and one without.
  • Prime signatures (unsolved problem 28). The table’s entry 1022303⁴ in cell (4,41) belongs in cell (4,411), where it is the smallest value (proved, using Ljunggren’s theorem on x² + 1 = 2y⁴); cell (4,41) is empty. The two open cases have no solution below 10¹⁷.
  • Optimal tournaments (unsolved problem 33). Exact optima over all adaptive schedules: ranking 3 players in 7 games succeeds with probability 2992/6561 (published 2944/6561), 4 players in 7 games with 1216/6561 (published 1136/6561); three entries for 5 players are corrected, and every other published value checked is optimal.

Preprint v2, 9 October 2026 (v1 the same day had the first four results), not peer reviewed. Some results are computer-certified and some proved by hand; the paper says which, and for each certified result what was computed, why it suffices, and how it was checked independently. No Lean formalisation: the certified results are exhaustive counts and solver refutations, checked by certificates and by programs written independently. The author used AI tools (Claude, Anthropic; Codex, OpenAI) in this work, as described in the paper, and is responsible for its content.

Chess capture patterns
Chess capture patterns. Friedman's capture digraphs q, r and s are not made by any chess position, on any board.
Kings that touch in a cycle
Kings that touch in a cycle. Kings with touch counts (1,3,6), (1,6,3) or (2,3,4) admit no finite arrangement: Friedman's unsolved problem 24.
A heptomino that never balances
A heptomino that never balances. This heptomino never fills a square with equal row and column counts, in any size: one shape of Friedman's problem 15.
A tournament schedule that beats the table
A tournament schedule that beats the table. Friedman's optimal tournaments: exact best schedules over all adaptive plans, two of them better than the published values.
A misfiled prime signature
A misfiled prime signature. In Friedman's prime-signature table, 1022303⁴ belongs in cell (4, 411), where it is the smallest; cell (4, 41) is empty.
Cubes cut into slabs
Cubes cut into slabs. A cube made of one k-slab for each k ≤ n has side at most n(n+1)/2, and none exists for n = 4, 5, 6; Friedman conjectures infinitely many.