PACKING PRIME-INDEXED SQUARES INTO A SQUARE: ELEMENTARY CONSTRUCTIONS AND A CERTIFIED THRESHOLD

Take one square of side p^{−t} for every prime p, where t > 1/2 is a real parameter.How small can a square box be and still hold all of them, without overlaps and without rotatingany square? The two largest squares, for p = 2 and p = 3, already force the box to have sideat least 2^{−t} + 3^{−t}, and the total area forces side at least √︁P(2t), where P is the prime zetafunction. These two bounds cross exactly once, at t_A = 0.6295834492..., which we locate withcertified interval arithmetic. Below tA the box of side 2^{−t} + 3^{−t} is too small. Our main resultis that above this point the first bound is in fact the right answer for every t ∈ [0.632,1]: thesquares can be packed into a box of side exactly 2^{−t} + 3^{−t}. The proof is computer-assistedand fully certified. It uses a guillotine packing for the squares with p < 2 · 10^4, shelf packingup to p < 10^7, the Rosser–Schoenfeld bound on π(x) for all remaining primes, and a simplemonotonicity argument that turns nine finite checks into a statement valid for all t in theinterval. This leaves open only a window of width less than 0.0025 above tA, and we conjecturethat the box 2^{−t} + 3^{−t} works all the way down to t_A. For t = 1 the answer 5/6 also follows fromthe classical Meir–Moser theorem on the squares 1/n. In this case we give short hand-madepackings for the primes and for the prime powers, which use nothing more than telescopingsums.