Other primes
The papers ask what happens if 5 is replaced by another prime p. This page explores two answers. The first shows that the formula itself cannot be copied: 5 is the only prime that works. The second shows that the palindrome and its counts survive for every prime, and that the triple 3² + 4² = 5² is the first of a family that continues with 5² + 12² = 13².
1. The formula works only for p = 5
Try the natural generalisation (n² + c·(n/p)) / p, with (n/p) the Legendre symbol modulo p. For which p and c is it a whole number for every n?
Map: rows are primes, columns are c = 0 … p − 1. A filled cell means whole numbers for every n. Only one cell is filled: p = 5, c = 4.
Why
Taking n = 1 forces 1 + c ≡ 0, so c ≡ −1 (mod p). Then every n with (n/p) = +1 needs n² ≡ 1, i.e. n ≡ ±1. So the only quadratic residues modulo p could be ±1, which leaves room for at most two of them. There are (p − 1)/2 quadratic residues, so p ≤ 5. For p = 3, n = 2 would need 4 − c ≡ 0, i.e. c ≡ 1, contradicting c ≡ −1. So p = 5 and c ≡ 4 (mod 5) is the only case: the formula of the cycle.
This short argument answers the first open question in the papers. It was checked by computer for every prime below 400.
2. The palindrome survives for every prime
Drop the correction and keep the whole part: Qp(n) = ⌊n²/p⌋ mod p. For p = 5 this has the same counts as the cycle (9 zeros, each other value 4 times). Choose a prime and see the pattern of one period of p² numbers, written in p rows of p.
Why the counts are always 2p − 1 and p − 1
Write n = pq + r with q and r between 0 and p − 1 (row q, column r in the grid). Then ⌊n²/p⌋ ≡ 2qr + ⌊r²/p⌋ (mod p). In column r = 0 every value is 0. In any other column, 2r is invertible modulo p, so as q runs through a column every value 0, …, p − 1 appears exactly once. That gives 0 a total of p + (p − 1) = 2p − 1 times and every other value p − 1 times, and (2p − 1) + (p − 1)² = p². The same argument, with the extra +1 when n ≡ ±1, gives the counts 9 and 4 of the Pentametric cycle.
Period p² and the mirror Qp(p² − n) = Qp(n) follow from expanding (n + p²)² and (p² − n)².
When is it a Pythagorean triple?
The identity (2p − 1) + (p − 1)² = p² becomes a Pythagorean triple m² + (p − 1)² = p² exactly when 2p − 1 is a perfect square m², that is when p = (m² + 1)/2. The primes of this form begin
5 (3, 4, 5) · 13 (5, 12, 13) · 41 (9, 40, 41) · 61 (11, 60, 61) · 113 · 181 …
So the split 9 + 16 = 25 of the cycle is the first case of a family. Whether there are infinitely many such primes is not known: it is a special case of a famous open question, whether a quadratic polynomial takes infinitely many prime values.
Why 5, also for the music? The same holds for the formula itself read with any modulus m, 1 + MOD(𝔓 − 1; m): every modulus makes a palindrome, but only up to 6 does every step sound consonant, and of those only 5 and 6 use all seven consonances, 5 with thirds and sixths, 6 with fourths and fifths. Hear them side by side on When silence becomes 5 · Why 5? (exploratory).