Two questions of Huguin on the multiplier polynomials of zᵈ+c, answered at twenty-five pairs (d,n)

The (d,n) cells the source itself settles or prints – n = 1 at every 2 <= d <= 10, d = 2 with n <= 4, d = 3 with n = 2 – re-derived from the definitions and certified: an exact Bezout identity u Qₙ + v Rₙ = m in Z[c] with m!= 0, hence Qₙ and Rₙ have no common root in C

Huguin's Question 1 answered NO at d = 2, n = 5 and n = 6: explicit Q₅ (deg 11), R₅ (deg 28), Q₆ (deg 20), R₆ (deg 85) in Z[c] with Bezout certificates u Qₙ + v Rₙ = m, m!= 0, so no c in C gives z² + c both a period-n cycle of multiplier 1 and two distinct period-n cycles with the same multiplier

Huguin's Question 1 answered NO at d = 3, n = 3 and n = 4: explicit Q₃ (deg 12), R₃ (deg 44), Q₄ (deg 40), R₄ (deg 350) in Z[c] with Bezout certificates

Huguin's Question 1 answered NO at d = 4, n = 2 and n = 3: explicit Q₂ (deg 6), R₂ (deg 18), Q₃ (deg 39), R₃ (deg 372) in Z[c] with Bezout certificates

Huguin's Question 1 answered NO at n = 2 for every degree 5 <= d <= 10; deg R₂ runs 62, 155, 327, 609, 1044, 1674 and the Bezout constants have 120 to 9465 decimal digits

The discriminant factorisation Deltaₙ = aₙ Qₙ Rₙ² verified with aₙ = +-1 in every cell the source covers: n = 1 at all 2 <= d <= 10, every n <= 6 at d = 2 (where the values are aₙ = -1 throughout), and d = 3 with n = 2, where a₂ = -1 is read off the Delta₂ the source prints

Huguin's Question 2 answered YES at d = 3, n = 3 (a₃ = +1) and n = 4 (a₄ = -1), by exhibiting Deltaₙ = aₙ Qₙ Rₙ² with aₙ = +-1; the question is open for every d >= 3

Huguin's Question 2 answered YES at d = 4, n = 2 and n = 3, both with aₙ = +1

Huguin's Question 2 answered YES at n = 2 for every degree 5 <= d <= 10: a₂ = -1, -1, -1, +1, -1, -1 for d = 5, 6, 7, 8, 9, 10, so a₂ = +1 exactly at d = 4 and d = 8 in the computed range

Negative control (too large): the coprimality of Qₙ and Rₙ is a characteristic-zero statement – explicit primes p and residues c₀ at which the SAME reductions do share a root, at (d,n) = (2,5) mod 3 and mod 11, (2,6) mod 5, (3,4) mod 31, (4,3) mod 11, (8,2) mod 7

Negative control (too large): aₙ = -1 is not universal – the (d,n) = (4,1) cell has a₁ = +1 and the -1 version of Deltaₙ = aₙ Qₙ Rₙ² is false there already over Z

Negative control (too small): Qₙ and Rₙ are NOT comaximal in Z[c] – no u, v in Z[c] give u Qₙ + v Rₙ = 1, at (d,n) = (2,4) and (2,5), because Rₙ(0) = 0 and Qₙ(0) is not a unit; so m!= 0, not m = +-1, is the right strength for the certificate

Negative control (not vacuous): Rₙ really has roots that Qₙ misses – the power-map parameter c = 0 is a root of Rₙ with Qₙ(0)!= 0 at d = 2, n = 4, 5, 6, and so is the Chebyshev parameter c = -2, where Qₙ(-2) = -17, -285417, -245246990625

Negative control (not vacuous): a neighbouring pair in the same circle of polynomials does share roots – Qₙ divides Deltaₙ, so every root of Q₅ is a root of Delta₅ at d = 2

Negative control (hypothesis not vacuous): the q(0)!= 0 hypothesis of the cell lemma – 'the power map has no parabolic cycle' – is not removable; explicit data satisfies every other hypothesis and has a common root at c = 0

Bad primes: for each computed cell, the primes p <= 10⁵ at which the reductions of Qₙ and Rₙ acquire a common factor, read off the Bezout constant m – e.g. 2,3,11,29,31,4217,86131 at (d,n) = (2,5) and 2,3,5,7,13,29,61,79,211,239,409,2693,3331 at (2,6), with an unfactored cofactor of 82 and 1231 bits

Cost of the (d,n) frontier: Mₙ dominates and scales roughly 55x per step in n at d = 2 (n = 5: 1.0 s, n = 6: 54 s, n = 7: > 25 CPU-min and unfinished); the whole d <= 10, n <= 2 plus d <= 4 sweep is about 6 CPU-minutes, and the three Lean modules check in 80 + 46 + 210 s at 7.3-8.4 GB peak RSS