check_note.py -- AMR-152-0049

Part A: Theorem 1.1(a), family A (delta != 1)
  [ok] (E1) b^2 + c - delta a = d
  [ok] (E2) d^2 + c - delta b = a
  [ok] (E3) a^2 + c - delta d = b
  [ok] a - b = b - d = -3(delta-1)/2
  [ok] c = -5/4 + 5t - 29t^2
  [ok] 29k^2 - 38k + 29 = 0 mod 16 exactly for k = 3 mod 4  (residues [13, 4, 5, 0, 5, 4, 13, 0, 13, 4, 5, 0, 5, 4, 13, 0])
  [ok] integers -201 <= delta <= 201: c_A in Z iff delta = 3 mod 4, and then the cycle is integral

Part B: Theorem 1.1(b), family B (delta = 1)
  [ok] (E1)-(E3) with delta = 1
  [ok] c = -5/4 + mu - mu^2
  [ok] a - b = -2 mu

Part C: Theorem 1.1(c), family C (delta != 0), and Theorem 1.1(d)
  [ok] x-sequence (p,q,r,r) satisfies the recurrence cyclically (4 identities)
  [ok] f^2(p,q) - (p,q) = (r - p, r - q) = (delta^2 + 1, (delta^2 + 1)/delta)
  [ok] p - q = -(delta - 1)(delta^2 + 1)/delta
  [ok] Remark 3.1: p = r + delta(q - r) for q = delta(delta-1) - r, p = (1-2delta) r + delta^2(delta-1)
  [ok] Remark 3.1: q - p - (p^2 - r^2) = -(delta - 1)(2r - delta^2 + delta)(2 delta r - delta^3 - 1)
  [ok] delta = 1: x-sequence (-nu,-nu,nu,nu) with c = -nu^2 (4 identities)

Part D: Lemma 2.1 (2-cycles; affine normal form)
  [ok] 2-cycles: b = -1 - delta - a, c = a(1 + delta) - b^2 give x-sequence (a, b)
  [ok] L^{-1} o g o L = f_{c,delta} with c = A(A m^2 + B m + E - delta m - m)

Part E: Theorem 1.1 on a grid (exact orbits of the map)
  [ok] family A: exact period 3 for all 1958 grid values delta != 1
  [ok] family B: exact period 3 for all 1958 grid values mu != 0
  [ok] family C: exact period 4, orbit (p,q)->(q,r)->(r,r)->..., for all 1958 grid values delta != 0
  [ok] delta = 1 family: exact period 4 for all 1958 grid values nu != 0

Part F: Lemma 4.1, Theorem 1.2(b) (the cubic and its node)
  [ok] D(u - 1/2, v - 1/2, w - 1/2) = -F(u, v, w)
  [ok] F is invariant under (u,v,w) -> (v,w,u)
  [ok] D(a,b,d) = (a - b)(b - a^2 - d + b^2) - (a - d)(a - d^2 - d + b^2) (rows 2, 3 minus row 1)
  [ok] formal partial derivatives: F_u = 3u^2 - 2uv - w^2, F_v = 3v^2 - 2vw - u^2, F_w = 3w^2 - 2wu - v^2
  [ok] Euler: u F_u + v F_v + w F_w = 3F (so the singular points of C are the common zeros of the partials)
  [ok] F(m + h, m + k, m) = 2m(h^2 - hk + k^2) + h^3 - h^2 k + k^3
  [ok] h^2 - hk + k^2 = (h - k/2)^2 + (3/4) k^2 (positive for real (h, k) != 0; tangent cone of the node)
  [ok] the tangent cone h^2 - hk + k^2 has discriminant -3k^2 in h (tangents conjugate over Q(sqrt(-3)))
  [ok] F_u(u, v, 1) = (-2u) v + (3u^2 - 1); F_u(0, v, 1) = -1
  [ok] 4u^2 F_v(u, v(u), 1) = 23u^4 - 12u^3 - 18u^2 + 4u + 3 = (u - 1)(23u^3 + 11u^2 - 7u - 3)
  [ok] 4u^2 F_w(u, v(u), 1) = -9u^4 - 8u^3 + 18u^2 - 1 = -(u - 1)(9u^3 + 17u^2 - u - 1)
  [ok] gcd of the two quartics is u - 1 (Euclid over Q), so a singular point with w = 1 has u = 1, v = 1  (gcd = ['-1', '1'] (constant term first); resultant of the two cubics = 331776)
  [ok] Res_v(F_u, F_v)|_{w=1} = (u - 1)(23u^3 + 11u^2 - 7u - 3)
  [ok] Res_v(F_u, F_w)|_{w=1} = -(u - 1)(9u^3 + 17u^2 - u - 1)
  [ok] on w = 0: F_u = u(3u - 2v), F_v = 3v^2 - u^2, F_w = -v^2 (only common zero u = v = 0)
  [ok] (1:1:1): F = 0 and grad F = 0
  [ok] special points (-1:1:1), (1:-1:1), (1:1:-1) lie on C; F(u,v,v) = (u-v)^2 (u+v); F(u,u,w) = (u-w)^2 (u+w)

Part G: Theorem 1.2(c), Corollary 1.3 (parametrisation, fibres)
  [ok] F(w0 + h, w0 + k, w0) = 0 identically
  [ok] (E1)-(E3) hold identically in (h, k) with delta(h,k) and c = d - b^2 + delta a
  [ok] (delta - 1)(v - u) = w^2 - v^2 (so delta = 1 + (w^2 - v^2)/(v - u) when u != v)
  [ok] 2w + k = -2h(h - k)^2/(2(h^2 - hk + k^2)) (step in the proof of the delta formula)
  [ok] (h, k) = (12t, 6t): (u, v, w) = (7t, t, -5t) and delta = 1 - 4t (family A)
  [ok] (h, k) = (-2mu, 0): (u, v, w) = (-mu, mu, mu), delta = 1 (family B)
  [ok] on the fibre: delta(h, k) = delta
  [ok] w = (delta - 1)(tau^3 - tau^2 + 1)/(2 tau (tau - 1))
  [ok] Corollary 1.3(a): a = -1/2 - eps(tau^3 - tau^2 + 2tau - 1), b = -1/2 + eps(tau^3 - 3tau^2 + 2tau - 1), d = -1/2 + eps(tau^3 - tau^2 + 1)
  [ok] c = c_delta(tau) = -N_delta(tau)/(4 tau^2 (tau - 1)^2)
  [ok] c_delta(1 - 1/tau) = c_delta(tau)
  [ok] N_delta(0) = N_delta(1) = (delta - 1)^2 (so c_delta has degree 6 for delta != 1)
  [ok] c_delta(2) = c_A(delta) = -(29 delta^2 - 38 delta + 29)/16
  [ok] c_0(tau) = -(tau^6 - 2tau^5 + 4tau^4 - 8tau^3 + 9tau^2 - 4tau + 1)/(4 tau^2 (tau - 1)^2) (delta = 0)
  [ok] fibre examples (delta in {3, -4, 2/7, 0}, tau in {2, 3, -5, 7/3}): exact period 3
  [ok] delta = 3: tau = 2 gives c = -11, tau = 3 gives c = -94/9  (c_3(3) = -94/9)

Part H: Proposition 1.5 (duality)
  [ok] f^{-1}(f(x, y)) = (x, y)
  [ok] T^{-1} o f_{c,delta}^{-1} o T = f_{c/delta^2, 1/delta}
  [ok] delta^2 c_A(1/delta) = c_A(delta) and delta^2 c_C(1/delta) = c_C(delta)
  [ok] the 5-cycle of f_{-469/16,-4}, reversed and divided by -4, is a 5-cycle of f_{-469/256,-1/4}  (x-sequence ['-9/16', '-19/16', '-9/16', '-29/16', '21/16'])

Part I: Remark 3.3 (the interpolation construction of the source, d = 2)
  [ok] 4-cycle (y0,y1,y2,y2): the four interpolation equations hold with these a0, a1, a2, delta
  [ok] determinant of the 4x4 interpolation matrix (y3 = y2) = -(y0 - y1)(y0 - y2)(y1 - y2)^2
  [ok] normal form of this map and cycle = family C at delta = (y0 - y2)/(y1 - y2)
  [ok] 3-cycle (y0,y1,y2), delta free: interpolation equations hold; a2 = -(delta-1)(...)/(...)

Part J: examples quoted in the note (exact orbits)
  [ok] A, delta=3, c=-11: point (-4, -1) has exact period 3  (found 3)
  [ok] A, delta=0, c=-29/16: point (5/4, -1/4) has exact period 3  (found 3)
  [ok] A, delta=-1, c=-6: point (3, 0) has exact period 3  (found 3)
  [ok] A, delta=2, c=-69/16: point (-9/4, -3/4) has exact period 3  (found 3)
  [ok] A, delta=7, c=-74: point (-11, -2) has exact period 3  (found 3)
  [ok] B, mu=1, c=-5/4: point (-3/2, 1/2) has exact period 3  (found 3)
  [ok] B, c=-1 (s=0): point (-1, 0) has exact period 3  (found 3)
  [ok] C, delta=2, c=-53/16: point (-11/4, -1/4) has exact period 4  (found 4)
  [ok] C, delta=-1, c=-2: point (-2, 2) has exact period 4  (found 4)
  [ok] C, delta=1, c=-1: point (-1, -1) has exact period 4  (found 4)
  [ok] delta=-4, c=-469/16: point (-21/4, 29/4) has exact period 5  (found 5)
  [ok] delta=-1/4, c=-469/256: point (-29/16, 21/16) has exact period 5  (found 5)
  [ok] delta=-4, c=-365/16: point (-15/4, 25/4) has exact period 4  (found 4)
  [ok] delta=-4, c=-685/16: point (-35/4, 35/4) has exact period 4  (found 4)
  [ok] delta=-4, c=-5749/144: point (-101/12, 103/12) has exact period 4  (found 4)
  [ok] delta=-1/4, c=-685/256: point (-35/16, 35/16) has exact period 4  (found 4)
  [ok] delta=-1/4, c=-365/256: point (-25/16, 15/16) has exact period 4  (found 4)
  [ok] delta=5/2, c=-2785/256: point (-69/16, -29/16) has exact period 4  (found 4)
  [ok] delta=-1, c=-7/9: point (-4/3, 4/3) has exact period 6  (found 6)
  [ok] delta=-1, c=-9/16 (Ingram): point (1/4, -3/4) has exact period 8  (found 8)
  [ok] delta=1, c=7/16: point (-1/4, 1/4) has exact period 5  (found 5)
  [ok] delta=1, c=-9/16: point (-3/4, -1/4) has exact period 7  (found 7)
  [ok] delta=1, c=-1/16: point (-3/4, -1/4) has exact period 9  (found 9)
  [ok] delta=1, c=-17/16: point (-5/4, -1/4) has exact period 10  (found 10)

Part K: Lemma 5.1, complete lists of rational periodic points (finite computations)
  [ok] f_(c=-469/16, delta=-4): cycle lengths [2, 5] = one 2-cycle and one 5-cycle, no fixed point  (box L=4, |X|<=33; x-sequences of length >= 3: [['-21/4', '29/4', '9/4', '19/4', '9/4']])
  [ok] f_(c=-469/256, delta=-1/4): cycle lengths [2, 5] = one 2-cycle and one 5-cycle, no fixed point  (box L=16, |X|<=33; x-sequences of length >= 3: [['-29/16', '21/16', '-9/16', '-19/16', '-9/16']])
  [ok] f_(c=-9/16, delta=-1): cycle lengths [1, 1, 2, 8] = two fixed points, one 2-cycle, one 8-cycle (Ingram's example)  (box L=4, |X|<=9; x-sequences of length >= 3: [['-5/4', '5/4', '-1/4', '3/4', '-1/4', '1/4', '-3/4', '1/4']])
  [ok] f_(c=-11, delta=3): cycle lengths [3] = one 3-cycle and nothing else  (box L=1, |X|<=5; x-sequences of length >= 3: [['-4', '-1', '2']])
  [ok] delta = 1: for all 295 values c = N/M, M in {1,4,9,16,25,36}, |c| <= 3, a rational 3-cycle exists iff -(1+c) is a square in Q  (18 values of c with a 3-cycle)

92 checks, 0 failed.
ALL CHECKS PASSED
(running time 1.3 s)
