[A] Theorem 1.2(a),(b): multi-affinity, degrees, certificate Q - T*P - sum (x_i y_i - 1)^2 = 0
  m= 1 n= 3: degP=2 degQ=3 multi-affine=True #terms(Q)=3 residual terms=0 -> ok
  m= 2 n= 5: degP=2 degQ=4 multi-affine=True #terms(Q)=6 residual terms=0 -> ok
  m= 3 n= 7: degP=2 degQ=4 multi-affine=True #terms(Q)=10 residual terms=0 -> ok
  m= 4 n= 9: degP=2 degQ=4 multi-affine=True #terms(Q)=15 residual terms=0 -> ok
  m= 5 n=11: degP=2 degQ=4 multi-affine=True #terms(Q)=21 residual terms=0 -> ok
  m= 6 n=13: degP=2 degQ=4 multi-affine=True #terms(Q)=28 residual terms=0 -> ok
  m= 7 n=15: degP=2 degQ=4 multi-affine=True #terms(Q)=36 residual terms=0 -> ok
  m= 8 n=17: degP=2 degQ=4 multi-affine=True #terms(Q)=45 residual terms=0 -> ok
  m= 9 n=19: degP=2 degQ=4 multi-affine=True #terms(Q)=55 residual terms=0 -> ok
  m=10 n=21: degP=2 degQ=4 multi-affine=True #terms(Q)=66 residual terms=0 -> ok
  m=11 n=23: degP=2 degQ=4 multi-affine=True #terms(Q)=78 residual terms=0 -> ok
  m=12 n=25: degP=2 degQ=4 multi-affine=True #terms(Q)=91 residual terms=0 -> ok
  m=13 n=27: degP=2 degQ=4 multi-affine=True #terms(Q)=105 residual terms=0 -> ok
  m=14 n=29: degP=2 degQ=4 multi-affine=True #terms(Q)=120 residual terms=0 -> ok
  m=15 n=31: degP=2 degQ=4 multi-affine=True #terms(Q)=136 residual terms=0 -> ok
  m=16 n=33: degP=2 degQ=4 multi-affine=True #terms(Q)=153 residual terms=0 -> ok
[B] Remark 3.2: Lemma 3.1 with f_i = x_i y_i - 1 and x0 -> x0 - m gives (P_m, Q_m - m P_m)
  m= 1: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 2: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 3: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 4: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 5: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 6: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 7: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 8: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m= 9: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m=10: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m=11: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
  m=12: P_block(x0-m) == P_m and Q_block(x0-m) == Q_m - m*P_m: True
[C] Lemma 3.1(a),(b) for random multi-affine blocks; Corollary 3.3 for f_j = z_j1...z_jd - 1
  random instances: 200/200 satisfy multi-affinity, degree bounds and Q = S*P + sum f_j^2
  d=2 k=1 n= 3: degP=2 degQ=3 MA=True certificate=True predicted b0=2^(1*1)=2 -> ok
  d=2 k=2 n= 5: degP=2 degQ=4 MA=True certificate=True predicted b0=2^(1*2)=4 -> ok
  d=2 k=3 n= 7: degP=2 degQ=4 MA=True certificate=True predicted b0=2^(1*3)=8 -> ok
  d=2 k=4 n= 9: degP=2 degQ=4 MA=True certificate=True predicted b0=2^(1*4)=16 -> ok
  d=2 k=5 n=11: degP=2 degQ=4 MA=True certificate=True predicted b0=2^(1*5)=32 -> ok
  d=3 k=1 n= 4: degP=3 degQ=4 MA=True certificate=True predicted b0=2^(2*1)=4 -> ok
  d=3 k=2 n= 7: degP=3 degQ=6 MA=True certificate=True predicted b0=2^(2*2)=16 -> ok
  d=3 k=3 n=10: degP=3 degQ=6 MA=True certificate=True predicted b0=2^(2*3)=64 -> ok
  d=3 k=4 n=13: degP=3 degQ=6 MA=True certificate=True predicted b0=2^(2*4)=256 -> ok
  d=3 k=5 n=16: degP=3 degQ=6 MA=True certificate=True predicted b0=2^(2*5)=1024 -> ok
  d=4 k=1 n= 5: degP=4 degQ=5 MA=True certificate=True predicted b0=2^(3*1)=8 -> ok
  d=4 k=2 n= 9: degP=4 degQ=8 MA=True certificate=True predicted b0=2^(3*2)=64 -> ok
  d=4 k=3 n=13: degP=4 degQ=8 MA=True certificate=True predicted b0=2^(3*3)=512 -> ok
  d=4 k=4 n=17: degP=4 degQ=8 MA=True certificate=True predicted b0=2^(3*4)=4096 -> ok
  d=4 k=5 n=21: degP=4 degQ=8 MA=True certificate=True predicted b0=2^(3*5)=32768 -> ok
  d=5 k=1 n= 6: degP=5 degQ=6 MA=True certificate=True predicted b0=2^(4*1)=16 -> ok
  d=5 k=2 n=11: degP=5 degQ=10 MA=True certificate=True predicted b0=2^(4*2)=256 -> ok
  d=5 k=3 n=16: degP=5 degQ=10 MA=True certificate=True predicted b0=2^(4*3)=4096 -> ok
  d=5 k=4 n=21: degP=5 degQ=10 MA=True certificate=True predicted b0=2^(4*4)=65536 -> ok
  d=5 k=5 n=26: degP=5 degQ=10 MA=True certificate=True predicted b0=2^(4*5)=1048576 -> ok
[D] Lemma 2.2: #{s in {+-1}^d : even number of -1} = 2^(d-1)
  d= 1: 1 = 2^0: True
  d= 2: 2 = 2^1: True
  d= 3: 4 = 2^2: True
  d= 4: 8 = 2^3: True
  d= 5: 16 = 2^4: True
  d= 6: 32 = 2^5: True
  d= 7: 64 = 2^6: True
  d= 8: 128 = 2^7: True
  d= 9: 256 = 2^8: True
  d=10: 512 = 2^9: True
[E] Proposition 4.1(b): affine rescaling preserves multi-affinity/degrees; 2^m zeros in the cube
  m=1: rho=4/23, deg(P o phi^-1)=2, deg(Q o phi^-1)=3, MA=True, zeros phi(m,s,s) in cube: 2/2 -> ok
  m=2: rho=1/45, deg(P o phi^-1)=2, deg(Q o phi^-1)=4, MA=True, zeros phi(m,s,s) in cube: 4/4 -> ok
  m=3: rho=2/81, deg(P o phi^-1)=2, deg(Q o phi^-1)=4, MA=True, zeros phi(m,s,s) in cube: 8/8 -> ok
  m=4: rho=8/41, deg(P o phi^-1)=2, deg(Q o phi^-1)=4, MA=True, zeros phi(m,s,s) in cube: 16/16 -> ok
  m=5: rho=1/47, deg(P o phi^-1)=2, deg(Q o phi^-1)=4, MA=True, zeros phi(m,s,s) in cube: 32/32 -> ok
[F] Remark 5.6: m = 1 is of elimination type with G = (x1 y1 - 1)^2 and degrees (2,3)
  Q_1 = T*P_1 + (x1 y1 - 1)^2 with deg P_1 = 2, deg Q_1 = 3: True; Z = {x0 = 1, x1 y1 = 1} has 2 components (Lemma 2.2 with d = 2)
ALL LEAD CHECKS PASSED
