If you factor the equations before passing them to solve then the solution
is
```
>>> fsol
[{a2: 0, a1: 0}, {a2: 0, a1: 0}, {a2: 0, b2: 0}, {a2: 0, b1: b2/2, a1:
0}, {a2: (64 + (-8 + (1 - sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(1 -
sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))**2/(36*(1 - sqrt(3)*I)**2*(37
+ 3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: (-64 + (1 - sqrt(3)*I)*(8 +
(-1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(37 +
3*sqrt(303)*I)**(1/3))/(6*(1 - sqrt(3)*I)*(37 +
3*sqrt(303)*I)**(1/3))}, {a2: (64 + (-8 + (1 + sqrt(3)*I)*(37 +
3*sqrt(303)*I)**(1/3))*(1 + sqrt(3)*I)*(37 +
3*sqrt(303)*I)**(1/3))**2/(36*(1 + sqrt(3)*I)**2*(37 +
3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: (-64 + (1 + sqrt(3)*I)*(8 - (1 +
sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(37 +
3*sqrt(303)*I)**(1/3))/(6*(1 + sqrt(3)*I)*(37 +
3*sqrt(303)*I)**(1/3))}, {a2: (16 + (4 + (37 +
3*sqrt(303)*I)**(1/3))*(37 + 3*sqrt(303)*I)**(1/3))**2/(9*(37 +
3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: 4/3 + 16/(3*(37 +
3*sqrt(303)*I)**(1/3)) + (37 + 3*sqrt(303)*I)**(1/3)/3}, {a2: a1**2,
b1: 0, b2: 0}]
```
and all of them are true solutions:
```
>>> [[e.subs(s).simplify() for e in Sys] for s in fsol]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0,
0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
```
On Sunday, December 5, 2021 at 8:35:28 AM UTC-6 Chris Smith wrote:
> `nonlinsolve(Sys, Unk)` gives
> ```
> {(0, 0, b1, b2), (0, 0, b2/2, b2), (a1, 0, b1, 0), (-sqrt(a2), a2, 0, 0),
> (sqrt(a2), a2, 0, 0)}
> ```
> On Saturday, December 4, 2021 at 4:36:04 PM UTC-6 emanuel.c...@gmail.com
> wrote:
>
>> A bug in Sympy’s solve
>>
>> Inspired by this ask.sagemath.orgquestion
>> <https://ask.sagemath.org/question/59063/weird-c-values-from-solving-system-of-equations/>
>> .
>>
>> This question was about the names of the symbolic variables created by
>> Maxima’s solve to denote arbitrary constants. Exploring the example used
>> in this questions revealed a non-trivial problem with sympy’s solve.
>> Problem
>>
>> Solve the system :
>> $$
>> \begin{align*}
>>
>> - a*{1}^{3} a*{2} + a*{1} a*{2}^{2} &= 0 \
>> - 3 a*{1}^{2} a*{2} b*{1} + 2 a*{1} a*{2} b*{2} - a*{1} b*{2} + a*{2}^{2}
>> b*{2} &= 0 \
>> - a*{1}^{2} a*{2}^{2} + a_{2}^{3} &= 0 \
>> - 2 a*{1}^{2} a*{2} b*{2} - 2 a*{2}^{2} b*{1} + 3 a*{2}^{2} b_{2} &= 0
>> \end{align*}
>> $$
>>
>> # Set up sympy (brutal version)import sympyfrom sympy import *
>> init_session()
>> init_printing(pretty_print=False)# System to solve
>> a1, a2, b1, b2 = symbols('a1 a2 b1 b2')
>> Unk = [a1, a2, b1, b2]
>> eq1 = a1 * a2**2 - a2 * a1**3
>> eq2 = 2*a1*a2*b2 + b2*a2**2 - 3*a2*a1**2*b1 - a1*b2
>> eq3 = a2**3 - a2**2*a1**2
>> eq4 = 3*a2**2*b2 - 2*a2*a1**2*b2 - 2*a2**2*b1
>> Sys = [eq1, eq2, eq3, eq4]
>>
>> IPython console for SymPy 1.9 (Python 3.9.9-64-bit) (ground types: gmpy)
>>
>> These commands were executed:
>> >>> from __future__ import division
>> >>> from sympy import *
>> >>> x, y, z, t = symbols('x y z t')
>> >>> k, m, n = symbols('k m n', integer=True)
>> >>> f, g, h = symbols('f g h', cls=Function)
>> >>> init_printing()
>>
>> Documentation can be found at https://docs.sympy.org/1.9/
>>
>> Attempt to use the “automatic” Sympy solver:
>>
>> Sol = solve(Sys, Unk)
>> DSol = [dict(zip(Unk, u)) for u in Sol]
>> DSol
>>
>> [{a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2:
>> 0, b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b2/2, b2: b2}, {a1: a1, a2: 0, b1:
>> b1, b2: 0}, {a1: -sqrt(a2), a2: a2, b1: 0, b2: 0}, {a1:
>> -2**(5/6)*sqrt(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
>> 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6, a2: 16/3 +
>> 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/3, b1: b2/2, b2: b2}, {a1: -sqrt(6)*sqrt(32 -
>> 2**(1/3)*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 +
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/6, a2: 2**(2/3)*(-208/3 + (1 +
>> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)),
>> b1: b2/2, b2: b2}, {a1: -sqrt(6)*sqrt(32 - 416*2**(2/3)/((1 -
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3))/6, a2: 2**(2/3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
>> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)),
>> b1: b2/2, b2: b2}]
>>
>> Something went sideways: the six first solutions are okay,but the last
>> three use expressions, some of them being polynomials in b2.
>>
>> Attempt to check them formally :
>>
>> Chk=[[u.subs(s) for u in Sys] for s in DSol]
>> Chk
>>
>> [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0,
>> 0, 0, 0], [-2**(5/6)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3))
>> + 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2*sqrt(48*2**(1/3) + 624/(1499
>> + 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6 +
>> sqrt(2)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 +
>> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))**(3/2)/54,
>> -2**(2/3)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 +
>> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/12 -
>> 2**(5/6)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*sqrt(48*2**(1/3) + 624/(1499 +
>> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/3 +
>> b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/3)**2 + 2**(5/6)*b2*sqrt(48*2**(1/3) + 624/(1499 +
>> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6,
>> -2**(2/3)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2*(48*2**(1/3) + 624/(1499 +
>> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/18 + (16/3
>> + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/3)**3, -2**(2/3)*b2*(16/3 + 104*2**(2/3)/(3*(1499 +
>> 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
>> 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/9 + 2*b2*(16/3 +
>> 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/3)**2], [2**(1/6)*sqrt(3)*(-208/3 + (1 +
>> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)*(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))**(3/2)/(18*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)) - 2**(5/6)*sqrt(3)*(-208/3 + (1 + sqrt(3)*I)*(32 +
>> (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)**2*(1499 +
>> 3*sqrt(303)*I)**(2/3)), -2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32
>> + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) -
>> 3*2**(2/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1
>> + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 +
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/(2*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)) + sqrt(6)*b2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499
>> + 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))/6 + 2*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 +
>> (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)**2/((1 + sqrt(3)*I)**2*(1499 +
>> 3*sqrt(303)*I)**(2/3)), -2*2**(1/3)*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 -
>> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)**2*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)) +
>> 4*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**3/((1 +
>> sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)), -2*2**(2/3)*b2*(-208/3 + (1 +
>> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 4*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 +
>> sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))], [-2**(5/6)*sqrt(3)*(-208/3 +
>> (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998
>> + 6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 416*2**(2/3)/((1 - sqrt(3)*I)*(1499
>> + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))
>> + 2**(1/6)*sqrt(3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(32 -
>> 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 -
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))**(3/2)/(18*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)), sqrt(6)*b2*sqrt(32 - 416*2**(2/3)/((1 -
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3))/6 + 2*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1
>> + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)**2/((1 - sqrt(3)*I)**2*(1499 +
>> 3*sqrt(303)*I)**(2/3)) - 3*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
>> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/6)/(2*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) -
>> 2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998
>> + 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*sqrt(32 -
>> 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 -
>> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)), 4*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
>> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)**3/((1 - sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)) -
>> 2*2**(1/3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2*(16/3 -
>> 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1
>> - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)**2*(1499 +
>> 3*sqrt(303)*I)**(2/3)), -2*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
>> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
>> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
>> 3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) +
>> 4*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
>> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 -
>> sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))]]
>>
>> Some of these expressions are polynomial in b2 whose coefficients of
>> not-null dregree are nort *obviously* null.
>> A bit of hit-and-miss trials leads to this attempt at *numerical* check :
>>
>> def chz(x):
>> r = x.factor().is_zero
>> if r is None: return x.coeff(b2).n()
>> return r
>> [[chz(u) for u in v] for v in Chk]
>>
>> [[True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, -223.427427555427 + 0.e-25*I, True, True],
>> [True, -0.388012232966408 - 5.25038392589403e-29*I, True, True],
>> [True, -0.e-137 + 1.12445821450677e-139*I, True, True]]
>>
>> Solution #8 may be exactr, but #6 and #7 cannot.
>> Manual solution.
>>
>> eq1 and eq3 give us solutions for a1 and a2 :
>>
>> S13=solve([eq3, eq1], [a2, a1], dict=True) ; S13
>>
>> [{a1: -sqrt(a2)}, {a1: sqrt(a2)}, {a2: 0}]
>>
>> which we prefer to rewrite as :
>>
>> S13 = [{a2:a1**2}, {a2:0}]; S13
>>
>> [{a2: a1**2}, {a2: 0}]
>>
>> Substituting these values in eq4 gives us :
>>
>> E4 = [eq4.subs(s) for s in S13] ; E4
>>
>> [-2*a1**4*b1 + a1**4*b2, 0]
>>
>> The second solution tells us that S13[1] is also,a solution to [eq1,
>> eq3, eq4].
>>
>> S134=S13[1:] ; S134
>>
>> [{a2: 0}]
>>
>> Substituting S13[0] ineq4and solving for the variables of the resulting
>> polynomial augments the setS134of solutions of[eq1, eq3`, eq4]’ :
>>
>> V4=E4[0].free_symbols
>> S4=[solve(E4[0], v, dict=True) for v in V4] ; S4for S in S4:
>> for s in S:
>> S0={u:S13[0][u].subs(s) for u in S13[0].keys()}
>> S134 += [S0.copy()|s]
>> S134
>>
>> [{a2: 0}, {a2: a1**2, b1: b2/2}, {a1: 0, a2: 0}, {a2: a1**2, b2: 2*b1}]
>>
>> Again, we prefer to rewrite it in a simpler (and shorter) fashion :
>>
>> S134=[{a2:0}, {a2:a1**2, b2:2*b1}] ; S134
>>
>> [{a2: 0}, {a2: a1**2, b2: 2*b1}]
>>
>> Substituting in eq3 gives :
>>
>> [eq2.subs(s) for s in S134]
>>
>> [-a1*b2, -a1**4*b1 + 4*a1**3*b1 - 2*a1*b1]
>>
>> Solving these equations for their free symbols a,d merging with the
>> previous partial solutions gives us the solutions of the full system :
>>
>> S1234=[]for S in S134:
>> # print("S=",S)
>> E=eq2.subs(S)
>> # print("E=",E)
>> S1=[solve(E, v, dict=True) for v in E.free_symbols]
>> # print("S1=",S1)
>> for s in flatten(S1):
>> # print("s=",s)
>> S0={u:S[u].subs(s) if "subs" in dir(S[u]) else S[u] for u in
>> S.keys()}
>> S1234+=[S0.copy()|s]
>> S1234
>>
>> [{a1: 0, a2: 0}, {a2: 0, b2: 0}, {a2: a1**2, b1: 0, b2: 0}, {a1: 0, a2: 0,
>> b2: 2*b1}, {a1: 4/3 + (-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) +
>> 16/(9*(-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)), a2: (4/3 + (-1/2
>> - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) + 16/(9*(-1/2 -
>> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)))**2, b2: 2*b1}, {a1: 4/3 +
>> 16/(9*(-1/2 + sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)) + (-1/2 +
>> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3), a2: (4/3 + 16/(9*(-1/2 +
>> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(37/27 +
>> sqrt(303)*I/9)**(1/3))**2, b2: 2*b1}, {a1: 4/3 + 16/(9*(37/27 +
>> sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3), a2: (4/3 +
>> 16/(9*(37/27 + sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3))**2,
>> b2: 2*b1}]
>>
>> Again, some solutions, substituted in eq2, give first-degree monomials
>> in b1 whose oefficient cannot be shownt to be null byis_zero :
>>
>> [[e.subs(s).is_zero for e in Sys] for s in S1234]
>>
>> [[True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, True, True, True],
>> [True, None, True, True],
>> [True, None, True, True],
>> [True, None, True, True]]
>>
>> But, this time, the numerical check points to a probably null result :
>>
>> [Sys[1].subs(S1234[u]).coeff(b1).n() for u in range(3,6)]
>>
>> [0, 0.e-125 + 0.e-127*I, 0.e-125 - 0.e-127*I]
>>
>>
>>
>
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