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Peter Mueller
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Added more details: From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

Added: If one fixes all but one of the parameters (answering$a,b,c,d$, and lets the questionremaining one run through some possibilities, one can guess the degrees of the dependencies of the coefficients of $f_{a,b,c,d}$ in $a,b,c,d$, and then solve the comment) Thecorresponding interpolation problem. For instance, the coefficient of $z^{14}$ seems to be $-3a/4+d-13/4$. The highest degree dependency seems to occur for the coefficient of $z^4$, of degrees $5,3,3,4$ in $a,b,c,d$, respectively. So in order to compute this coefficient, one has to compute $6\cdot 4\cdot 4\cdot 5=480$ examples, and solve the corresponding linear system of equations in $480$ unknowns. Don't know if that is possible, but I would expect it can be done.

The Sage code for the computation of $f_{a,b,c,d}$ is:

From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

Added: (answering the question in the comment) The Sage code

Added more details: From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

If one fixes all but one of the parameters $a,b,c,d$, and lets the remaining one run through some possibilities, one can guess the degrees of the dependencies of the coefficients of $f_{a,b,c,d}$ in $a,b,c,d$, and then solve the corresponding interpolation problem. For instance, the coefficient of $z^{14}$ seems to be $-3a/4+d-13/4$. The highest degree dependency seems to occur for the coefficient of $z^4$, of degrees $5,3,3,4$ in $a,b,c,d$, respectively. So in order to compute this coefficient, one has to compute $6\cdot 4\cdot 4\cdot 5=480$ examples, and solve the corresponding linear system of equations in $480$ unknowns. Don't know if that is possible, but I would expect it can be done.

The Sage code for the computation of $f_{a,b,c,d}$ is:

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Peter Mueller
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while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})
while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.<u,v,w,x,y,z>=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})

while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})
while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.<u,v,w,x,y,z>=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})
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Peter Mueller
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From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

Added: (answering the question in the comment) The Sage code


while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})

produces


(9, 4, 9, 5) z^16 - 5*z^14 + 47/4*z^12 - 833/64*z^10 + 2263/256*z^8 + 3973/1024*z^6 - 27697/4096*z^4 + 5775/4096*z^2 + 625/4096
(-7, 7, 5, -1) z^16 + z^14 + 13/8*z^12 - 23/8*z^10 + 3011/256*z^8 - 1193/128*z^6 + 16581/2048*z^4 + 2303/4096*z^2 + 2401/65536
(2, -7, 3, 1) z^16 - 15/4*z^14 + 29/4*z^12 - 305/32*z^10 + 75/8*z^8 - 1703/256*z^6 + 453/128*z^4 - 1233/1024*z^2 + 81/256
(-2, -5, -10, 2) z^16 + 1/4*z^14 - 17/8*z^12 - 41/32*z^10 + 457/256*z^8 + 1133/1024*z^6 - 605/1024*z^4 - 21/64*z^2 + 1/16
(-10, -3, -9, 4) z^16 + 33/4*z^14 + 219/8*z^12 + 1023/64*z^10 + 4017/256*z^8 - 6919/1024*z^6 - 1301/1024*z^4 - 138985/16384*z^2 + 130321/65536
(-4, 5, -4, 7) z^16 + 27/4*z^14 + 61/4*z^12 + 79/64*z^10 - 1293/64*z^8 + 4333/1024*z^6 + 24929/2048*z^4 - 40071/4096*z^2 + 130321/65536
(-7, -6, -9, 0) z^16 + 2*z^14 + z^12 - 23/4*z^10 + 643/128*z^8 - 331/128*z^6 + 171/256*z^4 - 243/256*z^2 + 6561/65536
... snip ...

From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

Added: (answering the question in the comment) The Sage code


while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})

produces


(9, 4, 9, 5) z^16 - 5*z^14 + 47/4*z^12 - 833/64*z^10 + 2263/256*z^8 + 3973/1024*z^6 - 27697/4096*z^4 + 5775/4096*z^2 + 625/4096
(-7, 7, 5, -1) z^16 + z^14 + 13/8*z^12 - 23/8*z^10 + 3011/256*z^8 - 1193/128*z^6 + 16581/2048*z^4 + 2303/4096*z^2 + 2401/65536
(2, -7, 3, 1) z^16 - 15/4*z^14 + 29/4*z^12 - 305/32*z^10 + 75/8*z^8 - 1703/256*z^6 + 453/128*z^4 - 1233/1024*z^2 + 81/256
(-2, -5, -10, 2) z^16 + 1/4*z^14 - 17/8*z^12 - 41/32*z^10 + 457/256*z^8 + 1133/1024*z^6 - 605/1024*z^4 - 21/64*z^2 + 1/16
(-10, -3, -9, 4) z^16 + 33/4*z^14 + 219/8*z^12 + 1023/64*z^10 + 4017/256*z^8 - 6919/1024*z^6 - 1301/1024*z^4 - 138985/16384*z^2 + 130321/65536
(-4, 5, -4, 7) z^16 + 27/4*z^14 + 61/4*z^12 + 79/64*z^10 - 1293/64*z^8 + 4333/1024*z^6 + 24929/2048*z^4 - 40071/4096*z^2 + 130321/65536
(-7, -6, -9, 0) z^16 + 2*z^14 + z^12 - 23/4*z^10 + 643/128*z^8 - 331/128*z^6 + 171/256*z^4 - 243/256*z^2 + 6561/65536
... snip ...
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Peter Mueller
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