The following Diophantine system
$25333-123\,a_{{2}}a_{{1}}-a_{{2}}a_{{1}}a_{{3}}-478\,a_{{2}}b_{{1}}-a_ {{2}}b_{{1}}c_{{3}}-223\,c_{{2}}a_{{1}}-c_{{2}}a_{{1}}b_{{3}}-589\,c_{ {2}}b_{{1}}-c_{{2}}b_{{1}}d_{{3}}-a_{{4}}=0 $ $ 29151-123\,b_{{2}}a_{{1}}-b_{{2}}a_{{1}}a_{{3}}-478\,b_{{2}}b_{{1}}-b_ {{2}}b_{{1}}c_{{3}}-223\,d_{{2}}a_{{1}}-d_{{2}}a_{{1}}b_{{3}}-589\,d_{ {2}}b_{{1}}-d_{{2}}b_{{1}}d_{{3}}-b_{{4}}=0 $ $ 54507-123\,a_{{2}}c_{{1}}-a_{{2}}c_{{1}}a_{{3}}-478\,a_{{2}}d_{{1}}-a_ {{2}}d_{{1}}c_{{3}}-223\,c_{{2}}c_{{1}}-c_{{2}}c_{{1}}b_{{3}}-589\,c_{ {2}}d_{{1}}-c_{{2}}d_{{1}}d_{{3}}-c_{{4}}=0 $ $ 62645-123\,b_{{2}}c_{{1}}-b_{{2}}c_{{1}}a_{{3}}-478\,b_{{2}}d_{{1}}-b_ {{2}}d_{{1}}c_{{3}}-223\,d_{{2}}c_{{1}}-d_{{2}}c_{{1}}b_{{3}}-589\,d_{ {2}}d_{{1}}-d_{{2}}d_{{1}}d_{{3}}-d_{{4}}=0 $
has a solution
$ [a_{{1}}=1,b_{{1}}=2,c_{{1}}=3,d_{{1}}=4,a_{{2}}=5,b_{{2}}=7,c_{{2}}= 13,d_{{2}}=14,a_{{3}}=11,b_{{3}}=21,c_{{3}}=31,d_{{3}}=41,a_{{4}}=21,b _{{4}}=31,c_{{4}}=41,d_{{4}}=51] $
It is known that in 1900, David Hilbert proposed the solvability of all Diophantine equations as the tenth of his fundamental problems. In 1970, Yuri Matiyasevich solved it negatively, by proving that a general algorithm for solving all Diophantine equations cannot exist.
Is there any mathematical way to find the solution of the above system without doing loop over $\mathbb Z_n$?
How many solutions in $\mathbb Z_n$?
Applying the condition, $3(a+c)=2(b+d)$ and substituting as below in the given equations we have:
($a_1,a_2,a_3,a_4$)=$(a,e,j,n)$
($b_1,b_2,b_3,b_4$)=$(b,f,k,p)$
($c_1,c_2,c_3,c_4$)=$(c,g,L,q)$
($d_1,d_2,d_3,d_4$)=$(d,h,m,r)$
$(s,t,u,v)=(25333,29151,54507,62645)$
$(x,y,z,w)=((L+478),(k+223),(j+123),(m+589))$
Because of common factor's the simultaneous equation given
by 'OP' simplifies to as shown below:
$2(s+t+u+v)=(a+c)[(e+f)(3x+2z)+(g+h)(2y+3w)]+2(n+p+q+r)$
But because the equation is cubic in nature & the number
of unknowns are sixteen which is more than number of equations
the problem is difficult. Unless the number of unknown's are reduced.