FeDeX_LaTeX said:
I found the solutions a = 2, b = 0, c = 0, d = 3 to this system of equations...
ab + c + d = 3
bc + d + a = 5
cd + a + b = 2
da + b + c = 6
But I don't know how to justify that those are the only solutions to this system, since it is non-linear. How can I prove that those are the only solutions? (Note: I'm not sure if those ARE the only solutions -- I just can't find any more.)
Thanks.
The solution is unique, and you have found it. However, justifying this involves tools way beyond pre-calculus algebra. In particular, I used the Groebner package in Maple 11 to compute a Groebner basis of the system, and here is what I got:
with(Groebner):
G:={a*b+c+d-3,b*c+d+a-5,c*d+a+b-2,d*a+b+c-6}; <--- these should = 0 at your solution
B:=Basis(G,tdeg(a,b,c,d));
B := [d - 3, c, b, a - 2] <--- this system is equivalent to G
In words: the elements of B are a basis for the ideal generated by G; that is, any multivariate polynomial of the form P = A(a,b,c,d)*(a*b+c+d-3) + B(a,b,c,d)*(b*c+d+a-5) + etc, with multivariate polynomial coefficients A, B, ... can be written as a linear combination of (d-3), c, b and (a-2), also with multivariate polynomial coefficients. So, if all elements of G = 0 (which is your system of equations) then each element of B is also zero, and vice-versa. Therefore, your equations necessarily imply d-3=0, c=0, b=0, a-2=0.
Note: Maple (or Mathematica, for example) use Buchberger's Algorithm to compute a Groebner basis.
RGV