Proving Uniqueness of Solutions in Non-Linear Systems of Equations

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The solutions a = 2, b = 0, c = 0, d = 3 have been identified for the given non-linear system of equations. To prove uniqueness, advanced techniques such as computing a Groebner basis using software like Maple or Mathematica are recommended, as they can confirm that the system's equations imply the found solution. The discussion highlights that without access to such tools during an exam, one must develop a convincing argument for uniqueness, though performing Buchberger's algorithm by hand may be impractical. The problem originates from the British Mathematical Olympiad, emphasizing the challenge of proving uniqueness in a limited time frame. Ultimately, the solution is confirmed as unique through algebraic methods, but practical exam strategies may vary.
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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.
 
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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
 
Thanks for the reply and confirming that this is the only solution. This particular problem was taken from the British Mathematical Olympiad, Round 1 (2003). Is this the only way to prove that this solution is unique? I won't have access to Maple/Mathematica in the exam, or a calculator. I've been told that if I find a solution, I have to make a convincing argument for why it is the only solution -- is it plausible to use Buchberger's algorithm by hand? (It is a 3 and a half hour paper with 6 questions.)

I just noticed that my solution is 2,0,0,3 -- the year the paper was taken from. Hmm, I did not notice that.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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