Suppose I have a system of nonlinear equations. Suppose also that I'm interested in knowing if a unique solution exists. Presumably, a quick check would be to ensure that the number of equations equals the number of unknowns. If the number of equations is less than the number of unknowns (at least in linear algebra), the solution won't be unique even if it does exist, so there's no point searching for a unique solution. Is there a similar necessary condition for systems of nonlinear equations? Put another way, is it possible to create a system of nonlinear equations in $n$ unknowns that contains $m \ne n$ equations for which a unique solution exists?

EDIT: As pointed out in an answer, there are examples of trivial solutions such as: $x^2 + y^2 = 0$, in which the only solution is $x = y = 0$. I am therefore interested in nontrivial solutions.

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    $\begingroup$ A linear system can have more equations than unknowns and a unique solution. $\endgroup$ Oct 16, 2017 at 15:16
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    $\begingroup$ How do you define trivial? Is $(x-1)^2+(y-5)^2=0$ trivial? $\endgroup$
    – Théophile
    Oct 16, 2017 at 15:17
  • $\begingroup$ Yes, you're right. I've edited it to indicate number of equations less than number of unknowns, although I'm starting to suspect that this question isn't precise enough to have a useful answer $\endgroup$ Oct 16, 2017 at 15:17
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    $\begingroup$ I think the spirit of the question is clear, and can be addressed. An equation in $n$ variables represents a set in $\Bbb{R}^n$. The number of equations required to pin down solutions depends on the nature of those solution sets. This is what we study in algebraic geometry. $\endgroup$ Oct 16, 2017 at 15:20
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    $\begingroup$ Even then, although it takes three independent planes in $\Bbb{R}^3$ to intersect at a point, you can get a point as the intersection of just two algebraic surfaces in $\Bbb{R}^3$, such as $z=x^2+y^2$ and $z=-x^2-y^2$... $\endgroup$ Oct 16, 2017 at 15:30

1 Answer 1


Yes, for example take the equation $$x^2+y^2=0$$ in $\mathbb{R}^2$.

  • $\begingroup$ Interesting, and as is always the case I didn't consider the trivial solution. Are there nontrivial cases as well? (I'll edit the question accordingly) $\endgroup$ Oct 16, 2017 at 15:15
  • $\begingroup$ Yes this is a clever solution, but is there a nontrivial solution. $\endgroup$
    – Gregory
    Oct 16, 2017 at 15:15
  • $\begingroup$ What do you mean by "non-trivial"? $\endgroup$ Oct 16, 2017 at 16:54

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