# Intersection of system of equations has finite number of solutions?

If $n$ is a positive integer and $b_1, b_2, \cdots, b_n$ are positive real numbers such that $b_i \not= b_{i+1}$ (with the convention that $b_{n+1} = b_1$), then can it be shown that the system of equations $a_i^2+a_ia_{i+1}+a_{i+1}^2 = b_i$ for $i=1,2,\cdots, n$ has a finite number of solutions? (EDIT: Gerry Myerson pointed out that I forgot to write $a_{n+1} = a_1.$)

I'm curious about this question because the corresponding questions for $x+y, xy, x^2+y^2, x^2+y^2+2xy$ all can have infinite numbers of solutions, while intuitively, this curve shouldn't because there aren't any obvious reductions to the linear case like in the previous examples.

I labeled this algebraic geometry because it seemed to be a problem that was perhaps solvable with the methods of algebraic geometry (In particular, it seems that if the variables are required to be complex rather than real, the statement should still hold.)

For n odd and all the variables positive reals, it can be shown there is at most 1 solution. If $a_1$ increases, then $a_2$ decreases, and $a_3$ increases, and so on until it can be shown that $a_1$ has both increased and decreased. A similar argument holds for when $a_1$ decreases, so $a_1$ must stay the same.

-

I'd be surprised if your proof for $x^2+y^2$ didn't extend to $x^2+xy+y^2$ which can, after all, be rewritten as $(x+(1/2)y)^2+((\sqrt3/4)y)^2$ and thus transformed to $u^2+v^2$ by a simple linear transformation.
I don't believe this works, because the proof working for $x^2+y^2$ requires that the y for one equation is equal to the x of the next equation, while this transform breaks the symmetry. – David Yang Mar 22 '12 at 3:30
To clarify my comment: For example, the system $a^2+b^2 = 3, b^2+c^2 = 4, c^2+d^2 = 5, d^2+a^2 = 4$ has an infinite number of solutions because the equations are linearly dependent. But I don't see a clear way to establish number of solutions to the system with the same equations but $(a+\frac{b}{2})^2 + (\frac{b\sqrt{3}}{4})^2$ instead... am I missing a way? – David Yang Mar 22 '12 at 3:35
I see your point. You didn't state $a_{n+1}=a_1$, but evidently you want that, which I missed. – Gerry Myerson Mar 22 '12 at 5:33