# Existence of solutions for system of equations

I have a system of equations and was wondering whether there is any obvious reason that you find solutions for $e,f,c$ given a fixed $a \in \mathbb{R}$(which is true). So I don't want to solve this system or apply any algorithm to it. I was just wondering whether it is possible to say a priori that there will be solutions to this system?

$$-e f+8 a f^2-4 a^2 f^2-4 c f^2=0$$ $$-6 f+2 e f+8 a e f-8 a^2 e f-8 c e f-16 a f^2=0$$ $$-3 e+e^2-4 a^2 e^2-4 c e^2+8 f-8 a^2 f-8 c f-24 a e f=0$$ $$-2+6 e-8 a e-8 a^2 e-8 c e-8 a e^2-16 a f=0$$ $$4-8 a-4 a^2-4 c-8 a e=0$$

As I am very unfamiliar with algebraic geometry this may be a stupid question, so don't be wondered if I do not see very obvious things. So $a$ is fixed and $f,e,c$ are unknown.

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It will be much easier to answer this question if you add some context. Now the polynomials look rather random. –  Fredrik Meyer Jun 6 '14 at 13:23
well, I made a guess to solve an ODE and this guess works if the equations that you see there have a solution for every $a$ cause that was a parameter of the ODE. So the question is now: Was it a priori obvious that I would find solutions? –  user66906 Jun 6 '14 at 13:25
This is a natural question, but one that I think is a little difficult to answer the way it's currently written. For example, does "a priori" mean that one should not use any properties of the actual equations you have written? If so, then I think the answer must be no; there's no reason for such a system to have a solution (especially not in $\mathbf R$). –  Asal Beag Dubh Jun 6 '14 at 13:36
(By the way, if $a$ is fixed, then the last equation is a linear equation relating $c$ and $e$, so you could eliminate one variable right away.) –  Asal Beag Dubh Jun 6 '14 at 13:37
by a priori I mean that you should not apply something like the Buchberger algorithm or other things that would solve this problem more or less. of course, you are allowed to work with the equations. –  user66906 Jun 6 '14 at 13:38

We have to be allowed to look at the equations, because otherwise we cannot say anything. However, if we look, then we can immediately exhibit a solution for any given $a$, without using any complicated algorithm, like the Buchberger algorithm. This goes as follows. Just put $f=0$. Then the first two equations are already satisfied, and the $5$-th equation gives $c= - a^2 - 2ae - 2a + 1$. Substituting this, we obtain two quadratic equations in $e$, namely equations $3$ and $4$. Both have a factor $(e+1)$. Hence we see that $e=-1$ is a solution. And we are done: $$f=0; e=-1, c=1-a^2$$ solves all given equations.