Let $a,b,c,d,e,f$ be real numbers. Solve the following system of equations:

$\begin{cases}a+b=-e\\ ab=f\\ c+d=-a\\ cd=b\\ e+f=-c\\ ef=d\end{cases}$

I got stuck trying to solve this problem by decreasing the number of variables using consecutive equations. What I tried next was to subtract the first equation from the second one, obtaining $ab-a-b=f+e=-c\implies (a-1)(b-1)=-(c-1)$. Analogously we can get $(c-1)(d-1)=-(e-1)$ and $(e-1)(f-1)=-(a-1)$. Now multiplying these equations all togethes gives, that either one of $a-1, c-1, e-1$ equals $0$ or $(b-1)(d-1)(f-1)=-1$ and here's where I got stuck not being able to deal with the last case.

By the way, all of the first case in all possibilities gives one solution $(1,-2,1,-2,1,-2)$. I feel like the other one should give the zeros-only solution.

Don't know how useful are these approaches... I'd be very grateful for any help :)

  • $\begingroup$ I think I do not understand what do you mean by that... $\endgroup$ – Thomas Gaffney Oct 14 '18 at 18:46
  • $\begingroup$ It is not the solution though. $\endgroup$ – Thomas Gaffney Oct 14 '18 at 18:50
  • $\begingroup$ I doesn't however prove that these are the only solutions. $\endgroup$ – Thomas Gaffney Oct 14 '18 at 18:54

Eliminating the variables $$b,c,d,e,f$$ we get for $a$ the equation



$$a=b=c=d=e=f=0$$ is also one solution.

  • $\begingroup$ I see, but would you like to show me a sketch of the process of elimination? $\endgroup$ – Thomas Gaffney Oct 14 '18 at 18:43
  • $\begingroup$ Is there any smart way to do this? $\endgroup$ – Thomas Gaffney Oct 14 '18 at 18:45
  • $\begingroup$ I'd be really grateful for any explanation $\endgroup$ – Thomas Gaffney Oct 17 '18 at 17:34

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