# System of six equations in real numbers

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 :)

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

Eliminating the variables $$b,c,d,e,f$$ we get for $$a$$ the equation
$$(a-1)(a^6+a^5+2a^4+3a^3+6a^2-a+1)=0$$
$$a=b=c=d=e=f=0$$ is also one solution.