System of four equations of four variables including second powers. I've been tasked with solving the following system of equations and it seems like I am stuck:
$$a-x^2=y$$$$a-y^2=z$$$$a-z^2=t$$$$a-t^2=x,$$where $a$ is a real number, for which $0\leq a\leq 1$. I thought the best way would be to subtract some equations from each other and the exploit $x^2-y^2=(x+y)(x-y)$. Even some estimates could be useful, since we have an estimate for $a$. However, I put this system of equations to WolframAlfa and the solutions (depending on $a$) looked very uneasy. In general, I have got very little experience in solving quadratic systems of equations like this one, could somebody please point me in the right direction? Thanks a lot! 

Because of the symmetry, it is natural to assume $x=y=z=t$, which gives you an easily solvable quadratic.  The solutions are $\frac 12(-1 \pm \sqrt{4a+1})$, which are real when $a \ge -\frac 14$, covering your region of interest.  Another approach is to assume $x=z, y=t$, which gives $a-(a-x^2)^2=x$ and the additional two solutions $x=\frac 12(1\pm \sqrt {4a-3}), y=\frac 12(1\mp \sqrt {4a-3})$, which are real when $a \ge \frac 34$.  If you don't assume the equality, you get the sixteenth degree polynomial of Dr. Sonnhard Graubner.  These will reduce the degree to $12$, but there is still a long way to go.
