System of four equations of four variables including second powers.

Question

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!

Answer 1

from the first equation we get $$a-(a-x^2)^2=z$$ then we obtain $$a-(a-(a-x^2)^2)^2=t$$ $$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)=t$$ and finally we obtain $$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)^2=x$$ and now you can try to get $x$ and this is what i'm get: $$a^8+a^7 \left(-8 x^2-4\right)+a^6 \left(28 x^4+24 x^2+6\right)+a^5 \left(-56 x^6-60 x^4-24 x^2-6\right)+a^4 \left(70 x^8+80 x^6+36 x^4+16 x^2+5\right)+a^3 \left(-56 x^{10}-60 x^8-24 x^6-16 x^4-8 x^2-2\right)+a^2 \left(28 x^{12}+24 x^{10}+6 x^8+8 x^6+4 x^4+1\right)+a \left(-8 x^{14}-4 x^{12}-2 x^8-1\right)=x \left(-x^{15}-1\right)$$

Answer 2

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.

Answer 3

Given the system, $$a-x^2=y\\a-y^2=z\\a-z^2=t\\a-t^2=x$$ Let $(x,\,y,\,z,\,t) = (-x_1,\,-x_2,\,-x_3,\,-x_4)$ and we get, $$x_1^2 = x_2+a\\ x_2^2 = x_3+a\\ x_3^2 = x_4+a\\ x_4^2 = x_1+a$$ This system was considered by Ramanujan, in particular for the case $a=5$, yielding the infinitely nested radical,

$$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$ and similar expressions for the three other $x_i$. However it can be solved in radicals for any $a$. See this post for more details.