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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!

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  • $\begingroup$ are $$x,y,z,t$$ the variables? $\endgroup$ – Dr. Sonnhard Graubner Nov 23 '16 at 16:47
  • $\begingroup$ Yes, $x,y,z,t$ are the variables. $\endgroup$ – pie314 Nov 23 '16 at 16:51
  • $\begingroup$ Is this in fact a contest problem? If so, one would expect a cute solution, but nobody has found one yet. $\endgroup$ – Ross Millikan Nov 23 '16 at 18:45
  • $\begingroup$ Yes it is, I used tag "contest-math". $\endgroup$ – pie314 Nov 23 '16 at 20:17
  • $\begingroup$ This system was considered by Ramanujan. I'm sure he would have won that contest blind-folded. See answer below. $\endgroup$ – Tito Piezas III Nov 25 '16 at 12:29
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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: $$a^8+a^7 \left(-8 x^2-4\right)+a^6 \left(28 x^4+24 x^2+6\right)+a^5 \left(-56 x^6-60 x^4-24 x^2-6\right)+a^4 \left(70 x^8+80 x^6+36 x^4+16 x^2+5\right)+a^3 \left(-56 x^{10}-60 x^8-24 x^6-16 x^4-8 x^2-2\right)+a^2 \left(28 x^{12}+24 x^{10}+6 x^8+8 x^6+4 x^4+1\right)+a \left(-8 x^{14}-4 x^{12}-2 x^8-1\right)=x \left(-x^{15}-1\right)$$

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    $\begingroup$ This doesn't seem like something from which I could get $x$ easily. $\endgroup$ – pie314 Nov 23 '16 at 17:09
  • $\begingroup$ you have asked how to solve the system, i think there is only a numerical way for a value of $a$ $\endgroup$ – Dr. Sonnhard Graubner Nov 23 '16 at 17:12
  • $\begingroup$ Yes, but there is that condition for $a$, it must simplify the solution somehow, like to prove that if $0\leq a\leq1$, then all other solutions but the trivial ones obtained from taking $x=y=z=t$ would not be real or something. $\endgroup$ – pie314 Nov 24 '16 at 13:03
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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.

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  • $\begingroup$ Thanks a lot, this is something I was looking for. Honestly it haven't cross my mind I could just take $x=y=z=t$. $\endgroup$ – pie314 Nov 23 '16 at 17:33
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    $\begingroup$ May be worth noting that symmetry arguments are somewhat speculative. For example the following also has the assymetric solutions $(0,1)$ and $(1,0)$: $$ \begin{cases} 1 - x^2 = y \\ 1 - y^2 = x \end{cases} $$ $\endgroup$ – dxiv Nov 23 '16 at 17:49
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    $\begingroup$ @dxiv: absolutely, but symmetric solutions are often easier to find. They are roots of the full polynomial, so can be divided out. Sometimes one can prove that the symmetric solutions are the only ones, sometimes not. Here they seem not to be. $\endgroup$ – Ross Millikan Nov 23 '16 at 18:38
  • $\begingroup$ @pie314: note that these are some solutions, not all solutions. With great luck, they may be the ones you want. $\endgroup$ – Ross Millikan Nov 23 '16 at 18:46
  • $\begingroup$ Yes, of course. I wrote that the solutions provided by Wolfram Alpha were uneasy looking, there indeed are some other very ugly solutions. I was thinking that because of condition for $a$, these might be the onle ones. $\endgroup$ – pie314 Nov 23 '16 at 19:08
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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.

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