Cyclical System of Quadratic Equations with Four Unknowns

Question

Solve this system of equations in ℝ (k ∈ [0,1]):

$\ k-x^2=\ y$

$\ k-y^2=\ z$

$\ k-z^2=\ u$

$\ k-u^2=\ x$

I saw the trivial solution (0,0,0,0), then I have tried substitiuin and got the a quation of fourth power with two variables, then I trie to put it into A*B=0 and got that x=(+-1+-sqrt(1+4k))/2. — LeoBern, Oct 22 '16 at 18:09
Presumably $k$ is a parameter, the remaining variables unknown. Then for $k\ge(-1/4)$ there will be solutions with $x=y=z=u$. Can you see that? — Oscar Lanzi, Oct 22 '16 at 19:46

score 1 · Answer 1 · answered Oct 22 '16 at 19:32

1

HINT:

When $\text{k}-x^2=y$ and $\text{k}-u^2=x$: $$\text{k}-\left(\text{k}-u^2\right)^2=y$$
When $\text{k}-z^2=u$ and $\text{k}-y^2=z$: $$\text{k}-\left(\text{k}-y^2\right)^2=u$$

So:

$$\text{k}-\left(\text{k}-\left(\text{k}-\left(\text{k}-u^2\right)^2\right)^2\right)^2=u$$

answered Oct 22 '16 at 19:32

Jan Eerland

It woul lead to the equation with u to the power more than ten. – LeoBern Oct 22 '16 at 19:43
@LeoBern For example, when $k=0$, then we get: $-u^{16}=u$. – Jan Eerland Oct 22 '16 at 19:57
@JanEerland: Interestingly, Ramanujan considered the related equation$$\text{k}+\left(\text{k}+\left(\text{k}+\left(\text{k}+x^2\right)^2\right)^2\right)^2=x$$ and solved it in radicals. Kindly see this post. – Tito Piezas III Oct 23 '16 at 13:30
@LeoBern: Of course, as one has to raise to the $2$nd power four times. Thus you need to solve a factorable $16$-deg equation. Fortunately, Ramanujan did related work. See the link I posted above. – Tito Piezas III Oct 23 '16 at 13:33

1 Answers1