hello, i have found $$x_1=x_2=x_3=x_4=0$$or $$x_1=x_2=x_3=x_4=2$$Sonnhard.

Yes. This is true. Can you write your solution?

Well, it has infinitely many solutions in real numbers

Emirhan wrote: Yes. This is true. Can you write your solution? hello, then tell us your proof! Sonnhard.

Wait. We can suppose $x_1 \geq x_2 \geq x_3 \geq x_4$ $?$ It looks like symmetrical, but its looks not (?)

The only solutions of the system of equations $(1) \;\; x_1 + x_2 = x_3^2$ $(2) \;\; x_2 + x_3 = x_4^2$ $(3) \;\; x_3 + x_4 = x_1^2$ $(4) \;\; x_4 + x_1 = x_2^2$ are $x_1=x_2=x_3=x_4=k$, where $k \in \{0,2\}$. Proof: Adding (1)-(4), the result is $(5) \;\; \sum_{i=1}^4 (x_i - 1)^2 = 4$. We may WLOG assume (due to the cyclic variables $x_1,x_2,x_3,x_4,x_5$ in (1)-(4)) $|x_1 - 1| \geq |x_i - 1|$ for all $i \in \{2,3,4\}$. Hence by (5) we have $|x_1 - 1| \geq 1$, i.e. $x_1 \leq 0$ or $x_1 \geq 2$. Subtracting (1) from (2) we obtain $(6) \;\; x_3 - x_1 = x_4^2 - x_3^2$. Moreover $x_4 = x_1^2 - x_3$ by (3), which inserted in (6) give us $x_3 - x_1 = (x_1^2 - x_3)^2 - x_3^2 = x_1^4 - 2x_1^2x_3$, yielding $(7) \;\; x_3 = \frac{x_1^4 + x_1}{2x_1^2 + 1} = f(x_1)$. Repeating this deduction, we obtain $x_1 = f(x_3)$. Hence by (7) we have $x_3 + x_1 = f(x_1) + f(x_3)$, i.e. $g(x_1) + g(x_3) = 0$, where $g(x) = f(x) - x = \frac{x^4 + x}{2x^2 + 1} - x = \frac{x^3(x - 2)}{2x^2 + 1}$. Observe that $g(x)<0$ iff $0 < x < 2$. With this fact in mind we consider the following two cases: Case 1: $x_1<0$. Then $g(x_3) = -g(x_1) < 0$, yielding $0 < x_3 < 2$. Hence $x_1 = f(x_3) = \frac{x_3^4 + x_3}{2x_3^2 + 1} > 0$. Case 2: $x_1>2$. Then $g(x_3) = -g(x_1) < 0$, yielding $0 < x_3 < 2$. Hence $x_1 = f(x_3) = \frac{x_3^4 + x_3}{2x_3^2 + 1} > 2$. Therefore $0 < x_3^4 + x_3 - 2(2x_3^2 + 1) = x_3^2(x_3 + 2)(x_3 - 2) + (x_3 - 2) = (x_3^3 + 2x_3^2 + 1)(x_3 - 2) < 0$ since $0 < x _3 < 2$. In both cases we end up with a contradiction. Consequently, since $x_1 \leq 0$ or $x_1 \geq 2$, we find that $x_1 = k \in \{0,2\}$. Thus $x_3 = f(x_1) = f(k) = k$, yielding $x_2 = x_4 = k$ by (1) and (3). Conclusion: The only solutions of the system of equations (1)-(4) are $x_1=x_2=x_3=x_4=0$ and $x_1=x_2=x_3=x_4=2$. q.e.d.