Solving $x_1+x_2=x_3^2, x_2+x_3=x_4^2, x_3+x_4=x_5^2,x_4+x_5=x_1^2, x_5+x_1=x_2^2$ in reals

There's probably a more clever solution.

Case 1: Suppose one of the $x_i$ is $0$: say $x_1$ is $0$. Then, $$ x_2-x_4=(x_2+x_3)-(x_3+x_4)=x_4^2-x_5^2=(x_4-x_5)(x_4+x_5)=(x_4-x_5)x_1^2=0 $$ so that $x_2=x_4$. This implies $$ -x_2=x_1-x_2=x_1-x_4=(x_1+x_5)-(x_4+x_5)=x_2^2-x_1^2=(x_2-x_1)(x_2+x_1)=x_2^2. $$ This means $x_2=0$ or $x_2=-1$. But $x_3^2=x_1+x_2=x_2$ so $x_2\geq 0$ therefore we infer that $x_2=x_4=0$. From here, it's easy to see that $x_3=x_5=0$ as well.

Case 2: each $x_i$ is nonzero. Label the given equations as (1)-(5). From (1) and (2), $$ x_1-x_3=(x_3-x_4)x_5^2. $$ Suppose $x_1>x_3$, then because $x_5^2>0$, we have $x_3>x_4$. In particular, we have $x_1>x_4$, which implies $x_2>x_1$ thanks to equations (4) and (5). So we have $x_2>x_1>x_3>x_4$. Equations (2) and (3) now imply that $x_4>x_5$. But then equations (3) and (4) imply $x_5>x_1$. So you have the impossibility $x_2>x_1>x_3>x_4>x_5>x_1$.

Supposing $x_1<x_3$ will lead to a similar contradiction as in the previous paragraph.

It must be that $x_1=x_3$ so that $x_3=x_4$. But then $x_1=x_4$ implies $x_1=x_2$. In turn, $x_2=x_4$ implies $x_4=x_5$. So $x_1=x_2=x_3=x_4=x_5=x$ for some nonzero $x$. You can now easily see that $x=2$.

Conclusion: 2 solutions $x_i=0$ $\forall i$ and $x_i=2$ $\forall$ $i$.

Assume, $x_1$ is the biggest of $x_1, x_2, x_3, x_4, x_5$. Then $x_1+x_2 = x_3^2 \ge 0$ and $x_1 \ge x_2$, therefore $x_1\ge 0$ and $x_1^2\ge x_2^2$.

Hence, $x_4+x_5 \ge x_5 + x_1$, therefore $x_4 \ge x_1$, or $x_4=x_1$. We obtained that $x_4$ is also the biggest of those numbers. In the same way you prove, that $x_4=x_2$, then $x_2 = x_5%$ and $x_5=x_1$.

So all numbers are equal to each other and they are solutions of $x^2-2x=0$.