Find all real solutions of the following equations: $$a+b=c^2$$ $$b+c=d^2$$ $$c+d=e^2$$ $$d+e=a^2$$ $$e+a=b^2$$
Solution: If $a=b=c=d=e$, then we have two real solutions as $$a=b=c=d=e=0$$ and $$a=b=c=d=e=2$$
Without loss of generality, assume $a\le b \le c \le d \le e$ and $a<e$. Additonally, $e\ge 0$, otherwise, $b^2=e+a<0$, implying that $b$ is not a real number.
Because $c\le d$ and $c+d=e^2$. Therefore $2c\le e^2$. We have $$c\le\dfrac{1}{2}e^2\tag{1}$$
Because $d\le e$ and $d+e=a^2$. Therefore $2d\le a^2$. We have $$d\le\dfrac{1}{2}a^2\tag{2}$$
Because $e^2=c+d$, applying inquality in $(1)$ and $(2)$, we have $$e^2\le\dfrac{1}{2}e^2+\dfrac{1}{2}a^2$$ Therefore $$e^2\le a^2\tag{3}$$
Because of the asumptions that $a<e$ and $e>=0$, we have $e\le-a$, and $a<0$. Therefore $b^2=e+a\le 0$. Because $b$ is real, we have $e+a=0$, and $b=0$.
However, because $c^2=a+b=a+0=a<0$, $c$ cannot be real. Therefore, the assumption of $a<e$ is false, we have $a=e$, implying $a=b=c=d=e$.
Therefore, the solutions to the problem are $$a=b=c=d=e=0$$ and $$a=b=c=d=e=2$$
Note: One strategy to solve a cyclic system of equations is to make certain assumption and then prove that the assumption would introduce a contradiction. In this problem, we think that all variable of real values are equal. We need to prove that if they are not equal, it would result in contradiction. Can you find real solutions of the following system of equations: $$3a=(b+c+d)^3$$ $$3b=(c+d+e)^3$$ $$3c=(d+e+a)^3$$ $$3d=(e+a+b)^3$$ $$3e=(a+b+c)^3$$