Find real solutions for the following equations: $$a+bcd = 2$$ $$b+cda=2$$ $$c+dab=2$$ $$d + abc=2$$
Solution: Because $a+bcd=2$, $b+cda=2$, we have $a+bcd=b+cda$. Factorizing it, we have $$(a-b)(cd-1)=0$$ Therefore either $a=b$ or $cd=1$.
Case 1: If $a=b$, we have $$a+ac^2=2$$ $$c+da^2=2$$ $$d+ca^2=2$$ We have $c+da^2=d+ca^2$. Factorizing it, we have $$(c-d)(a^2-1)=0$$ Therefore $c=d$ or $a^2=1$.
Case 1.1: if $c=d$, we have $c+ca^2=2$. Because $a+ac^2=2$, we have $a+ac^2=c+ca^2$. Factorizing it, we have $$(a-c)(ac-1)=0$$ Therefore, $a=c$ or $ac=1$.
Case 1.1.1 If $a=c$, then $a+a^3=2$. Factorizing it, we have $$(a-1)(a^2+a+2)=0$$ The only real solution to the above equation is $a=1$, implying $a=b=c=d=1$.
Cass 1.1.2: if $ac=1$, then $a+a=2$, we have $a=1$. Then $c=1$, implying $a=b=c=d=1$.
Case 1.2: If $a^2=1$, then $a=1$ or $a=-1$.
Case 1.2.1: If $a=1$, then, $b=1$, and $1+cd=2$ and $c+d=2$. implying $a=b=c=d=1$.
Case 1.2.2: if $a=-1$, then, $b=-1$, and $-1-cd=2$ and $c+d=2$, implying $a=b=c=-1$, $d=3$ or $a=b=d=-1$ and $c=3$.
Case 2: if $cd=1$, we have $a+b=2$. Therefore $c+da(2-a)=2$ and $d+ca(2-a)=2$. We have $c+da(2-a)=d+ca(2-a)$. Factorizing it, we have $$(a-1)^{2}(c-d)=0$$ We have $a=1$ or $c=d$.
Case 2.1: If $a=1$, then $b=1$, and $1+cd=2$ and $c+d=2$, implying $a=b=c=d=1$.
Case 2.2: If $c=d$, therefore $c^2=1$, which is similar to Case 1.2, with $a=b=c=d=1$; $a=3$, $b=c=d=-1$; or $a=c=d=-1$, $b=3$.
Therefore there are 5 solutions of (a, b, c, d) as $(1,1,1,1), (-1,-1,-1,3), (-1,-1,3,1), (-1,3,-1,-1), (3,-1,-1,-1)$.