Positive reals $a, b, c$ satisfies the following system of equations : $a=\frac{b+1}{c-1}, b=\frac{c-1}{a-1}, c=\frac{a-1}{b+1}$
Now prove or disprove that $a-b+c=4$.
This problem came out when I tried to prove the following APMO 2012 - inspired problem.
This answer proves that $a-b+c=4$.
Proof :
Since we have $a-1=\frac{c-1}{b}$ and $a-1=c(b+1)$, we get $$\frac{c-1}{b}=c(b+1)\implies \frac 1c=-b^2-b+1\implies c=\frac{1}{-b^2-b+1}$$ So, $$a=\frac{b+1}{c-1}=\frac{b+1}{\frac{1}{-b^2-b+1}-1}=\frac{(b+1)(-b^2-b+1)}{1-(-b^2-b+1)}=\frac{-b^2-b+1}{b}$$ So, we have $$\begin{align}&b=\frac{c-1}{a-1}=\frac{\frac{1}{-b^2-b+1}-1}{\frac{-b^2-b+1}{b}-1}=\frac{b(b^2+b)}{(-b^2-b+1)(-b^2-2b+1)} \\\\&\implies b=\frac{b(b^2+b)}{(-b^2-b+1)(-b^2-2b+1)} \\\\&\implies (-b^2-b+1)(-b^2-2b+1)=b^2+b \\\\&\implies b^4+3b^3-b^2-4b+1=0 \\\\&\implies (b-1)(b^3+4b^2+3b-1)=0 \\\\&\implies b=1\quad\text{or}\quad b^3+4b^2+3b-1=0\end{align}$$ If $b=1$, then we get $c=-1$ which is negative, so $b\not=1$.
Let $f(b)=b^3+4b^2+3b-1$. Then, we have $$f(-3)\lt 0\lt f(-2),f(-1)\lt 0,f(0)\lt 0\lt f\left(\frac 12\right)$$ So, $f(b)=0$ has the only one real root $b=b_0$ satisfying $0\lt b_0\lt\frac 12$.
Hence, the system has only one solution $$(a,b,c)=\left(\frac{-b_0^2-b_0+1}{b_0},b_0,\frac{1}{-b_0^2-b_0+1}\right)$$
Using $b_0^3+4b_0^2+3b_0-1=0$, we get $$\begin{align}a-b+c&=\frac{-b_0^2-b_0+1}{b_0}-b_0+\frac{1}{-b_0^2-b_0+1} \\\\&=\frac{(-b_0^2-b_0+1)^2-b_0^2(-b_0^2-b_0+1)+b_0}{b_0(-b_0^2-b_0+1)} \\\\&=\frac{2b_0^4+3b_0^3-2b_0^2-b_0+1}{-b_0^3-b_0^2+b_0} \\\\&=\frac{2b_0(-4b_0^2-3b_0+1)+3b_0^3-2b_0^2-b_0+1}{-b_0^3-b_0^2+b_0} \\\\&=\frac{-5b_0^3-8b_0^2+b_0+1}{-b_0^3-b_0^2+b_0} \\\\&=\frac{-5(-4b_0^2-3b_0+1)-8b_0^2+b_0+1}{(4b_0^2+3b_0-1)-b_0^2+b_0} \\\\&=\frac{12b_0^2+16b_0-4}{3b_0^2+4b_0-1} \\\\&=4\end{align}$$