[tex]Avem~b= \frac{a+c}{2}. \\ \\ \frac{a}{b+1}= \frac{b}{c+1}= \frac{c}{2} \Leftrightarrow \frac{a}{ \frac{a+c}{2}+1 } = \frac{ \frac{a+c}{2} }{c+1}= \frac{c}{2} \Leftrightarrow \frac{a}{ \frac{a+c+2}{2} }= \frac{a+c}{2(c+1)}= \frac{c}{2} \Leftrightarrow \\ \\ \Leftrightarrow \frac{2a}{a+c+2}= \frac{a+c}{2(c+1)}= \frac{c}{2} . \\ \\ Din~\frac{a+c}{2(c+1)} = \frac{c}{2} ~rezulta~a+c= \frac{2c(c+1)}{2} \Leftrightarrow a+c=c^2+c \Rightarrow a=c^2. [/tex]
[tex]Daca~c=0 \Rightarrow a=b=0. \\ \\ Daca~c \neq 0,~atunci,~din~ \frac{2a}{a+c+2}= \frac{c}{2}~rezulta~a+c= \frac{4a}{c}-2 \Leftrightarrow \\ \\ c^2+c= \frac{4c^2}{c}-2 \Leftrightarrow c^2+c =4c-2 \Leftrightarrow c^2=3c-2. \\ \\ Deci~c^2-3c+2=0 \Leftrightarrow (c-1)(c-2)=0 \Rightarrow c \in \{1;2 \}.[/tex]
[tex]c=1 \Rightarrow a=1 ~si~b= \frac{1+1}{2}=1. \\ \\ c=2 \Rightarrow a= 2^2=4 ~si~b= \frac{ 4+2}{2}=3. \\ \\ Solutie:~(a,b,c) \in \{(0,0,0);(1,1,1);(4,3,2)\}.[/tex]
