Can you solve this RMO question? Part 2

$Subtracting\quad the\quad 2\quad equations\quad gives:\\ { a }^{ 2 }-{ b }^{ 2 }=c(b-a)\Longrightarrow \boxed { a+b=-c\quad or\quad a=b } \\ \\ Case\quad 1\\ If\quad a=b,\quad then\quad plugging\quad into\quad the\quad first\quad equation\quad we\quad get,\\ { a }^{ 2 }=ac+1\Longrightarrow c=\frac { { a }^{ 2 }-1 }{ a } =\quad a\quad -\frac { 1 }{ a } \\ But\quad c\quad is\quad an\quad integer\quad which\quad implies\quad a=\pm 1=b\quad and\quad c=0\\ Thus\quad 2\quad ordered\quad triplets\quad are\quad (1,1,0)\quad and\quad (-1,-1,0)\\ \\ Case\quad 2\\ Plugging\quad the\quad value\quad of\quad c\quad into\quad the\quad first\quad equation,\quad we\quad get\\ { a }^{ 2 }+{ b }^{ 2 }+ab=1\Longrightarrow { (a+b) }^{ 2 }=1+ab\Longrightarrow ab+1\ge 0\Longrightarrow ab\ge -1\quad (A)\\ Now\quad note\quad that\quad 1-ab={ a }^{ 2 }+{ b }^{ 2 }\Longrightarrow 1-ab\ge 0\Longrightarrow ab\le 1(B)\\ \\ (A),(B)\Longrightarrow ab\in \left\{ -1,0,1 \right\} \Longrightarrow a=0,b=\pm 1,\quad b=0,a=\pm 1,\quad a=-1,b=1,\quad a=1,b=-1,\quad a=1,b=1\quad and\quad a=-1,b=-1\\ Recall\quad c=-(a+b)\\ Thus\quad all\quad ordered\quad triplets\quad are\quad as\quad follows:-\\ 1)(1,1,0)\\ 2)(-1,-1,0)\\ 3)(0,-1,1)\\ 4)(0,1,-1)\\ 5)(1,0,-1)\\ 6)(-1,0,1)\\ 7)(-1,1,0)\\ 8)(1,-1,0)\\ Thus\quad desired\quad answer=\quad \boxed { 8 } \\ \\ \\ \\ \\$

Subtracting the equations yields $a^{2} - b^{2} = c(b - a) \begin{cases}{a + b + c = 0} \\ {a = b} \end{cases}$

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$a = b \Rightarrow \boxed{(a , b , c) = (1 , 1 , 0) , (-1 , -1 , 0)}$ .

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Adding the two equations gives $a^{2} + b^{2} = c(a + b) + 2$ .

$a + b + c = 0 \Rightarrow a + b = -c$ ,

Replacing gives $a^{2} + b^{2} + c^{2} = 2$

Now this gives us the set of solutions as $\boxed{(0 , 1 , - 1)}$

So from the set we can obtain $6$ solutions namely $\boxed{(0 , -1 , 1) , (0 , 1 , -1) , (1 , 0 , -1) , (1 , -1 , 0) , (-1 , 0 ,1) , (-1 , 1 , 0)}$ and we had two solutions earlier so altogether $\boxed{8}$ solutions.