Bulgaria National Olympiad Problem 1

$Let\quad a,b\quad be\quad the\quad roots\quad to\quad the\quad first\\ quadratic\quad and\quad a,c\quad be\quad the\quad roots\quad to\quad \\ the\quad second\quad quadratic.\\ \therefore \quad { a }^{ 2 }-2ma-4({ m }^{ 2 }+1)=0\\ { and\quad a }^{ 2 }-4a-2m({ m }^{ 2 }+1)=0.\quad \\ { a }^{ 2 }=\frac { \begin{vmatrix} -2m & -4({ m }^{ 2 }+1) \\ -4 & -2m({ m }^{ 2 }+1) \end{vmatrix} }{ \begin{vmatrix} 1 & -2m \\ 1 & -4 \end{vmatrix} } =\frac { 4({ m }^{ 2 }+1)({ m }^{ 2 }-4) }{ 2(m-2) } \\ \Rightarrow a^{ 2 }=2({ m }^{ 2 }+1)(m+2).\\ a=\quad -\quad \frac { \begin{vmatrix} 1 & -4({ m }^{ 2 }+1) \\ 1 & -2m({ m }^{ 2 }+1) \end{vmatrix} }{ \begin{vmatrix} 1 & -2m \\ 1 & -4 \end{vmatrix} } =\frac { 2({ m }^{ 2 }+1)(m-2) }{ 2(m-2) } ={ m }^{ 2 }+1.\\ \therefore \quad ({ m }^{ 2 }+1)^{ 2 }=2({ m }^{ 2 }+1)(m+2)\\ \Rightarrow { m }^{ 2 }-2m-3=0.\\ m=3,-1.\quad \\ If\quad we\quad put\quad m=-1,\quad we\quad can\quad see\quad that\quad a\quad is\quad \\ repeated\quad twice:\quad In\quad the\quad second\quad quadratic,\\ a=c=2\quad for\quad m=-1.\quad If\quad we\quad put\quad m=3,\quad we\quad \\ can\quad see\quad that\quad a\quad is\quad repeated\quad as\quad 10\quad only\quad once\\ in\quad each\quad quadratic\quad and\quad b=-4;\quad c=-6.\quad \\ \therefore \quad m=\boxed { 3 } \quad$