FelixD wrote: Let $ a$, $ b$, $ c$ be real numbers such that for every two of the equations \[ x^2 + ax + b = 0, \quad x^2 + bx + c = 0, \quad x^2 + cx + a = 0\] there is exactly one real number satisfying both of them. Determine all possible values of $ a^2 + b^2 + c^2$. If I'm not mistaken,the possible values of $ a^2 + b^2 + c^2$ are $ 0$ and $ 6$.If my answer is correct ( however,I'm not sure about it,because I'm a bit tired now and I did all the work in mind ),I may post my solution tomorrow, since It's already time to sleep now

FelixD wrote: Let $ a$, $ b$, $ c$ be real numbers such that for every two of the equations \[ x^2 + ax + b = 0, \quad x^2 + bx + c = 0, \quad x^2 + cx + a = 0\] there is exactly one real number satisfying both of them. Determine all possible values of $ a^2 + b^2 + c^2$. We have two possibilities : 1) All three equations share a common root and so : $ x^2+ax+b=(x-u)(x-v)$ $ x^2+bx+c=(x-u)(x-w)$ $ x^2+cx+a=(x-u)(x-t)$ Solving this implies $ v=w=t=\frac{-u^3+u^2-u}{u^3+1}$ ad so no solution because we want exactly one common root between any two equations and not two). 2) All three equation share three roots in a circular manner and so : $ x^2+ax+b=(x-u)(x-v)$ $ x^2+bx+c=(x-v)(x-w)$ $ x^2+cx+a=(x-w)(x-u)$ Solving this gives $ u^4+u^3-4u^2-3u+2=0$ $ \iff$ $ (u+2)(u^3-u^2-2u+1)=0$ $ u=-2$ implies $ u=v=w=-2$ and does not fit the requirement "$ u,v,w$ pairwise different". And so $ u,v,w$ are the three roots of $ u^3-u^2-2u+1=0$ So $ u+v+w=1$ and $ uv+vw+wu=-2$ Then $ a^2+b^2+c^2=(u+v)^2+(v+w)^2+(w+u)^2$ $ =2(u+v+w)^2-2(uv+vw+wu)$ $ =6$ Hence the result : If such $ a,b,c$ exist, then $ \boxed{a^2+b^2+c^2=6}$ (and, indeed, they exist : choose $ (u,v,w)=(2\sin(\frac{\pi}{14}),2\sin(\frac{5\pi}{14}),-2\sin(\frac{3\pi}{14}))$) Notice also that $ a^2+b^2+c^2=0$ cant be an answer since then $ a=b=c=0$ and all equation have two roots in common ($ 0$ and $ 0$) and not exactly one

pco wrote: Notice also that $ a^2 + b^2 + c^2 = 0$ cant be an answer since then $ a = b = c = 0$ and all equation have two roots in common ($ 0$ and $ 0$) and not exactly one My solution is very similar to yours pco ( a bit simpler,I must say ).However,I still can't understand this last point.If $ a^2 + b^2 + c^2 = 0$ ( $ \Rightarrow a = b = c = 0$ ),then for every two of the equations : $ x^2 = 0,x^2 = 0$ and $ x^2 = 0$ there is exactly one real number which satisfies both of them,namely $ 0$

Then $ a=b=c=4$ works also, which means $ a^2+b^2+c^2=48$ is also a solution

Rafikafi wrote: Then $ a = b = c = 4$ works also, which means $ a^2 + b^2 + c^2 = 48$ is also a solution Quite good response, thanks The problem is always the same : does $ x^2 = 0$ have one root or two roots (one double). If you agree that $ x^2 = 0$ has two roots (as all quadratic, if we consider them in $ \mathbb C$), then all pair of equations have both roots equal.

Rafikafi wrote: Then $ a = b = c = 4$ works also, which means $ a^2 + b^2 + c^2 = 48$ is also a solution You're right,thank you ( BTW,$ a^2 + b^2 + c^2 = 48$ is the only left solution ) ! Indeed,in my solution I used uncarefully $ a = b = c \Rightarrow a = b = c = 0$ which is not true ( Note that I did all the work in mind and I was really tired as noted above ),because $ a = b = c \Rightarrow$ The equation $ x^2 + ax + a = 0$ has only one real solution $ \Rightarrow$ The discriminant $ = a^2 - 4a = 0 \Rightarrow a = 0$ or $ a = 4 \Rightarrow a = b = c = 0$ or $ a = b = c = 4$. In conclusion,the only possible values of $ a^2 + b^2 + c^2$ are $ 0,6$ and $ 48$ ( I consider that the equation $ (x - a)^2 = 0$ has only one solution $ a$ )