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Find all real numbers $a_1, a_2, a_3, b_1, b_2, b_3$ such that for every $i\in \lbrace 1, 2, 3 \rbrace$ numbers $a_{i+1}, b_{i+1}$ are distinct roots of equation $x^2+a_ix+b_i=0$ (suppose $a_4=a_1$ and $b_4=b_1$).

There are many ways to do it but I've really wanted to finish the following idea:

From Vieta's formulas we get:

\begin{align} \begin{cases} a_1+b_1=-a_3 \ \ \ \ \ \ \ \ (a) \\a_2+b_2=-a_1\ \ \ \ \ \ \ \ (b)\\a_3+b_3=-a_2\ \ \ \ \ \ \ \ (c)\\a_1b_1=b_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (d)\\a_2b_2=b_1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (e)\\a_3b_3=b_2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (f)\end{cases} \end{align} First we notice that each $b_i$ is nonzero. Indeed, suppose $b_1=0$. Then from (d) and (f) we deduce that $b_3=0$ and $b_2=0$, so from (a), (b), (c) we get $a_1=-a_3=-(-a_2)=-(-(-a_1))=-a_1$, hence $a_1=0$ which is impossible.

Now, from (a), (b), (c), (d), (e), (f) we obtain: \begin{align} \begin{cases} a_1+b_1-a_1b_1=-a_3-b_3 \ \ \ \ \ \ \ \ \\a_2+b_2-a_2b_2=-a_1-b_1\ \ \ \ \ \ \ \ \\a_3+b_3-a_3b_3=-a_2-b_2\end{cases}, \end{align} so: \begin{align} \begin{cases} (b_1-1)(a_1-1)=1-a_2 \ \ \ \ \ \ \ \ \\(b_2-1)(a_2-1)=1-a_3\ \ \ \ \ \ \ \ \\(b_3-1)(a_3-1)=1-a_1\end{cases}. \end{align} Therefore: \begin{align*} (b_1-1)(b_2-1)(b_3-1)(a_1-1)(a_2-1)(a_3-1)=(1-a_1)(1-a_2)(1-a_3), \end{align*} which implies: \begin{align*} \bigl((a_1-1)(a_2-1)(a_3-1)\bigr)\bigl((b_1-1)(b_2-1)(b_3-1)+1\bigr)=0. \end{align*}

I got stuck here. Is it possible to prove that in this case $b_i=0$ is the only solution to equation $(b_1-1)(b_2-1)(b_3-1)=-1$ or maybe get contradiction in some other way? If so, we can assume that $a_1=1$ and from here we can easily show that also $a_2=a_3=1$, so $b_1=b_2=b_3=-2$.

Since $(b_1-1)(b_2-1)(b_3-1)>-1$ for every $b_i>0$ and $(b_1-1)(b_2-1)(b_3-1)<-1$ for every $b_i<0$, it suffices to prove that the signs of $b_1, b_2, b_3$ can't be different but I don't know how to do it. I also found out that $(b_1+1)^2+(b_2+1)^2+(b_3+1)^2=3$, so $b_i\in [-\sqrt{3}-1, \sqrt{3}-1]$ but I don't know if we can use it somehow.

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Well, here is a way to proceed after you note none of the $b_i$ are zero. Hints:

1) Multiplying the last three equations, $(d)\times (e)\times (f)$ gives $a_1a_2a_3=1$.

2) Now, multiplying just any two among those, e.g. $(d)\times (e)$ and using the result 1) above gives $a_2b_2=a_3b_3$, by symmetry and simplification using $(d), (e), (f)$ gives $b_i = b$, for some non-zero constant $b$, and hence $a_i=1$.

3) Now it is easy to conclude $b=-2$ from any of the first three equations. Hence $(a_i, b_i)=(1, -2)$.

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  • $\begingroup$ Yes I've also done this but I was specifically asking about the idea with equation $(b_1-1)(b_2-1)(b_3-1)=-1$. Does it mean it's not possible to finish that solution? $\endgroup$ – glopf Jun 22 at 23:02
  • $\begingroup$ By itself, $(b_1-1)(b_2-1)(b_3-1)=-1$ does not allow one to conclude anything useful, you have to use some among equations $(a)-(f)$ in addition to draw a contradiction. Then why not solve $(a)-(f)$, which is what is really needed anyway? $\endgroup$ – Macavity Jun 23 at 16:57

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