# Find all real numbers $a_1, a_2, a_3, b_1, b_2, b_3$.

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.

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 $$(a_i, b_i)=(1, -2)$$.
• 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? – glopf Jun 22 at 23:02
• 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? – Macavity Jun 23 at 16:57