A question on algebra

Question

Given that $$a+{1\over b}=b+{1\over c}=c+{1\over a}=p$$ $a,b,c$ are distinct real numbers What is the value of $p$ in terms of $a,b,c$?

1. $abc$
2. $-abc$
3. $a+b+c$
4. $ab+bc+ca$

Answer 1

I was trying to solve for all possible $(p,a,b,c)\in\mathbb{C}^4$ (not requiring that $a$, $b$, and $c$ be distinct, nor real). This is a long solution, but maybe there is a simpler way to deal with the problem.

Let $a,b,c\in\mathbb{C}$ be such that there exists $p\in\mathbb{C}$ for which $$pb=ab+1\,,$$ $$pc=bc+1\,,$$ and $$pa=ca+1\,.$$ Write $x:=a+b+c$, $y:=bc+ca+ab$, and $z:=abc$. Adding all three equations above yields $$px=y+3\,.\tag{1}$$ Observe that $$pbc=abc+c\,,$$ $$pca=abc+a\,,$$ and $$pab=abc+b\,.$$ Adding these three equation leads to $$py=3z+x\,.\tag{2}$$ Finally, we also have $$pabc=a^2bc+ca=az+ca\,,$$ $$pabc=ab^2c+ab=bz+ab\,,$$ and $$pabc=abc^2+bc=cz+bc\,.$$ Sum these three equations to get $$3pz=xz+y\,.\tag{3}$$ From (1) and (2), $$y(y+3)=pxy=x(3z+x)\,.\tag{*}$$ From (1) and (3), $$3z(y+3)=3pxz=x(xz+y)\,.\tag{#}$$

If $x=0$, then $y=-3$ by (1). Hence, (2) implies $z=-p$. Then, (3) gives $$-3p^2=3pz=xz+y=-3$$ or $p=\pm1$. That is, $$(p,x,y,z)=(\pm1,0,-3,\mp1)\,.$$

If $x\neq 0$ and $y=0$, then $x=-3z$ by (2). Using (1), we get $px=3$ or $pz=-1$. By (3), we see that $$-3=3pz=xz=-3z^2$$ or $z=\pm 1$. That is, $$(p,x,y,z)=(\pm1,\pm3,0,\mp1)\,.$$

If $x\neq 0$ and $y=-3$, then (1) shows that $p=0$. Thus, (2) gives $x=-3z$, and (3) gives $$0=3pz=xz+y=-3z^2-3\,.$$ That is, $z=\pm\text{i}$. This means $$(p,x,y,z)=(0,\pm3\text{i},-3,\mp\text{i})\,.$$

Suppose now that $x\neq 0$ and $y\notin\{-3,0\}$. Using (*) and (#), we have $$x(3z+x)\cdot3z(y+3)=y(y+3)\cdot x(xz+y)\,.$$ As $x\neq 0$ amd $y\neq -3$, we obtain $$3z(3z+x)=y(xz+y)\text{ or }z(y-3)x=-(y-3z)(y+3z)\,.$$ If $y=3$, then $y-3z=0$ or $y+3z=0$, which gives $z=\pm 1$. That is, $$(p,x,y,z)\in\big\{(2,3,3,1), (-1,-6,3,1), (1,6,3,-1),(-2,-3,3,-1)\big\}\,.$$ If $y\neq 3$, then $$x=-\frac{(y-3z)(y+3z)}{z(y-3)}\,,$$ where we note that $z\neq 0$.

Plug this into (*) to get $$y(y+3)=-\frac{(y-3z)(y+3z)}{z(y-3)}\,\left(3z-\frac{(y-3z)(y+3z)}{z(y-3)}\right)\,.$$ That is, $$y(y+3)=\frac{y(y-3z)(y+3z)(y-3z^2)}{z^2(y-3)^2}\,.$$ Because $y\neq 0$, we obtain $$z^2(y-3)^2(y+3)=(y-3z)(y+3z)(y-3z^2)\,.$$ This is equivalent to $$(z-1)(z+1)(y^3-27z^2)=0\,.$$ If $z=-1$, then we obtain the family $$(p,x,y,z)=(+1,t,-3-t,-1)\,,$$ where $t\in\mathbb{C}\setminus\{0,\pm3\}$. If $z=-1$, then we obtain the family $$(p,x,y,z)=(-1,t,-3+t,+1)\,,$$ where $t\in\mathbb{C}\setminus\{0,\pm3\}$. If $y^3=27z^2$, then we obtain the family $$(p,x,y,z)=\left(t+\frac{1}{t},3t,3t^2,t^3\right)$$ for $t\in\mathbb{C}\setminus\left\{0,\dfrac{\pm1\pm\sqrt{3}\text{i}}{2},\pm1\right\}$.

In conclusion, all possible $(p,x,y,z)\in\mathbb{C}^4$ are

1. $(+1,t,-3-t,-1)$ with $t\in\mathbb{C}\setminus\left\{\dfrac{+3\pm3\sqrt{3}\text{i}}{2}\right\}$,
2. $(-1,t,-3+t,+1)$ with $t\in\mathbb{C}\setminus\left\{\dfrac{-3\pm3\sqrt{3}\text{i}}{2}\right\}$, and
3. $\left(t+\dfrac{1}{t},3t,3t^2,t^3\right)$ with $t\in\mathbb{C}\setminus\{0\}$.

Note that in the third case, we obtain $$(p,x,y,z,a,b,c)=\left(t+\dfrac{1}{t},3t,3t^2,t^3,t,t,t\right)\text{ where }t\in\mathbb{C}\setminus\{0\}\,.$$ If we demand that $a$, $b$, and $c$ be distinct, then the only possibilities are the first two cases, where $p=-z=-abc$.

The first two cases yield pairwise distinct values of $a$, $b$, and $c$. Therefore, the solutions to the original problem are $$(p,x,y,z,a,b,c)=(s,st,t-3,-s,s\alpha_t,s\beta_t,s\gamma_t)$$ with $s\in\{\pm1\}$ and $t\in\mathbb{C}\setminus\left\{\dfrac{3\pm3\sqrt{3}\text{i}}{2}\right\}$, where $\alpha_t,\beta_t,\gamma_t$ are the roots of the polynomial $$Q_t(u):=u^3-tu^2+(t-3)u+1\,.$$ The discriminant of $Q_t(u)$ is $$D(t):=t^4-6t^3+27t^2-54t+81\,,$$ and we have $$D(t)=\left(t^2-3t+9\right)^2=\left(\left(t-\frac{3}{2}\right)^2+\frac{27}{4}\right)^2>0$$ for $t\in\mathbb{R}$. Hence, when $t\in\mathbb{R}$, we are guaranteed that $\alpha_t,\beta_t,\gamma_t$ are three distinct real numbers. (Indeed, the four complex roots of $D(t)$ are $\dfrac{3\pm3\sqrt{3}\text{i}}{2}$, each with multiplicity $2$, and so $Q_t(u)$ has no repeated roots when $t$ is not a root of $D(t)$.)