(For those not in the know, CCC is complex/complex/complex)

We show $(p,q) = (-3,3)$ is the only possible ordered pair. Setup for proof Let us denote $a = y/x$, $b = z/y$, $c = x/z$, where $x$, $y$, $z$ are nonzero complex numbers. Then \begin{align*} p + 3 &= 3 + \sum_{\text{cyc}} \left( \frac xy + \frac yx \right) = 3 + \frac{x^2(y+z) + y^2(z+x) + z^2(x+y)}{xyz} \\ &= \frac{(x+y+z)(xy+yz+zx)}{xyz}. \\ q - 3 &= -3 + \sum_{\text{cyc}} \frac{y^2}{zx} = \frac{x^3+y^3+z^3-3xyz}{xyz} \\ &= \frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{xyz}. \end{align*}It follows that \begin{align*} {\mathbb R} &\ni 3(p+3) + (q-3) \\ &= \frac{(x+y+z)(x^2+y^2+z^2+2(xy+yz+zx))}{xyz} \\ &= \frac{(x+y+z)^3}{xyz}. \end{align*}Now, note that if $x+y+z = 0$, then $p = -3$, $q = 3$ so we are done. Main proof We will prove that if $x+y+z \neq 0$ then we contradict either the hypothesis that $a,b,c \notin {\mathbb R}$ or that $a$, $b$, $c$ do not have absolute value $1$. Scale $x$, $y$, $z$ in such a way that $x+y+z$ is nonzero and real; hence so is $xyz$. Thus, as $p+3 \in {\mathbb R}$, we conclude $xy+yz+zx \in {\mathbb R}$ as well. Hence, $x$, $y$, $z$ are the roots of a cubic with real coefficients. Thus, either all three of $\{x,y,z\}$ are real (which implies $a,b,c \in {\mathbb R}$), or two of $\{x,y,z\}$ are a complex conjugate pair (which implies one of $a$, $b$, $c$ has absolute value $1$). Both of these were forbidden by hypothesis. Construction As we saw in the setup, $(p,q) = (-3,3)$ will occur as long as $x+y+z = 0$, and no two of $x$, $y$, $z$ to share the same magnitude or are collinear with the origin. This is easy to do; for example, we could choose $(x, y, z) = (3, 4i, -(3+4i))$. Hence $a = \frac{3}{4i}$, $b = -\frac{4i}{3+4i}$, $c = -\frac{3+4i}{3}$ satisfies the hypotheses of the problem statement.

Nice to see this eldritch horror of a problem statement finally make it out of hiding. (It's been five years!) Though the authorship credits are a bit generous xD. I'm certainly responsible for 90% of the problem, but my original statement was different: Quote: Let $a$, $b$, and $c$ be three nonreal complex numbers with $abc=1$. Suppose that \[ \left(a + \frac 1a\right) + \left(b + \frac1b\right) + \left(c + \frac1c\right)\quad\text{and}\quad \frac ab + \frac bc + \frac ca \]are both real numbers. Show that at least one of $a$, $b$, or $c$ has magnitude equal to $1$. As you might suspect, this problem is not true; I asserted that a certain matrix has rank $2$ when, in fact, it does not always have rank $2$. Evan then modified the problem statement to reflect this extra case.

I have been bothering everyone who will listen at MOP with my thoughts on this problem, so I suppose it is only fair that I post them here as well: In some sense the main difficulty of this problem is that the relation between $p, q$ and $a, b, c$ is a bit convoluted; if $p$ and $q$ were symmetric in $a, b, c$, then the problem would be easy since retrieving three complex numbers from their symmetric sums is rather well-understood (just solve a cubic). Instead, the symmetries fixing $p, q$ consist of cyclically permuting $a, b, c$, and swapping two of $a, b, c$ and then reciprocating each of $a, b, c$. In fact, it is true that $\mathbb{Q}(a, b)/\mathbb{Q}(p, q)$ (here $c = (ab)^{-1}$) is Galois of degree $6$ with Galois group $S_3$. This observation motivates two broad classes of solutions to the problem: find some substitution that turns the $S_3$-symmetry into a permutation symmetry. Some ways to do this include $a = y/x$ and cyclic variants, or $x = a + b^{-1}$ and cyclic variants. These operations are sort of inverse to each other, but the solutions they induce look somewhat different. expand one's view to the larger symmetry group $S_3 \times C_2$, where the factors are given by permutations of $a, b, c$ and the reciprocation symmetry. The fixed field of this is $\mathbb{Q}(p, f)$, where $f = (a + b + c)(a^{-1} + b^{-1} + c^{-1})$. Then, one can come up with an abstract procedure to "solve for $a, b, c$" by writing $f$ in terms of $p$ and $q$, eliminating the $C_2$ symmetry by finding $e_1 = a + b+ c$ and $e_2 = a^{-1} + b^{-1} + c^{-1}$ in terms of $p$ and $f$, and finally solving a cubic to yield $a, b, c$. Analyzing this idea allows one to extract the answer (for more details see the second solution in Evan's packet). This idea can be viewed as proceeding counterclockwise in the following diagram of field extensions: [asy][asy] unitsize(2.5cm); usepackage("amsfonts"); label("$\mathbb{Q}(a,b)$",(0,1)); label("$\mathbb{Q}(p,f)$",(0,-1)); label("$\mathbb{Q}(e_1,e_2)$",(1,0)); label("$\mathbb{Q}(p,q)$",(-1,0)); draw((0.85,0.15)--(0.15,0.85)); draw((-0.85,0.15)--(-0.15,0.85)); draw((0.85,-0.15)--(0.15,-0.85)); draw((-0.85,-0.15)--(-0.15,-0.85)); draw((0,0.85)--(0,-0.85)); label("\footnotesize $S_3$", (0.6,0.6)); label("\footnotesize $S_3$", (-0.6,0.6)); label("\footnotesize $C_2$", (0.6,-0.6)); label("\footnotesize $C_2$", (-0.6,-0.6)); label("\footnotesize $S_3 \times C_2$", (0.3,0)); [/asy][/asy] As a further sidenote, the group $S_3 \times C_2$ happens to be isomorphic to the dihedral group on $12$ elements, which is the group of symmetries of a regular hexagon. It is possible to use this to visualize how the group acts on $a,b,c$ and $x,y,z$, if one considers the following "magic hexagon": [asy][asy] unitsize(3cm); label("$a$",dir(120),dir(120)); label("$b$",dir(240),dir(240)); label("$c$",dir(360),dir(360)); label("$a^{-1}$",dir(300),dir(300)); label("$b^{-1}$",dir(60),dir(60)); label("$c^{-1}$",dir(180),dir(180)); label("$x$",sqrt(3)/2*dir(-30),dir(-30)); label("$y$",sqrt(3)/2*dir(90),dir(90)); label("$z$",sqrt(3)/2*dir(210),dir(210)); label("$x^{-1}$",sqrt(3)/2*dir(150),dir(150)); label("$y^{-1}$",sqrt(3)/2*dir(270),dir(270)); label("$z^{-1}$",sqrt(3)/2*dir(30),dir(30)); draw(rotate(30)*polygon(3),red+linewidth(1pt)); draw(rotate(90)*polygon(3),red+linewidth(1pt)); draw(rotate(0)*scale(sqrt(3)/2)*polygon(3),blue+linewidth(1pt)); draw(rotate(60)*scale(sqrt(3)/2)*polygon(3),blue+linewidth(1pt)); draw(polygon(6),linewidth(2pt)); [/asy][/asy] The $S_3$ permuting $a, b, c$ is the subgroup fixing the red triangles, while the "twisted $S_3$" permuting $x,y,z$ is the subgroup fixing the blue triangles.

Sorry I posted my solution under the wrong thread it looks like. I'll move it here.

The answer is \((-3,3)\) only. Let \(a=\frac xy\), \(b=\frac yz\), \(c=\frac zx\). Of course \(x\), \(y\), \(z\) are nonzero, and we also are given the restriction that no two share the same magnitude or argument (modulo \(180^\circ\)). Observe that \begin{align*} p+3&=3+\sum_\mathrm{cyc}\frac xy+\sum_\mathrm{cyc}\frac yx\\ &=\left( \frac1x+\frac1y+\frac1z \right)(x+y+z)\\ \text{and}\quad p+q&=\sum_\mathrm{cyc}\frac{xz}{y^2}+\sum_\mathrm{cyc}\frac xy+\sum_\mathrm{cyc}\frac yx\\ &=\left( \frac1x+\frac1y+\frac1z \right) \left( \frac{yz}x+\frac{zx}y+\frac{xy}z \right).\end{align*}We additionally compute \begin{align*} 3p+q+6&=\left( \frac1x+\frac1y+\frac1z \right) \left[ 2(x+y+z)+\left( \frac{yz}x+\frac{zx}y+\frac{xy}z \right) \right]\\ &=xyz\left( \frac1x+\frac1y+\frac1z \right) \left[ 2\left( \frac1{yz}+\frac1{zx}+\frac1{xy} \right) +\left( \frac1{x^2}+\frac1{y^2}+\frac1{z^2} \right)\right]\\ &=xyz\left( \frac1x+\frac1y+\frac1z \right)^3. \end{align*} Now, any choice of \(\frac1x\), \(\frac1y\), \(\frac1z\) summing to 0 and satisfying the restrictions listed above necessarily gives \(p+3=p+q=0\) and thus \((-3,3)\). It is clear to see this is possible, thus achieving \((-3,3)\). Further, it will suffice to show we must have \(\frac1x+\frac1y+\frac1z=0\). Assume for contradiction \(\frac1x+\frac1y+\frac1z\ne0\), and scale \(x\), \(y\), \(z\) so that we may assume without loss of generality \(\frac1x+\frac1y+\frac1z=1\). We then have \[x+y+z=p+3\quad\text{and}\quad xy+yz+zx=xyz=3p+q+6.\]It follows that \(x\), \(y\), \(z\) are roots of a polynomial \[f(t)=t^3-(p+3)t^2+(3p+q+6)t-(3p+q+6).\]with real coefficients. But then, one of the following two cases must hold: \(f\) has three real roots, in which case \(x\), \(y\), \(z\) all have the same argument; or \(f\) has exactly one real root, in which the other two nonreal roots are complex conjugates and thus have the same magnitude. Both cases contradict our restrictions.

The answer is $(p,q)=(-3,3)$ only. We will provide a construction later. Since $abc=1$, we may substitute $a=\tfrac{x}{y},b=\tfrac{y}{z},c=\tfrac{z}{x}$, where $x,y,z \in \mathbb{C}$. Furthermore, since multiplying $x,y,z$ by the same (nonzero) constant doesn't change the corresponding values of $a,b,c$, we may assume WLOG that $xyz=1$. Then, we can write $$p=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3,$$and $$q=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}.$$Noting that $\tfrac{x}{y}$ can be written as $\tfrac{xy}{y^2}$ and similar things are true for the other $5$ terms in $p$, we have the amazing identity $$p+q=(xy+yz+zx)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$Now suppose that $x,y,z$ are roots of the polynomial $f(t):=t^3+ut^2+vt-1$. By Vieta's, we have $x+y+z=-v$ and $xy+yz+zx=u$. Further, by considering the "reverse polynomial" $f^*(t):=-t^3+vt^2+ut^2+1$, which has roots $\tfrac{1}{x},\tfrac{1}{y},\tfrac{1}{z}$, we find that $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=v \text{ and } \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=v^2-2(-u)=v^2+2u,$$hence $p=uv$ and $p+q=v(v^2+2u)=v^3+2uv$. Thus we find that $v^3 \in \mathbb{R}$. We now have the following key claim. Claim: $v$ is $0$. Proof: Suppose otherwise. Then $g(t):=f(\tfrac{t}{v})$ is a polynomial with real coefficients, since $\tfrac{1}{v^3},\tfrac{u}{v^2},1,-1$ are all real. Therefore, since it has degree $3$, it either has $1$ or $3$ real roots. If all $3$ roots of $g$ are real, then the roots of $f$ have the same complex argument (modulo $180^\circ$), so their ratios, i.e. $a,b,c$, are real: illegal. If exactly $1$ root of $g$ is real, then the two nonreal roots are complex conjugates, say $\omega$ and $\overline{\omega}$. Then one of $a,b,c$ is $\tfrac{\omega/v}{\overline{\omega}/v}$, which has magnitude $1$: also illegal. $\blacksquare$ Therefore it follows that $p=-3$ and $p+q=0$, so $q=3$. Thus $(p,q)=(-3,3)$ is the only possibility. We now provide a construction. Let $\omega$ be a complex number such that $\omega^2=\tfrac{1}{64}+\tfrac{i}{2}$. Let $x=2i,y=\tfrac{1}{8}+\omega,z=\tfrac{1}{8}-\omega$. It is easy to check that $xy+yz+zx=0$, so this construction in fact gives $(p,q)=(-3,3)$. It remains to check that none of $a,b,c$ are real or have magnitude $1$. First, since $y$ and $z$ aren't pure imaginary, $a$ and $c$ are nonreal. Also, since the line joining $y$ and $z$ in the complex plane intersects the real axis at $\tfrac{1}{8}$ (and not $0$), $y$ and $z$ are not real multiples of each other. Finally, since $\omega$ is not pure imaginary, $|b| \neq 1$, and since $|\omega|<1$ we have $|y|,|z|<|x|$, so $|a|,|c| \neq 1$. Therefore this construction satisfies every necessary condition, and we are done. $\blacksquare$

Here's a nice solution, mostly based on the submission of Chris Bao. Let $x=a+\frac{1}{c}$, $y=b+\frac{1}{a}$, $ z=c+\frac{1}{b}$. Then we can check that \begin{align*} x+y+z &= p \\ xy+yz+zx &= p+q+3 \\ xyz &= p+2. \end{align*}Therefore $x$, $y$, $z$ are the roots of a cubic with real coefficients. As in the first official solution, we note that this cubic must either have all real roots, or a complex conjugate pair of roots. We also have the relation $a(y+1)=ab+a+1=x+1$, and likewise $b(z+1)=y+1, c(x+1)=z+1$. This means that if any of $x$, $y$, $z$ are equal to $-1$, then all are equal to $-1$. Assume for the sake of contradiction that none are equal to $-1$. In the case where the cubic has three real roots, $a=\frac{x+1}{y+1}$ would be real. On the other hand, if there is a complex conjugate pair (without loss of generality, $x$ and $y$) then $a$ has magnitude $1$. Therefore this cannot occur. We conclude that $x=y=z=-1$, so $p=-3$ and $q=3$. The solutions $(a, b, c)$ can then be parameterized as $(a, -1-\tfrac{1}{a}, -\tfrac{1}{1+a})$. To construct a solution, we need to choose a specific value of $a$ such that none of the wrong conditions hold. When $a=2i$ we obtain the solution $(2i, -1+\tfrac{i}{2}, \tfrac{-1+2i}{5})$.

Let $a=x/y, b=z/x$ and $c=y/z$. We claim that $x+y+z=0$. Note that (all sums are cyclic) $p=(a+b+c)+(1/a+1/b+1/c)=\dfrac{\displaystyle \sum x^2y+\sum x^2z}{xyz},$ and so $p+3=\dfrac{\displaystyle \sum x \sum xy}{xyz}$ Moreover, $q=a/b+b/c+c/a=\dfrac{x^3+y^3+z^3}{xyz},$ and so $q-3=\dfrac{(x+y+z)^3-3(x+y)(y+z)(z+x)-3xyz}{xyz}=\dfrac{(x+y+z)^3}{xyz}-\dfrac{\displaystyle 3\sum x \sum xy}{xyz}=\dfrac{(x+y+z)^3}{xyz}-3(p+3),$ hence we obtain $\dfrac{(x+y+z)^3}{xyz}=3p+q+6=M^3$, for some $M \in \mathbb{R}$. Hence, if $M \neq 0$ (this is equivalent to $x+y+z \neq 0$), there is an $s$ such that $xyz=s^3$ and $x+y+z=sM$. Moreover, $p+3=\dfrac{\displaystyle \sum x \sum xy}{xyz}$ is real, and so $\dfrac{\displaystyle K\sum xy}{s^2}$ is real, hence there is an $N \in \mathbb{R}$ such that $\sum xy=Ns^2$. Now, let $x/s=k, y/s=\ell,z/s=m$. Then, $k \ell m=1, k+\ell+m=M$ and $k \ell +\ell m +mk=N,$ with $M,N$ being both real. Consider $f(t)=t^3-Mt^2+Nt-1$. Then, $k,\ell,m$ are all roots of $f$. Hence, we have two cases to consider. Case 1: All of $k,\ell,m$ are real. Then, $x/s,y/s,z/s$ are real, and so $a=x/y=(x/s)/(y/s)$ is real, too, a contradiction. Case 2: Only one of $k,\ell,m$ is real, let it be $k$. Then, $|\ell|=|m|,$ and so $|c|=|y/z|=|y/s|/|z/s|=|\ell|/|m|=1,$ and so $|c|=1,$ a contradiction. Hence, in both cases we obtain a contradiction, and so $x+y+z=0$ must be true. In this case, by the equalities in the beggining we obtain that $(p,q)=(-3,3)$. In order to prove this is indeed possible, we need three complex numbers $x,y,z$ such that $x+y+z=0$, $|x| \neq |y| \neq |z| \neq |x|,$ and $x/y,z/x$ and $y/z$ are not real. To that end we may, for example, take $x=10+5i,y=2-3i$ and $z=-12-2i$, hence we may finish.

Fairly straightforward for 2/5, but a great and innovative problem. Substitute $a = \frac yx, b = \frac zy, c = \frac xz$, so that \begin{align*} p &= \frac{x^2y+y^2z+z^2x+xy^2+yz^2+zx^2}{xyz} \\ q &= \frac{x^3+y^3+z^3}{xyz}. \end{align*}Note that the given condition implies that $x, y, z$ have different magnitudes and directions. On the other hand, notice that $$q+3p+6 = \frac{(x+y+z)^3}{xyz}$$is also real. Assume $x+y+z \neq 0$. We may assume up to scaling in $\mathbb C$ that $x+y+z = xyz = 1$ are real, and also $$x^3+y^3+z^3-3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))$$is real, hence $xy+yz+zx$ is also real. Hence $x, y, z$ are the roots of a polynomial with real coefficients, which contradicts at least one of the given conditions. Hence the only possibility is $x+y+z = 0$. Then \begin{align*} p+3 &= \frac{(x+y+z)(xy+yz+zx)}{xyz} = 0 \\ q - 3 &= \frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{xyz} = 0. \end{align*}So $(p, q) = (-3, 3)$. Construction is easy simply by taking any $a, b, c$ that satisfy the given condition.

djmathman wrote: Nice to see this eldritch horror of a problem statement finally make it out of hiding. (It's been five years!) I'm glad too. It's not the kind of thinking people are used to at all any more, so I specifically have been waiting a long time for this problem to finally get some airtime.

numbersandnumbers wrote: In fact, it is true that $\mathbb{Q}(a, b)/\mathbb{Q}(p, q)$ (here $c = (ab)^{-1}$) is Galois of degree $6$ with Galois group $S_3$. If $G$ is the group of operations generated by $(a,b)\to (b, 1/ab)$ and $(a,b)\to (1/a, ab)$, then we know $\mathbb{Q}(p,q)$ is a subfield of $\mathbb{Q}(a,b)^G$. How do I show they are the same field? Hmm One way is to write an element of $\mathbb{Q}(a,b)$ as $f(a,b)/g(a,b)$ where $f,g$ are polynomials in $\mathbb{Q}[X_1,X_2]$. Then I can consider $\frac{f(a,b) g(b, 1/ab) g(1/ab, a) g(1/a, ab) g(ab, 1/b) g(1/b, 1/a)}{g(a,b) g(b, 1/ab) g(1/ab, a) g(1/a, ab) g(ab, 1/b) g(1/b, 1/a)}$ and reduce the problem for polynomials. That still seems difficult.

we have $p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ $q=\frac{a}{b}+\frac{a}{c}+\frac{c}{a}, abc=1$ Consider $r=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ We notice $qr=3+\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+a^3+b^3+c^3$ So $p^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)+\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)+3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(p)$ so we have $p^3=qr-3+6(q+r+2)+3(3+q+r)(p) \implies r(q+6+3p) \in \mathbb{R}$ as $p^3 \in \mathbb{R}$ Claim:- $r \notin \mathbb{R}$ pf:- FTSOC consider $r \in \mathbb{R}$ Consider $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+q+r$ So since $r \in \mathbb{R}$ we have $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ and $(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \in \mathbb{R}$ hence we have two cases: Case 1:- $(a+b+c) , \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=(ab+bc+ca) \in \mathbb{R}$ If this is the case then we have the equation $z^3-(a+b+c)z^2+(ab+bc+ca)z-1=0$ has at least one real root $\implies$ at least one of $a,b,c$ is real which is a contradiction. Hence this case don't works. Case 2:- $\overline{a+b+c}=ab+bc+ca=\Omega$ then since $abc=1$ we notice if $\omega$ is a root of $z^3-\overline{\Omega}z^2+\Omega z-1=0$ then $\frac{1}{\overline{\omega}}$ is also a root , which is contradiction as $|a|,|b|,|c| \neq 1$ and $\Im(a), \Im(b), \Im(c) \neq 0$. Hence this case also not works So overall we observe no such case works and hence $r \notin \mathbb{R}$. So claim follows $\square$ so we must have $q+6+3p=0 \implies p^3-3pq-9p-6q-9=0$ , plugging $q=-3p-6$ we get $(p+3)^3=0 \implies p=-3, q=3$ hence we have $\boxed{(p,q)=(-3,3)}$ as the only pair. $\blacksquare$ Construction: $(a,b,c) =\left(-2i , \frac{-(i+2)}{2}, \frac{-(2i+1)}{5}\right)$ works.

numbersandnumbers wrote: In fact, it is true that $\mathbb{Q}(a, b)/\mathbb{Q}(p, q)$ (here $c = (ab)^{-1}$) is Galois of degree $6$ with Galois group $S_3$. Let's prove this claim. Define $G$ as I did in PM #13. First, clearly, $\mathbb{Q}(a,b)^G \supset \mathbb{Q}(p,q)$, with $[\mathbb{Q}(a,b)\colon \mathbb{Q}(a,b)^G] = (G\colon 1) = 6$. Therefore, it suffices to show that $$[\mathbb{Q}(a,b) \colon \mathbb{Q}(p,q)] = 6$$ Quote: find some substitution that turns the $S_3$-symmetry into a permutation symmetry. Let $a = \frac xz, b = \frac yx, c= \frac zy$ where $xyz=1$. Then we can check that $y^3 = b/c, z^3 = c/a, x^3 = a/b$. We consider $\mathbb{Q}(x,y,z)$ as a field extension of $\mathbb{Q}(a,b,c)$. We can check that $[\mathbb{Q}(a,b,c,y) \colon \mathbb{Q}(a,b,c)]=3$ since the minimal polynomial of $y$ in $\mathbb{Q}(a,b,c)$ is $y^3 - b/c = 0$. Furthermore, since $\frac xy = b^{-1} \in \mathbb{Q}(a,b,c)$, it follows that $\mathbb{Q}(x,y,z) = \mathbb{Q}(a,b,c,x,y,z) = \mathbb{Q}(a,b,c,y)$ is a field extension of $\mathbb{Q}(a,b,c)$ of degree 3. It remains to show that $[\mathbb{Q}(x,y,z) \colon \mathbb{Q}(p,q)] = 18$. Note $$p = a+1/a+b+1/b+c+1/c = \sum_{sym} xy^{-1} \qquad q=a/b+b/c+c/a=x^3+y^3+z^3$$ Therefore, we want to compute $$[\mathbb{Q}(x,y,z) \colon \mathbb{Q}(x^3+y^3+z^3, \sum_{sym} x^2y, xyz)]$$ Note $\mathbb{Q}(x,y,z)$ is Galois over $\mathbb{Q}(x+y+z, xy+yz+zx, xyz)$ with Galois group $Sym\{x,y,z\}$, so $[\mathbb{Q}(x,y,z) \colon \mathbb{Q}(x+y+z, xy+yz+zx, xyz)] = 6$. Thus it remains to show that $[\mathbb{Q}(x+y+z, xy+yz+zx, xyz) \colon \mathbb{Q}(x^3+y^3+z^3, \sum_{sym} x^2y, xyz)] = 3$ Let $s_1 = x+y+z, s_2=xy+yz+zx, s_3 = xyz$ (Since I'm lazy I will not bash but do this) Let $V$ be the vector space of symmetric polynomials of $x,y,z$ of degree 3 over $\mathbb{Q}$. Note $V$ has a basis $s_1^3, s_1s_2, s_3$, so $\dim V = 3$. Note $x^3+y^3+z^3, \sum_{sym} x^2y, xyz$ also form a basis of $V$. Therefore, $$\mathbb{Q}(s_1,s_2,s_3)= \mathbb{Q}(s_1, s_1s_2, s_3)$$ Is a vector space over $\mathbb{Q}(s_1^3, s_1s_2,s_3)$ of degree 3 (with basis $1,s_1,s_1^2$), and therefore a field extension of $\mathbb{Q}(s_1^3, s_1s_2,s_3)$ of degree 3. We are done. Update Actually $\mathbb{Q}(x,y,z)$ is Galois over $\mathbb{Q}(x^3+y^3+z^3, \sum_{sym} x^2y, xyz)$, with galois group $Sym\{x,y,z\} \times \langle t\rangle$, where $t(x,y,z)=(\omega x, \omega y, \omega z)$ where $\omega $ is a primitive 3rd root of unity.

In some sense, turning the symmetry from inverting + cyclic shifts into permutation symmetry can be expanded into this fix field diagram: [asy][asy] unitsize(2.5cm); usepackage("amsfonts"); label("$\mathbb{Q}(x,y,z)$",(0,2)); label("$\mathbb{Q}(p,q)$",(0,-2)); label("$\mathbb{Q}(a,b,c)$",(2,0)); label("$\mathbb{Q}(x+y+z, xy+yz+zx)$",(-2,0)); draw((1.7,0.3)--(0.3,1.7)); draw((-1.7,0.3)--(-0.3,1.7)); draw((1.7,-0.3)--(0.3,-1.7)); draw((-1.7,-0.3)--(-0.3,-1.7)); draw((0,1.7)--(0,-1.7)); label("\footnotesize $\langle \tau \rangle$", (1.2,1.2)); label("\footnotesize $Sym\{x,y,z\}$", (-1.2,1.2)); label("\footnotesize $Sym\{x,y,z\}$", (1.2,-1.2)); label("\footnotesize $\langle \tau \rangle$", (-1.2,-1.2)); label("\footnotesize $\langle \tau, Sym\{x,y,z\} \rangle$", (0.6,0)); [/asy][/asy] Here $\tau$ turns $(x,y,z)$ into $(\omega x, \omega y, \omega z)$ where $\omega$ is a 3rd root of unity. I am just taking fixed fields in two different orders. Going back how $\mathbb{Q}(a,b,c)$ is Galois over $\mathbb{Q}(p,q)$, I can create an isomorphism between $\pi \colon \{1,2,3\} \to \{x,y,z\}$ to $\left(\frac{\pi(1)}{\pi(2)}, \frac{\pi(2)}{\pi(3)}, \frac{\pi(3)}{\pi(1)}\right)$ (each of these is either a,b,c or 1/a,1/b,1/c)

me not scaling wasted 2 hours of my life Let $a = \frac{y}{x}$, $b = \frac{z}{y}$, $c = \frac{x}{z}$. It will help to consider \begin{align*} p + 3 = \left( x + y + z \right) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \\ q - 3 = \frac{(x + y + z)(x^2+y^2+z^2-xy-yz-zx)}{xyz} \end{align*}instead of $p$, $q$. The main claim of the problem is the following: Claim: $x + y + z = 0$. Proof. Assume otherwise and scale so that $x + y + z$ is real. Now $\frac{xy + yz + zx}{xyz}$ is also real, and by dividing $p+3$ and $q - 3$, it follows that $xy + yz + xz$ is also real. Therefore, $x$, $y$, $z$ are the roots of a polynomial with real coefficients. Now either all of $x$, $y$, $z$ are real (contradiction, since then $a$, $b$, $c$ are real) or (WLOG) $x$ is real and $y$, $z$ are complex conjugates, which is also a contradiction since then $\lvert b \rvert = \left\lvert \frac{z}{y} \right\rvert$ has absolute value $1$. $\square$ Now from the expressions for $p + 3$ and $q - 3$, it follows that $p + 3 = 0$ and $q - 3 = 0$, so $p = -3$ and $q = 3$. A construction is to take $(x, y, z) = (2, i+1, -i - 3)$, from which $a$, $b$, $c$ readily follows.

Hello. \[x=a+\frac{1}{c}\]\[y=b+\frac{1}{a}\]\[z=c+\frac{1}{b}\]Then $x+y+z$ is real, $xyz$ is real, and $xy+yz+zx$ is real. Hence $x,y,z$ are roots of a cubic. If these are all real then taking \[a=\frac{u}{w}\]\[b=\frac{v}{u}\]\[c=\frac{w}{v}\]gives \[x=\frac{u+v}{w}\]and cyclically. Then the complex numbers $0,u,v,w,u+v+w$ collinear hence $a,b,c$ are real. So one of $x,y,z$ is complex, then WLOG $x$ and $y$ are conjugates. Hence \[\frac{u+v+w}{w},\frac{u+v+w}{u}\]should be conjugates. Hence $|w|=|u|$ hence $|a|=1$. Bad. Course there are people who fakesolved by saying at this point there are no solutions. I almost made the same mistake when resolving. You only run into issues with the proof when $u+v+w=0$ then $x=y=z=-1$. So $p=-3$. Now \[3=xy+yz+zx=3+p+q\implies q=3\]hence done.

What. Write $a=\frac{x}{y},b=\frac{z}{x},c=\frac{y}{z}$. Then $p = \sum_{\text{sym}} \frac{x}{y}$ and $q = \sum_{\text{cyc}} \frac{x^2}{yz}$. Since $p$ and $q$ are homogenous expressions, we can WLOG $xyz=1$. Now write $(T-x)(T-y)(T-z) = T^3 - rT^2 + sT - 1$. After some work with symmetric sums we have $$p = (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) - 3 = rs - 3,\:\:\:\:\:\:\: q = \sum_{\text{cyc}} x^3 = r^3 - 3rs + 3$$ so $rs$ and $r^3$ are all real. Now, if $r$ is nonzero the polynomial $(T-\frac{x}{r})(T-\frac{y}{r})(T-\frac{z}{r})$ has real coefficients, so two of $rx,ry,rz$ are real or complex conjugates, a contradiction. Hence $r=0$, giving $(p,q) = (-3,3)$. This is easily achieved by most solutions to $xyz=1,x+y+z=0$, for example $x=-i,y=\frac{i+\sqrt{-1-4i}}{2},z=\frac{i-\sqrt{-1-4i}}{2}$.

Substitute $a = \frac{x}{y}$, $b = \frac{y}{z}$ and $c = \frac{z}{x}$. Define the elementary sums $u = x + y + z$, $v = xy + yz + zx$ and $w = xyz$. Then we may find, \begin{align*} p &= \frac{(xy + yz + xz)(x + y + z) - 3xyz}{xyz} = \frac{uv}{w} - 3\\ q &= \frac{(xy + yz + zx)^3 - 3xyz((xy + yz + xz)(x + y + z) - xyz)}{x^2y^2z^2} = \frac{v^3 - 3uvw}{w^2} + 3 \end{align*}Now consider the following real quantities; \begin{align*} (q - 3) + 3(p + 3) &= \frac{v^3}{w^2}\\ \frac{(q - 3) + 3(p + 3)}{(p + 3)^2} &= \frac{v}{u^2}\\ \frac{(q - 3)}{(q - 3) + 3(p + 3)} &= 1 - \frac{3uw}{v} \end{align*}Clearly then we find, \begin{align*} \frac{uw}{v} \cdot \frac{uv}{w} = u^2 \in \mathbb{R} \end{align*}From here we may notice that $\frac{v}{u^2} \in \mathbb{R}$ implies that $v \in \mathbb{R}$, and similarly $\frac{v^3}{w^2} \in \mathbb{R}$ implies that $w^2 \in \mathbb{R}$. Finally note that $\frac{uw}{v} \in \mathbb{R}$ forces $uw \in \mathbb{R}$. Thus we have the system, \begin{align*} (x + y + z)^2 \in \mathbb{R}\\ xy + yz + zx \in \mathbb{R}\\ (xyz)^2 \in \mathbb{R}\\ xyz(x + y + z) \in \mathbb{R} \end{align*}Note that the first condition simplifies to $x^2 + y^2 + z^2 \in \mathbb{R}$. Then squaring the second, and subtracting two times the last we find $x^2y^2 + y^2z^2 + z^2x^2 \in \mathbb{R}$. Thus we may reduce our system to, \begin{align*} x^2y^2z^2 \in \mathbb{R}\\ x^2y^2 + y^2z^2 + z^2x^2 \in \mathbb{R}\\ x^2 + y^2 + z^2 \in \mathbb{R} \end{align*}Consider now the polynomial with roots $x^2$, $y^2$ and $z^2$; note that it has real coefficients, which forces at least one of $x^2$, $y^2$ or $z^2$ to be real - without loss of generality assume that $x^2 \in \mathbb{R}$. Then we may reduce our system to, \begin{align*} y^2z^2 \in \mathbb{R}\\ y^2 + z^2 \in \mathbb{R} \end{align*}Say $y^2 = r_1 + s_1i$ and $z^2 = r_2 + s_2i$, and observe that the conditions imply $s_1 = -s_2$, and that $r_1 = r_2$. Therefore $y^2$ and $z^2$ are in fact conjugates! Namely they must have the same magnitude, contradicting the problem assumption that $\frac{y}{z}$ does not have magnitude $1$. Note that this argument fails exactly when $u$ is $0$. In this case we find, \begin{align*} p &= \frac{uv}{w} - 3 = -3\\ q &= \frac{v^3 - 3uvw}{w^2} + 3 = 3 \end{align*}which leads to the solution set $(p, q) = \boxed{(-3, 3)}$. For a construction take $(x, y, z) = (2, i+1, -i - 3)$ and we're done.

The answer is $(-3,3)$ with a construction of $(a,b,c) = ( \tfrac{3+i}{2}, \tfrac{-8+i}{5}, \tfrac{-5+i}{13} )$. Since we're given that $abc = 1$, there exist nonzero complex numbers $d$, $e$ and $f$ such that $a = \tfrac{d}{e}$, $b = \tfrac{e}{f}$ and $c = \tfrac{f}{d}$. The problem conditions guarantee that no two of $d$, $e$ or $f$ have the same argument or magnitude. Let $J = d+e+f$, $K = de+ef+fd$ and $L = def$. After some manipulation, we may rewrite our given equations as \begin{tabular}{c c c} $p=\dfrac{JK}{L} - 3$ & \text{and} & $q= \dfrac{K^3}{L^2} - \dfrac{3JK}{L} + 3$. \end{tabular}Now we split by cases: If $K = 0$: Then, $(p,q) = (-3,3)$ automatically. If $J = 0$ and $K \neq 0$: Then, $\tfrac{K^3}{L^2}$ must be real. We may scale $d$, $e$ and $f$ WLOG by some complex number such that $K = 1$, so it follows that $L$ is also real. But this implies that $d$, $e$ and $f$ are the roots of a polynomial $P(x)$ with no real coefficients. This contradicts the fact that no two of $d$, $e$ or $f$ have the same magnitude. If $J \neq 0$ and $K \neq 0$: we may scale $d$, $e$ and $f$ WLOG by some complex number such that $J = 1$. Since $p$ is real, $\tfrac{K}{L}$ is real; since $q$ is real, $\tfrac{K^3}{L^2}$ is real. So, \[\frac{\left( \frac{K^3}{L^2} \right)}{\left( \frac{K}{L} \right)^2} = K\]is real, and so is $L$. Similar to the previous case, it follows that $d$, $e$ and $f$ are the roots of a polynomial with real coefficients, yielding a contradiction.

The answer is $\boxed{(-3,3)}$ only, with construction $a=\frac12 i$, $b=2i-1$, $c=\frac25 (i-2)$. It is not hard to check that this works. Substitute $a=\frac xy$, $b=\frac zx$, $c=\frac yz$. Then \[p=(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)-3\]\[q=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y+z)(xy+yz+zx)}{xyz}+3.\]Note that if $x+y+z=0$, then we get $(p,q)=(-3,3)$. So assume $x+y+z\ne 0$. Then we can scale them so that $x+y+z=1$, so that \[\frac{xy+yz+zx}{xyz}\]and \[\frac{x^2+y^2+z^2-xy-yz-zx}{xyz}\]are real. Add $3$ times the first expression to the second expression to get that $xyz$ is real, then we also get that $xy+yz+zx$ is real. So $x,y,z$ are the roots of a real polynomial. Note that the restrictions of the problem imply that no two of $x,y,z$ has the same magnitude and no two have the same argument (mod $\pi$). Thus they can't all be real, so there has to be one conjugate pair, which has the same magnitude, contradiction. $\blacksquare$

The answer is $(-3,3)$ only. Substitute $(a,b,c)=(\tfrac xy,\tfrac zx,\tfrac yz)$ for complex $x,y,z$ with distinct magnitudes and arguments. WLOG $xyz$ is real, which we can do by rotating $x,y,z$. Then setting $x+y+z=r,xy+yz+zx=s$ we get $rs=p+3,r^3-3rs=q-3$. Thus $r^3,rs$ are real. If $r$ is nonzero then $r$ is a real multiple of the third roots of unity $1,\omega,\omega^2$. If $r$ is real then $s$ is real and $x,y,z$ are roots of a real polynomial, so either all three roots are real and have the same argument, or two are conjugates and have the same magnitude. If $r$ is $\omega$ times a real number, then $\omega^2 x,\omega^2 y,\omega^2 z$ are the roots of a real polynomial and similarly fail, and similarly if $r$ is $\omega^2$ times a real number. Thus $r$ is zero, so $(p,q)=(-3,3)$. To construct just take $x,y,z$ summing to zero with distinct magnitudes and arguments, for example by taking an arbitrary triangle with distinct median lengths and translating its centroid to the origin.