Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $1/x+1/y+1/z=0$ 
Prove that there are no integers $x, y, z$ such that $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0$

My thinking was that since the numbers are integers, then there can't be $2$ negative values that cancel out the positive or $2$ positive numbers to cancel the negative. One integer would have to cancel out the second, and the third wouldn't make the sum zero. Am I right?
 A: The statement remains to be true even when one allow $x, y, z$ to be real numbers.
In order for the second condition $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ to make sense, none of $x, y, z$ can be zero.
If the first condition $x + y + z = 0$ is also satisfied, we will have
$$x^2 + y^2 + z^2 = (x+y+z)^2 - 2\left(\frac1x + \frac1y + \frac1z\right)xyz = 0^2 - 0 = 0$$
As long as $x,y,z$ are real numbers, this forces $x = y = z = 0$ and makes second condition absurd.
A: We can assume that $gcd(x,y,z)=1$;
$(1/x+1/y+1/z)xyz=xy+xz+yz=x(y+z)+yz$, $x+y+z=0$ implies that $y+z=-x$,  we have $x(y+z)+yz=-x^2+yz=0$. We deduce that $x^2=yz$, let $p$ be a prime which divides $x$, $p$ divides $yz$ implies that $p$ divides $y$ or $z$, suppose that $p$ divides $y$, $-x/p=-y/p+z/p$ implies $p$ divides $z$, contradiction. We deduce that $x=1$ or $x=-1$, the same argument shows that $y,z\in\{-1,1\}$. This is impossible.
A: Another proof using symmetric polynomials/Vieta's formulas. This generalizes to any field.
Let $\sigma_1=x+y+z$, $\sigma_2=xy+yz+xz$, and $\sigma_3=xyz$ be the elementary symmetric polynomials. In order to make sense of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$, we can assume that $xyz\ne 0$. Then
$\sigma_3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\sigma_2$, so $\sigma_2=0$.
Thus we know $\sigma_2=\sigma_3=0$. Now by Vieta's formulas, $x$, $y$, and $z$ are the roots of the cubic polynomial $t^3-\sigma_1t^2+\sigma_2t-\sigma_3=t^3-xyz$. Hence $x^3-xyz=0$, and similarly for $y$ and $z$, so we get 
$$x^3=y^3=z^3=xyz.$$ 
Dividing this equation by $x^3$, we get
$$\newcommand{\of}[1]{\left({#1}\right)}1=\of{\frac{y}{x}}^3=\of{\frac{z}{x}}^3=\frac{y}{x}\frac{z}{x}.$$
Now there are two cases. The first case is that $y/x=z/x=1$, in which case $x=y=z$. Together with $x\ne 0$ and $x+y+z=3x=0$, we see that $3=0$, so this case can only happen if the characteristic of the field is 3. In that case, $x=y=z=c$ for any nonzero constant $c$ gives a solution. In particular this can't happen for $\Bbb{Q}$, so there are no integral solutions to the equation in this case.
The other case is that $y/x=\omega$ where $\omega\ne 1$, but $\omega^3=1$. This can only happen when the characteristic of $K$ is not three, since in characteristic three $(t^3-1)=(t-1)^3$, so in characteristic three, the only cube root of unity is 1. (Note that $\omega^3=1$ and $\omega\ne 1$, together imply that $\omega$ satisfies the polynomial $1+t+t^2$.)
Then for any nonzero constant $c$, $x=c$, $y=\omega c$, $z=\omega^2 c$ gives a valid solution and these are the only solutions, since $x+y+z=c+\omega c+\omega^2 c= c(1+\omega+\omega^2)=0$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = \frac{1}{c}+\frac{1}{c\omega}+\frac{1}{c\omega^2}=\frac{1}{c}\of{1+\omega^2+\omega}=0$. However, once again, there are no primitive cube roots of one in $\Bbb{Q}$, so this case cannot yield any integral solutions.
Note however that there are solutions over $\Bbb{C}$ (namely $1,\omega,\omega^2$, where $\omega=e^{2\pi i/3}$).
To summarize, we have a complete description of solutions to these equations for any field $K$.
The form of the solutions depends on  the characteristic of $K$. If the characteristic is 3, the solutions are of the form $x=y=z=c\ne 0$ for any nonzero constant $c$.
Otherwise, there are solutions in $K$ if and only if the equation $1+t+t^2$ has a solution $\omega$ in $K$, in which case the solutions are precisely of the form $x=c$, $y=c\omega$, $z=c\omega^2$ for any nonzero constant $c$.
A: Assume there are. W.l.o.g. $y$ and $z$ are positive. Then 
$$x=-y-z \implies 0 = \frac{-1}{y+z} +\frac{1}{y}+\frac{1}{z} = \frac{-yz+z(y+z) +y(y+z)}{yz(y+z)} = \frac{z^2+y^2+yz}{yz(y+z)} >0,$$
contradiction. Note that this works for $x,y,z\in \Bbb R$.
A: Let $x,y,z \not = 0$ be integers. 
$1$) Multiply $\dfrac1x+\dfrac1y+\dfrac1z=0$ by $xyz:$
$$zy + zx + yx=0.$$
$2$) Square $x+y +z = 0:$
$$x^2+y^2+z^2 + 2(xy +xz +yz) =0$$
Second term is $0$, see $1$).
Hence : $$x^2+y^2+z^2= 0.$$
A contradiction, we assumed  $x,y,z \not = 0$ are integers.
A: We can't have any of $x,y$ or $z$ equal to zero, because we need their reciprocals, so there must be either two positives and a negative, or one positive and two negatives. Since you can multiply both sides by $-1$, these are actually the same case. Then you can just think about how, when $2+3-5=0$, we don't have $\frac12+\frac13-\frac15=0$, and even though $\frac1{30}+\frac1{15}-\frac1{10}=0$, we don't have $30+15-10=0$
A: Hint: if $x+y+z=0$, then
$$
\begin{align}
\frac1x+\frac1y+\frac1z
&=\frac1x+\frac1y-\frac1{x+y}\\
&=\frac{(x+y)^2+x^2+y^2}{2xy(x+y)}\\
&=-\frac{x^2+y^2+z^2}{2xyz}
\end{align}
$$
A: We have that $$x+y=-z\quad\text{ and }\quad\frac1x+\frac1y=-\frac1z\\\implies\frac1{x+y}=\frac1x+\frac1y\implies (x+y)^2=xy\\\implies x^2+y^2=-xy\implies-\frac{x}y-\frac{y}x=1\tag1$$ for some $x,y,z\in\Bbb Z\setminus\{0\}$. Setting $r:=-\frac{x}y$ the last identity result in
$$r+r^{-1}=1\tag2$$
for some $r\in\Bbb Q\setminus\{0\}$, what is not possible because
$$r+r^{-1}=\frac{r^2+1}r=1\implies r^2-r+1=0\tag3$$
what doesnt have real solutions.
A: Alternatively, denote: $y=ax,z=abx$, then:
$$\begin{cases}x+ax+abx=0 \\ \frac{1}{x}+\frac{1}{ax}+\frac{1}{abx}=0\end{cases} \Rightarrow \begin{cases}1+a+ab=0 \\ 1+\frac{1}{a}+\frac{1}{ab}=0\end{cases} \Rightarrow \begin{cases}a=-\frac{1}{b+1} \\ b^2+b+1=0\end{cases} \Rightarrow \emptyset.$$
