Determining the number of integer solutions $(x,y,z,w)$, each value between $-10$ and $10$, satisfying $x+y+z=w$ and $\frac1x+\frac1y+\frac1z=\frac1w$

Question

I need help moving forward with this question:

Determine the number of solutions $(x, y, z, w)$ of integers with each of $x$, $y$, $z$, and $w$ between $-10$ and $10$ inclusive, using these equations: $x + y + z = w; \frac1x + \frac1y + \frac1z = \frac1w.$

So far, I have gotten rid of the denominators and substituted $w = x + y + z$ in and have $2xyz + x^2z + x^2y + y^2x + y^2x + z^2x + z^2y$ and I have no idea where to go next.

Answer 1

building off of the second equation we have :

$$ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{x+y+z} \Rightarrow (yz+xz+xy)(x+y+z) = xyz $$ above was attained by multiplying both sides by $xyz(x+y+z)$ we have: $$ 2xyz + yz(y+z) + xz(x+z) + xy(x+y) = 0 $$

note that: $$ 2xyz + yz(y+z) + xz(x+z) + xy(x+y) = x(2yz + z^2 + y^2 + x(y+z)) + yz(y+z) = x((y+z)^2 + x(y+z)) + yz(y+z) = (y+z)(xy + xz + x^2 + yz) $$

so : $$ (y+z)(x^2 + xy + yz + xz) = 0 $$ Also note that : $$ x^2 + (y+z)x + yz = (x+y)(x+z) $$

so this simplifies to : $$ (x+y)(y+z)(x+z) = 0 $$ this means that the answer is : $$ A = \{(x, y, z, w) | (x+y)(y+z)(x+z) = 0, x, y, z\neq 0,-10\leq x, y, z \leq 10\} $$

there are many possible answers. including but not limited to : $$ (1, 1, -1, 1), (1, 2, -2, 1), (1, 3, -3, 1) ... $$

Answer 2

From your equations we have $$\begin{align}x+y+z &= w\\ w(xy+yz+zx) &= xyz \end{align}$$ Eliminating $w$, $$(x+y+z)(xy+yz+zx)=xyz$$ Compare with the well-known identity $(x+y+z)(xy+yz+zx)=xyz+(x+y)(y+z)(z+x)$ which is true for all values of $x,y,z$. It follows that our system is equivalent to the following equation:

$$\begin{align}(x+y)(y+z)(z+x)&=0 && \text{where }0\notin \{x,y,z\}\end{align}$$

Thus, $w$ is redundant (ie, always fixed after determining $x,y,z$) and any solution is just an ordered $3$-tuple of form $(x,y,z)$. Thus we have to construct triples with each element as a non-zero integer in $[-10,10]$ s.t. some two elements must add to $0$.

For this count, we first consider tuples of the form $(a,-a,b)$ in some order where $|b| \neq |a|$. First we choose the $\color{blue}{\text{position and value of the $b$ element}}$; then $\color{orange}{\text{count all posible permutations of $(a,-a)$}}$. Thus there are $\color{blue}{3\cdot 20}\cdot\color{orange}{9\cdot 2}=\color{magenta}{1080}$ such tuples. Then we consider tuples of form $(a,-a,a)$ in some order. We first $\color{blue}{\text{choose the value of $a$}}$ and then $\color{orange}{\text{choose the position of the negative sign}}$. Thus there are $\color{blue}{20}\cdot \color{orange}{3} = \color{magenta}{60}$ such tuples.

Thus, there are exactly $\color{magenta}{1080+60} = \color{teal}{1140}$ solutions.