Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
Best Answer
They are the only possible solution. Proof is as follows:
Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime. Substituting in the original equation we get $$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$ Solving for $d$: $$d=\frac{\left( y_0+x_0\right) \,z}{x_0\,y_0\,\left( z-1\right) }$$ Since $x_0$ and $y_0$ are co-prime, for $d$ to be an integer, $x_0\,y_0$ should divide $z$. Hence we require $$ z= k ~x_0 ~y_0$$ Substituting in the equation for $d$ and solving for $k$: $$ k=\frac{d}{d\,x_0\,y_0-y_0-x_0}$$ This shows that $d$ is a multiple of $k$. Let $$ d= \mu k$$. Then $$k=\frac{k\,\mu}{k\,\mu\,x_0\,y_0-y_0-x_0}$$ Solving for $k$: $$k=\frac{1}{x_0\,y_0}+\frac{1}{\mu\,y_0}+\frac{1}{\mu\,x_0}$$ This implies that $1 \le x_0 \le 3$, $1 \le y_0 \le 3$, $1 \le \mu\le 3$
It is possible to eliminate some of the 27 possible values since $k$ has to be an integers this will result in 12 possible values for $(x_0,y_0,\mu)$ and two of the solutions are repetitions giving the 10 solutions mentioned in the problem.