Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$

Question

I need to determine all positive integer solutions for the equation: $$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$

This is how I have tried to do it:

Mulitiplied both sides by $xyz$ to get $$yz+xz+xy+z+x+y+1=xyz.$$

Factor it: \begin{align} x(y+z+1-yz)+yz+y+z &= -1 \\ x(y(1-z)+z+1)+y(1+z)+z &= -1 \end{align}

If $z=0$, we get \begin{align} x(y+1)+y=-1 &\iff xy+x+y=-1 \\ &\iff (x+1)(y+1)=0, \end{align} which gives us $x=y=-1$.

Is this all positive integer solutions? Or have I missed something?

EDIT: I am stupid.

New attempt.

If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.

Then there is no solutions of positive integers for both $x$ and $y$ at the same time.

If I try for $z=2$, I get \begin{align} x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1 \\ &\iff 3x+3y+3-xy=0 \end{align} and I won't get a solution where all the variables are positive integers.

Answer 1

Hint: We can write the equation as $$ \left(1+\frac1 x\right)\left(1+\frac1 y\right)\left(1+\frac1 z\right)=2. $$ Assume without loss of generality $x\le y\le z$ by rearranging. Observe that $ 1+\frac 1 x<2\le \left(1+\frac1 x\right)^3 \implies x\in \{2,3\} $ and using this information, investigate each case $x=2,3$ and so on.

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Addendum, Solution.

1. For $x=2$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 43$. Then $y>3$ and only $(y,z)=(4,15), (5,9), (6,7)$ work. (Check this using $\frac{z+1}z =\frac{4}{3}\frac{y}{y+1}$.)
   
2. For $x=3$, we have that $\frac{y+1}{y}\cdot\frac{z+1}{z}=\frac 32$. Using $y\ge x=3$ and $\frac{z+1}z =\frac{3}{2}\frac{y}{y+1}$, check that only $(y,z)=(3,8), (4,5)$ work.
   

By rearranging, every solution is a permutation of such solutions with $x\le y\le z$.

Answer 2

Assume, without loss of generality, that $x\leq y\leq z$. We see from the original equation that $x>1$ (since $x = 1$ means $\frac1x + \cdots > 1$). At the same time, we must have $x<4$, as otherwise the sum is clearly less than $1$.

So, where does $x = 2$ actually take us? We insert and get $$ \frac12 + \frac1y + \frac1z + \frac1{2y} + \frac1{2z} + \frac1{yz} + \frac1{2yz} = 1\\ \frac3{2y} + \frac3{2z} + \frac3{2yz} = \frac12\\ 3z + 3y + 3 = yz\\ 12 = yz - 3y - 3z + 9 = (y-3)(z-3) $$ Since $y$ and $z$ are integers, this is an easy solve.

What about $x = 3$? We get $$ \frac13 + \frac1y + \frac1z + \frac1{3y} + \frac1{3z} + \frac1{yz} + \frac1{3yz} = 1\\ \frac4{3y} + \frac4{3z} + \frac4{3yz} = \frac23\\ 4z + 4y + 4 = 2yz\\ 2z + 2y + 2 = yz\\ 6 = yz - 2y-2z + 4 = (y-2)(z-2) $$ which, again, is an easy solve using the fact that $y, z$ are integers.

Answer 3

This thing factors nicely after adding 1 to both sides. Not that it helps us much, though.

$$\left(1+{1\over x}\right)\cdot\left(1+{1\over y}\right)\cdot\left(1+{1\over z}\right)=2$$

Basically, it all boils down to brute force search which can easily be done by hand.

Let's assume WLOG $x\leqslant y\leqslant z$. Then just run through the possible values:

• $x=1$: LHS is too big regardless of $y$ and $z$.
• $x=2$:
  • $y=2$: LHS is too big regardless of $z$.
  • $y=3$: same as above.
  • $y=4$: need to check.
  • $y=5$: need to check.
  • $y=6$: need to check.
  • $y\geqslant7$: LHS is too small regardless of $z$.
• $x=3$:
  • $y=3$: need to check.
  • $y=4$: need to check.
  • $y\geqslant5$: LHS is too small regardless of $z$.
• $x\geqslant4$: LHS is too small regardless of $y$ and $z$.

After checking each of the cases marked "need to check", you'll end up with quite a bunch of nice positive solutions.

Answer 4

Rewrite the equation as $$ z( - xy + x + y + 1) + (x+1)(y+1)=0. $$ Suppose first that $xy=x+y+1$, i.e., $y(1-x)=x+1$. Then $x\neq 1$, so that $y=\frac{x+1}{1-x}$. Substituting this into the original equation gives $$ x(x+1)=0, $$ a contradiction. Now assume that $- xy + x + y + 1\neq 0$. Then $$ z=\frac{(x+1)(y+1)}{xy-x-y-1}. $$ When can this be a positive integer? For $x=1$ we have $z=-y-1<0$, a contradiction. For $x=2$ we have $$ z=\frac{3y+3}{y-3}, $$ which has solutions for $y=4,5, 6,7,9,15$. For $x=3$ we obtain $$ z=\frac{2y+2}{y-2}. $$ This is an integer for $y=3,4,5,8.$ For $x\ge 4$ it is easy to see that we have no solution in positive integers.

Answer 5

Let $x\geq y\geq z$.

Thus, $$1\leq\frac{1}{z^3}+\frac{3}{z^2}+\frac{3}{z}$$ or $$z^3\leq1+3z+3z^2$$ and we got some values of $z$: $$1\leq z\leq3.$$ Now, for all value of $z$ we can make the similar thing with $y$.