Consider the equation $(x+1)(y+1)(z+1)=3xyz$. I want to find all positive integer solutions such that $x\le y\le z$ to this equation using only pen and paper and mathematical techniques. How can I do this rigorously(ie also showing that the solutions you come up with are the only solutions)? An at least semiformal solution would be appreciated.


2 Answers 2


Dividing by $xyz$, we are led to $$\tag1\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)=3,$$ hence (assuming $0<x\le y\le z$), we need that the biggest factor $1+\frac1x$ is at least $\sqrt[3]3$, i.e. $x\le 2$.

Case (i): $x=1$. Then the equation $(1)$ becomes $$\tag2\left(1+\frac1y\right)\left(1+\frac1z\right)=\frac32$$ and we conclude $1+\frac1y\ge\sqrt{\frac32}$, i.e. $y\le4$. Thus the allowed cases $y=3$ and $y=4$ can be easily checked by hand: $y=3$ leads to $1+\frac1z = \frac98$, i.e. $z=8$, whereas $y=4$ leads to $1+\frac1z = \frac{6}{5}$, i.e. $z=5$.

Case (ii): $x=2$. Then the equation $(1)$ becomes $$\left(1+\frac1y\right)\left(1+\frac1z\right)=2$$ and we conclude $1+\frac1y\ge\sqrt{2}$, i.e. $y\le2$ and together with $y\ge x$ this means $y=2$ and finally $1+\frac1z=\frac43$, i.e. $z=3$.

In summary, the only solutions $(x,y,z)\in \mathbb N$ with $x\le y\le z$ are $(1,3,8)$, $(1,4,5)$, and $(2,2,3)$.


The LHS has $xyz$ as its highest power term and the RHS has $3xyz$ so if we consider large $x$ the RHS should dominate the LHS. That would let us eliminate large values of $x$ and only have to deal with small $x$.

To do this we need to find some lower bound on $x$ that $$(x+1)(y+1)(z+1)-xyz = (xy + xz + yz) + (x + y + z) + 1$$ cannot beat $2xyz$. Because $x \le y \le z$ we can bound $$(xy + xz + yz) + (x + y + z) + 1 \le 2xz + yz + 2x + z + 1$$ whereas $$2xyz > 2yz + yz + yz + 2yz + 2yz \ge 2xz + yz + 2x + z + 1$$ as long as $x > 4$ (compare term by term using $x \le y \le z$ to see the last inequality). This shows the Diophantine equation cannot hold for any $x > 4$ (and in fact you could make this bound tighter then you'd have less cases to do).

Now we only have to characterize the solutions to any of:

  • $2(y+1)(z+1) = 3yz$
  • $3(y+1)(z+1) = 6yz$
  • $4(y+1)(z+1) = 9yz$
  • $5(y+1)(z+1) = 12yz$

these are all easy to do I'll show how to do $5(y+1)(z+1) = 12yz$: it's just finding positive integer solutions of $$y = \frac{5z + 5}{7z - 5}$$ and that demands $5z+5 \ge 7z-5$ which only happens when $z \le 5$ so just check each one and we get solutions $(4,5,1)$ but obviously discard that since it's not in order (they will all come up in the correct order).


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