Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)
If $\displaystyle x,y\geq 4$ then $\displaystyle 4xy > 5(x+y+4)$.
We will show that if $\displaystyle x,y,z\geq 4$ then $\displaystyle xyz > 5(x+y+z)$. We will use induction starting at $\displaystyle z=4$. At $\displaystyle z=4$ this statement is true by above. Say $\displaystyle z$ is true i.e. $\displaystyle xyz > 5(x+y+z)$ then $\displaystyle xy(z+1) = xyz + xy > 5(x+y+z) + 5 = 5(x+y+(z+1))$. This completes induction.
This tells us that if $\displaystyle z\geq 4$ then it must mean that $\displaystyle x=1,2,3$ and $\displaystyle y=1,2,3$ for there to be a possible solution. And if $\displaystyle z=1,2,3$ then there is a chance to be a solution. Since this is a symmetric equation it means if $\displaystyle (a,b,c)$ is a solution then any permutation is a solution. Therefore, it is sufficient to check whether this equation has integer solution for $\displaystyle z=1,2,3$.
If $\displaystyle z=1$ then the equation is $\displaystyle xy = 5(x+y+1)$. We can write $\displaystyle xy-5x-5y=5$. Then add and subtract $\displaystyle 25$ to get, $\displaystyle xy-5x-5y+25 = 30$, finally factor, $\displaystyle (x-5)(y-5)=30$. The possible solutions are $\displaystyle x-5=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$ and $\displaystyle y-5 = \pm 30,\pm 15,\pm 10, \pm 6, \pm 5,\pm 3,\pm 2,\pm 1$. But because $\displaystyle x,y>0$ we cannot choose the negative solutions. Thus, in fact the solutions are $\displaystyle x=6,7,8,10,11,15,,20,35$ and $\displaystyle y=35,20,15,11,10,8,7,6$ (in that order). Because of permuting solutions we can ignore some and write: $\displaystyle (6,35,1),(7,20,1),(8,15,1),(10,11,1)$.
If $\displaystyle z=2$ then the equation is $\displaystyle 2xy = 5(x+y+2)$. Rewrite as $\displaystyle (2x-5)(2y-5) = 45$. The solutions are (no negatives) $\displaystyle 2x-5=1,3,5,9,15,45$ and $\displaystyle 2y-5=45,15,9,5,3,1$. Thus, $\displaystyle x=3,4,5,7,10,25$ and $\displaystyle y=25,10,7,5,4,3$. This gives us $\displaystyle (3,25,2),(4,10,2),(5,7,2)$.
If $\displaystyle z=3$ then equation is $\displaystyle 3xy = 5(x+y+3)$. Rewrite as $\displaystyle (3x-5)(3y-5)=70$. Thus, $\displaystyle 3x-5=1,2,5,7,10,14,35,70$ and $\displaystyle 3x-5=70,35,14,10,7,5,2,1$. This gives $\displaystyle x=2,4,5,25$ and $\displaystyle y=25,5,4,2$. This gives us $\displaystyle (2,25,3),(4,5,3)$.
And if $\displaystyle z\geq 4$ then as explained above there are no other solutions if we think of permutations.