# Thread: Find all positive integers such that...?

1. ## Find all positive integers such that...?

Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)

2. (3,4,5)
(2,5,7)
(2,4,10)
(2,3,25)

I believe these are the only solutions. My method of solving was nothing to speak of, aside from starting with the recognition that at least one of the variables had to be a multiple of five.

3. The solution itself does not carry any importance. Does anybody know the method of solving this problem?

4. Originally Posted by fardeen_gen
Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)
If $x,y\geq 4$ then $4xy > 5(x+y+4)$.

We will show that if $x,y,z\geq 4$ then $xyz > 5(x+y+z)$. We will use induction starting at $z=4$. At $z=4$ this statement is true by above. Say $z$ is true i.e. $xyz > 5(x+y+z)$ then $xy(z+1) = xyz + xy > 5(x+y+z) + 5 = 5(x+y+(z+1))$. This completes induction.

This tells us that if $z\geq 4$ then it must mean that $x=1,2,3$ and $y=1,2,3$ for there to be a possible solution. And if $z=1,2,3$ then there is a chance to be a solution. Since this is a symmetric equation it means if $(a,b,c)$ is a solution then any permutation is a solution. Therefore, it is sufficient to check whether this equation has integer solution for $z=1,2,3$.

If $z=1$ then the equation is $xy = 5(x+y+1)$. We can write $xy-5x-5y=5$. Then add and subtract $25$ to get, $xy-5x-5y+25 = 30$, finally factor, $(x-5)(y-5)=30$. The possible solutions are $x-5=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$ and $y-5 = \pm 30,\pm 15,\pm 10, \pm 6, \pm 5,\pm 3,\pm 2,\pm 1$. But because $x,y>0$ we cannot choose the negative solutions. Thus, in fact the solutions are $x=6,7,8,10,11,15,,20,35$ and $y=35,20,15,11,10,8,7,6$ (in that order). Because of permuting solutions we can ignore some and write: $(6,35,1),(7,20,1),(8,15,1),(10,11,1)$.

If $z=2$ then the equation is $2xy = 5(x+y+2)$. Rewrite as $(2x-5)(2y-5) = 45$. The solutions are (no negatives) $2x-5=1,3,5,9,15,45$ and $2y-5=45,15,9,5,3,1$. Thus, $x=3,4,5,7,10,25$ and $y=25,10,7,5,4,3$. This gives us $(3,25,2),(4,10,2),(5,7,2)$.

If $z=3$ then equation is $3xy = 5(x+y+3)$. Rewrite as $(3x-5)(3y-5)=70$. Thus, $3x-5=1,2,5,7,10,14,35,70$ and $3x-5=70,35,14,10,7,5,2,1$. This gives $x=2,4,5,25$ and $y=25,5,4,2$. This gives us $(2,25,3),(4,5,3)$.

And if $z\geq 4$ then as explained above there are no other solutions if we think of permutations.