Find $x, y, z$ given $xyz = 60$, $x+y+z = 12$

Question

If this can be solved in a way so that there is a unique value for $x$, $y$, and $z$, is there a name for the method that can be used to solve this problem?

$$x+y+z = 12$$

$$xyz = 60$$

For this example, the values can be $3$,$4$,$5$, but I'm wondering if there is a formula or algorithm I can use to solve this problem. Lets say for our formula or algorithm, x y z can be any unique integer so that the following can be solved:

x+y+z = some constant

xyz = some other constant

Answer 1

GENERAL ALGORITHM:

In that case, we can see that we have three variables and two equations. To find values of three variables, we require three equations, but since we don't have that, we must try and reduce the problem to one of two variables and two equations.

To do that, in case of this question, would be to use the equation $xyz=60$

$$60=2^2\cdot 3 \cdot 5$$ The different factors of $60$ are:$$1,2,3,4,5,6,10,12,15,20,30,60$$

But since $x+y+z=12$, $x$ can't be greater than , or equal to $12$.

So, $x$ can take the values: $1,2,3,4,5,6,10$

We now come up with different possibilities of two variable equations. $$[(yz=60,y+z=11),(yz=30,y+z=10),(yz=20,y+z=9),(yz=15,y+z=8),(yz=12, y+z=7),(yz=10,y+z=6),(yz=6,y+z=2)]$$

Obviously, a lot of these combinations won't make sense, so you can easily eliminate them.

Agreed, this will be very lengthy for larger numbers, but this is the general idea behind solving such questions.

Answer 2

Presumably you are restricting $x,y,z$ to be natural numbers. You then need to look at factorization. We have $60=2^2\cdot 3 \cdot 5$ so $x,y,z$ can only have those factors. There are not many possibilities to look at. In larger Diophantine problems you can exclude possibilities based on the factors, but it is not worthwhile here.