I don’t know how exhaustive your search is going to be, but just as a result of the algebra, if a,b,c is a solution, so is na, nb, nc, where n is any integer.

yes am aware of that

To find **x,y,z** as positive integer where following equation is valid:

(1) x/y + y/z + z/x = R (R is real num)

(2) x^2/y + x*y/z + z = R*x

(3) x^2/y + x*(y/z-R) + z = 0

we have to solve quadratic equation (3) (find x1,x2) for given **y** and **z**.

If we select any z and try to find all pairs {x,y} for which (1) is valid we can do it by iterating over y to find solution for x that satisfy quadratic equation (3)

For quadratic equation a* x^2 + b* x + c = 0 we have solutions:

(4) x1= -b/2a + Sqrt(D)/2a

x2= -b/2a - Sqrt(D)/2a

where D is discriminant of quadratic equation:

(5) D= b^2 – 4ac

It is obvious from (5) that quadratic equation has real solution if D>=0 and if D=0 it has only one solution, x1 = x2 = -b/2a.

Code for iterating over **y** for given z is in c# component.

findingXY_Z.gh (6.9 KB)

You are awesome… Thank you very much for this.

So here a = 1/y and b = (y/z - R) and C = z and y iterates from 1 to n (value whatever we set for y max) and we find x value and it is substituted to quadratic equation to make it satisfy to zero … Is that right?

Correct. And you are looking just for solution where x, y and z are integeres.

For some combination of **y** and **z** you will have two soultion of your quadratic eqution.

Generaly, for example if eqution is:

x^2 - 16 = 0

it is obvious that solution is valid for two values of x, x1= -4 and x2= +4.

Quadratic equation may have none solution (solution is not in real numbers).

For example if quadratic equation is:

x^2 + 16 = 0

then solution is complex number x1= 4**i** and x2= -4**i**

( **i** is imaginary number, by definition ** i**^2= -1, see https://en.wikipedia.org/wiki/Imaginary_number )

R

Thank you, Its like middle school all over again.