Factoring Polynomials with four terms and three variables

Asked Feb 16 '18 at 14:55

Active Feb 16 '18 at 15:43

Viewed 200 times

If $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 17,$$ then solve for $x$, $y$ and $z$, where $x$, $y$, and $z$ are positive integers.

I got a common denominator of $xyz$, but then after that I was stuck:

$$x^2z + xy^2 + yz^2 = 17xyz$$

I also tried the quadratic formula, solving for $x$, but that was also a dead end:

$$x = \frac{-y^2+17zy \pm \sqrt{(y^2-17zy)^2-4z^3y}}{2z}.$$

I even wrote a rudimentary program to test a few numbers:

// index.js
function solve() {
    const limit = 10000;
    for (x = 1; x < limit; x++) {
        for (y = 1; y < limit; y++) {
            for (z = 1; z < limit; z++) {
                if (x*x*z + x*y*y + y*z*z === 17*x*y*z)
                    return { x, y, z };
            }
        }
    }
}

console.log(solve());

Getting undefined. Any help would be appreciated.

edited Feb 16 '18 at 15:43

asked Feb 16 '18 at 14:55

azizj

1

$x=6845,y=450,z=11988$. If $x,y,z$ is a solution, then so is $nx,ny,nz$ so we need only consider cases where $x,y,z$ have no common factor. It is easy to show that if $p^n$ is highest power of prime $p$ dividing $x$, then $p^{2m}$ is highest power dividing $y$. Similarly, for other pairs, so we put $x=uw^2,y=u^2v,z=v^2w$ and then $u^3+v^3+w^3=17uvw$. – almagest Feb 16 '18 at 16:35
3

Related Integer values of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$? – kingW3 Feb 16 '18 at 16:48
Damn. Thanks @almagest. I don't really understand the "it is easy to show that if ..." can you elaborate? – azizj Feb 16 '18 at 17:11
and how'd you get $2m$ from $n$? – azizj Feb 16 '18 at 17:16
And why bring power primes into this? – azizj Feb 16 '18 at 17:16

Factoring Polynomials with four terms and three variables

0 Answers0