0
$\begingroup$

A serious challenge:

Can someone find 3 positive whole numbers that solve this equation?
$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4$$

The numbers must be whole!

$\endgroup$
  • $\begingroup$ Do you have the answer? $\endgroup$ – CY Aries Jun 14 '17 at 17:22
  • 2
    $\begingroup$ what have you tried so far? $\endgroup$ – Dando18 Jun 14 '17 at 17:22
  • 2
    $\begingroup$ @soccerStack If you already know the answer, then this isn't the website to post a puzzle. You should post it on the puzzling SE website. $\endgroup$ – Χpẘ Jun 14 '17 at 18:20
  • 2
    $\begingroup$ Possible duplicate of Find answer of $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=4$ $\endgroup$ – Dietrich Burde Jun 14 '17 at 18:54
  • 1
    $\begingroup$ @soccerStack There is one given in the MO question which is linked at the duplicate, in the article "An Unusual Cubic Representation Problem" by Andrew Bremner. It has " truly enormous size". $\endgroup$ – Dietrich Burde Jun 14 '17 at 19:15
6
$\begingroup$

An answer has been given by Michael Stoll at this MO-question, linked at this MSE-duplicate, using the elliptic curve

$$E_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr) =: x(x^2 + Ax + B),$$

where in our case $n=4$. Then the curve is known to have rank $1$, and thus there is a solution in positive integers of "truly enormous size", see the article "An Unusual Cubic Representation Problem" by Andrew Bremner. It is given by $$ x = 437361267792869725786125260237139015281653755816161361862143‌​7993378423467772036; $$

$$ y = 368751317941299998271978115652254748254929799689719709962831‌37471637224634055579‌​; $$

$$ z = 154476802108746166441951315019919837485664325669565431700026‌63489825320203527799‌​9 $$ There are other solutions in positive integers, of course, but this is the smallest one.

$\endgroup$
  • 2
    $\begingroup$ Typical of the expected enormity of whole solutions, when they exist, in some elliptic curves. $\endgroup$ – Piquito Jun 14 '17 at 19:26
  • $\begingroup$ Is this solution possibly a multiple of Eric Lee's partly negative solution $(11,9,-5)$? $\endgroup$ – Hagen von Eitzen Jun 14 '17 at 19:56
  • $\begingroup$ I have a negative number in my solution, I didn't realize they had to be positive $\endgroup$ – Eric Lee Jun 14 '17 at 21:22
-1
$\begingroup$

$x=11$, $y=9$, $z=-5$

I brute forced it.

$\endgroup$
  • 2
    $\begingroup$ The question has more detail than the title, and requires positive integer solutions. $\endgroup$ – user326210 Jun 14 '17 at 17:57
  • 2
    $\begingroup$ Still, it's good to know that there are any integer solutions at all. $\endgroup$ – user49640 Jun 14 '17 at 17:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.