What are $x$, $y$ and $z$ if $\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y} = 4$ and $x$, $y$ and $z$ are whole numbers?

Question

What are $x$, $y$ and $z$ if $$\dfrac{x}{y + z} + \dfrac{y}{x + z} + \dfrac{z}{x + y} = 4$$ and $x$, $y$ and $z$ are whole numbers?

MY ATTEMPT

Let $u = x + y + z$. Then the equation can be rewritten as $$\dfrac{x}{u - x} + \dfrac{y}{u - y} + \dfrac{z}{u - z} = 4$$

Suppose I set $$1 = \dfrac{x}{u - x} = \dfrac{y}{u - y}$$ and $$2 = \dfrac{z}{u - z}.$$

Then I get $$x = y + z$$ $$y = x + z$$ $$z = 2(x + y),$$ so that $$z = 0 = x + y,$$ which is impossible.

Next, suppose I set $$\dfrac{4}{3} = \dfrac{x}{u - x} = \dfrac{y}{u - y} = \dfrac{z}{u - z}.$$

Then I get $$4(u - x) = 3x$$ $$4(u - y) = 3y$$ $$4(u - z) = 3z$$ so that $$12u - 4(x + y + z) = 3(x + y + z)$$ which implies that $$12u = 7(x + y + z) = 7u$$ from which it follows that $$u = 0.$$ This is, again, impossible.

Alas, here is where I get stuck. Any hint(s) will be appreciated.

Answer 1

If we set $$x+y=u, y+z= v, z+x=w $$

We have $$x=\frac{u+w-v}{2}, y=\frac{u+v-w}{2}, z=\frac{w+v-u}{2}$$

Now $$2\sum_{cyc} \frac{z}{x+y}=\sum_{cyc}\frac{u+v-w}{w}=8 \Rightarrow \sum_{cyc} \frac{u+v}{w}=11$$ However, one can prove that the only naturals that can be expressed of the $$\sum_{cyc} \frac{a+b}{c}$$

Where $a,b,c \in \mathbb{N}$ are $6,7,8$. This is relatively simple, as we can assume $\text{WLOG}$ $$ a \ge b \ge c, \dfrac{b+c}{a} \in \mathbb{N}$$so I leave this to you.

EDIT

The previous answer here is, unfortunately, incorrect. This is because I got confused and accidentally assumed that $(x,y,z)$ are all relatively coprime. However, as this is not implied in the question, my answer written here is incorrect.

The correct answer is written here, where the user @Next gives us a solution for $$\sum_{cyc} \frac{a}{b+c}=4$$ It is the following.

$a=$2332797891204725453580403814216955612718693675609518139813675622446336 8530351921955206357565424226029748329737767516130520072674084336131550 2597616224970927979227396663481447506019173462295157784788781420305046 5201815993661680059006448575315523206103260762210944137954571975497854 9786027663601160534574317253280344812956727894696796553762212813889660 9065956718516224446015577143267128739011935697434909021669583635379832 35022557869209259

$b=$1161640217306132458900911441651415023972393417197892812143262449233898 8034221463466278254018560734492913221738943224762433374574861704275058 0062902808034990817009121975186967451351814311101112040391014295321972 8784138582766210837461563508481437266175417187186208008663435889653439 7066554486263784443013141020886435995672339322997499528376940620045001 1919735272479457688230567501843839892799164246003766614214017398378635 0444307965016411

$c=$5054729227475450427274369484803239479825091305751388135572448603576037 6549781961422098862259430557133842304461180359698183208339647924784255 6816542651386138853492649101592171641096957016404851774814750638840260 3496289958758089911825477669004739864966841494437579004665357462952425 4130327474390635537868978719887059697148297723373566417781389238382736 3204638301684342182024187145267526992579708085994452308601529371953916 7125415529515145

And there are infinitely many other solutions. I apologize for my previously incorrect answer. There does exist such $x,y,z$.