# How I can prove or disprove that $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=1$ has solutions in rationals? [duplicate]

The motivation of this question is to look if there is such solution in rational number to the identity which mentioned here, I have done many attempts using Wolfram Alpha to find such pairs of rationals $$(x,y,z)$$ for which $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=1$$ but I failed even I believed that there are no such solutions?

Let, $$(x,y,z)$$ be real. Then $$a=x+y, b=y+z, c=z+x$$ all are reals.

Now, $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1$$

Or, $$(x+y+z)(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})=4$$

Or,

$$((x+y)+(y+z)+(z+x))(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})=8$$

Or, $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=8$$

Or,

$$(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})^2+(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}})^2+(\sqrt{\frac{c}{a}}-\sqrt{\frac{a}{c}})^2=-1$$

So, $$a,b,c$$ shouldn't be all positive or all negative.

Suppose, $$c$$ is negative then we change the equation as below

$$(a+b-c)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})=8$$ taking $$c>0$$.

Let, $$a-c=d>0, a>c>b$$. Then the equation becomes $$(d+b)(\frac{1}{b}-\frac{d}{m}), m=ac$$...$$(1)$$

Or, let $$c-a=d>0, a>c>b$$. Or, $$(b-d)(\frac{1}{b}+\frac{d}{m})$$....$$(2)$$

All other cases are symmetrically equivalent.

These two cases similarly implies if $$\sqrt{(\frac{b}{m}-\frac{1}{b})^2-\frac{28}{m})}$$ is rational, then there are rational solutions. It is only left to show whether there exists non zero rational $$b,m$$ such that $$\sqrt{((b²-m)^2-28b^2m)}=\text{rational}$$.

Now, $$(b²-m)^2-28b^2m)= {m}^2(t^2-30t+1)={m}^2(t-\alpha)(t-\alpha*)$$

where, $$l, t$$rational and $$t=\frac{b^2}{m}$$. And $$\alpha=15+\sqrt{224}$$ and $$\alpha*=15-\sqrt{224}$$.

Its our next job to find whether there is such $$t$$ such that $$(t-\alpha)(t-\alpha*)=r^2$$ for some rational $$r$$....$$(3)$$

1. $$t=30, r=1$$ is a trivial solution, $$b^2=30m=30ac$$ , but this doesn't give any solution as $$b=\sqrt{30ac} \nless \text{both} a,c$$ which is required in $$(1)$$ or $$(2)$$.

$$(1), (2)$$ and $$(3)$$ requires $$t<15-\sqrt{224}$$.

• square root of a real number is not necessarily real. – GTA May 2 '20 at 4:18