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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?

Glorfindel
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1 Answers1

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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 $t<15-\sqrt{224}$.

Alapan Das
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