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Question Number 206879 by mr W last updated on 29/Apr/24
solve for positive integers (x/(y+z))+(y/(z+x))+(z/(x+y))=4
$${solve}\:{for}\:{positive}\:{integers} \\ $$$$\frac{{x}}{{y}+{z}}+\frac{{y}}{{z}+{x}}+\frac{{z}}{{x}+{y}}=\mathrm{4} \\ $$
Commented by Frix last updated on 30/Apr/24
I think it′s impossible.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{impossible}. \\ $$
Commented by mr W last updated on 30/Apr/24
https://www.quora.com/a-b-c-b-a-c-c-a-b-4-What-will-be-values-of-a-b-c
Commented by mr W last updated on 30/Apr/24
https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
Commented by Frix last updated on 30/Apr/24
Ok, just impossible for ordinary mortals like me. Interesting links, thank you!
$$\mathrm{Ok},\:\mathrm{just}\:\mathrm{impossible}\:\mathrm{for}\:\mathrm{ordinary}\:\mathrm{mortals} \\ $$$$\mathrm{like}\:\mathrm{me}.\:\mathrm{Interesting}\:\mathrm{links},\:\mathrm{thank}\:\mathrm{you}! \\ $$
Commented by mr W last updated on 30/Apr/24
me too!
$${me}\:{too}! \\ $$
Answered by Rasheed.Sindhi last updated on 29/Apr/24
(x/(y+z))+1+(y/(z+x))+1+(z/(x+y))+1=4+3 ((x+y+z)/(y+z))+((x+y+z)/(z+x))+((x+y+z)/(x+y))=7 (x+y+z)((1/(x+y))+(1/(y+z))+(1/(z+x)))=7 3≤x+y+z∈Z^+ 7=k×(7/k)=7k×(1/k)=... x+y+z=k ∧ (1/(x+y))+(1/(y+z))+(1/(z+x))=(7/k)...(i) OR x+y+z=7k ∧ (1/(x+y))+(1/(y+z))+(1/(z+x))=(1/k)...(ii) (i)⇒x+y+z=k ∧ (1/(k−x))+(1/(k−y))+(1/(k−z))=(7/k)...(i) ∧ ((k−x+k−y)/(k^2 −kx−ky+xy))+(1/(k−z))=(7/k) ∧ ((2k−(x+y))/(k^2 −k(x+y)+xy))+(1/(k−z))=(7/k) ∧ ((2k−z)/(k^2 −kz+xy))+(1/(k−z))=(7/k) ...
$$\frac{{x}}{{y}+{z}}+\mathrm{1}+\frac{{y}}{{z}+{x}}+\mathrm{1}+\frac{{z}}{{x}+{y}}+\mathrm{1}=\mathrm{4}+\mathrm{3} \\ $$$$\frac{{x}+{y}+{z}}{{y}+{z}}+\frac{{x}+{y}+{z}}{{z}+{x}}+\frac{{x}+{y}+{z}}{{x}+{y}}=\mathrm{7} \\ $$$$\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}+{y}}+\frac{\mathrm{1}}{{y}+{z}}+\frac{\mathrm{1}}{{z}+{x}}\right)=\mathrm{7} \\ $$$$\mathrm{3}\leqslant{x}+{y}+{z}\in\mathbb{Z}^{+} \\ $$$$\mathrm{7}={k}×\frac{\mathrm{7}}{{k}}=\mathrm{7}{k}×\frac{\mathrm{1}}{{k}}=… \\ $$$${x}+{y}+{z}={k}\:\wedge\:\frac{\mathrm{1}}{{x}+{y}}+\frac{\mathrm{1}}{{y}+{z}}+\frac{\mathrm{1}}{{z}+{x}}=\frac{\mathrm{7}}{{k}}…\left({i}\right) \\ $$$$\mathrm{OR} \\ $$$${x}+{y}+{z}=\mathrm{7}{k}\:\wedge\:\frac{\mathrm{1}}{{x}+{y}}+\frac{\mathrm{1}}{{y}+{z}}+\frac{\mathrm{1}}{{z}+{x}}=\frac{\mathrm{1}}{{k}}…\left({ii}\right) \\ $$$$\left({i}\right)\Rightarrow{x}+{y}+{z}={k}\:\wedge\:\frac{\mathrm{1}}{{k}−{x}}+\frac{\mathrm{1}}{{k}−{y}}+\frac{\mathrm{1}}{{k}−{z}}=\frac{\mathrm{7}}{{k}}…\left({i}\right) \\ $$$$\:\:\:\:\:\:\wedge\:\frac{{k}−{x}+{k}−{y}}{{k}^{\mathrm{2}} −{kx}−{ky}+{xy}}+\frac{\mathrm{1}}{{k}−{z}}=\frac{\mathrm{7}}{{k}} \\ $$$$\:\:\:\:\:\:\wedge\:\frac{\mathrm{2}{k}−\left({x}+{y}\right)}{{k}^{\mathrm{2}} −{k}\left({x}+{y}\right)+{xy}}+\frac{\mathrm{1}}{{k}−{z}}=\frac{\mathrm{7}}{{k}} \\ $$$$\:\:\:\:\:\:\wedge\:\frac{\mathrm{2}{k}−{z}}{{k}^{\mathrm{2}} −{kz}+{xy}}+\frac{\mathrm{1}}{{k}−{z}}=\frac{\mathrm{7}}{{k}} \\ $$$$… \\ $$

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