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Question Number 194158 by Frix last updated on 28/Jun/23

$$\mathrm{Find}\mathrm{all}\mathrm{possible}\mathrm{solutions}:$$$$\frac{1}{s}+\frac{1}{t}+\frac{1}{u}+\frac{1}{v}=1$$$$\mathrm{With}s,t,u,v\in \mathbb{N}\mathrm{and}s<t<u<v$$

Answered by deleteduser1 last updated on 29/Jun/23

$$s<t\Rightarrow \frac{1}{s}>\frac{1}{t}\Rightarrow \frac{4}{s}>\frac{1}{s}+\frac{1}{t}+\frac{1}{u}+\frac{1}{v}=1\Rightarrow s<4$$$$s=1givesabsurdresult,sos=2or3$$$$whens=2,\frac{1}{t}+\frac{1}{u}+\frac{1}{v}=\frac{1}{2}\Rightarrow \frac{3}{t}>\frac{1}{2}\Rightarrow t<6$$$$t=1,2givesabsurdresults..Soconsider3,4or5$$$$t=3\Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{6}\Rightarrow u=\frac{6(v-6)+36}{v-6}=6+\frac{36}{v-6}<v$$$$\Rightarrow 6v<v^2-6v\Rightarrow v^2-12v>0\Rightarrow v>12$$$$\Rightarrow v-6\mid 36\Rightarrow v=15,18,24,42(sincev>12)$$$$Foreachofthevalues,wegetanintegervalue$$$$foru=10,9,8,7consecutively$$$$Similarly,wecangetvaluesfort=4or5$$$$t=4\Rightarrow \frac{2}{u}>\frac{1}{4}\Rightarrow u<8..Checkingfor5,6,7$$$$wegetu,v=(5,20);(6,12)$$$$t=5\Rightarrow \frac{2}{u}>\frac{3}{10}\Rightarrow u<\frac{20}{3}\Rightarrow u=6givesv=\frac{15}{2}\notin \mathbb{N}$$$$So,considers=3;weget\frac{1}{t}+\frac{1}{u}+\frac{1}{v}=\frac{2}{3}$$$$\frac{3}{t}>\frac{2}{3}\Rightarrow t<4.5\Rightarrow t=4\Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{5}{12}$$$$\frac{2}{u}>\frac{5}{12}\Rightarrow u<4.8\Rightarrow nosolutionsinceu>t=4$$$$s=2\Rightarrow t=3\Rightarrow (u,v)=(10,15),(9,18),(8,24),(7,42)$$$$s=2\Rightarrow t=4\Rightarrow (u,v)=(5,20);(6,12)$$

Commented by Frix last updated on 29/Jun/23

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