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Find-all-possible-solutions-1-s-1-t-1-u-1-v-1-With-s-t-u-v-N-and-s-lt-t-lt-u-lt-v-

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Question Number 194158 by Frix last updated on 28/Jun/23
Find all possible solutions: (1/s)+(1/t)+(1/u)+(1/v)=1 With s, t, u, v ∈N and s<t<u<v
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}: \\ $$$$\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1} \\ $$$$\mathrm{With}\:{s},\:{t},\:{u},\:{v}\:\in\mathbb{N}\:\mathrm{and}\:{s}<{t}<{u}<{v} \\ $$
Answered by AST last updated on 29/Jun/23
s<t⇒(1/s)>(1/t)⇒(4/s)>(1/s)+(1/t)+(1/u)+(1/v)=1⇒s<4 s=1 gives absurd result,so s=2 or 3 when s=2,(1/t)+(1/u)+(1/v)=(1/2)⇒(3/t)>(1/2)⇒t<6 t=1,2 gives absurd results.. So consider 3,4 or 5 t=3⇒(1/u)+(1/v)=(1/6)⇒u=((6(v−6)+36)/(v−6))=6+((36)/(v−6))<v ⇒6v<v^2 −6v⇒v^2 −12v>0⇒v>12 ⇒v−6∣36⇒v=15,18,24,42(since v>12) For each of the values,we get an integer value for u=10,9,8,7 consecutively Similarly,we can get values for t=4 or 5 t=4⇒(2/u)>(1/4)⇒u<8..Checking for 5,6,7 we get u,v=(5,20);(6,12) t=5⇒(2/u)>(3/(10))⇒u<((20)/3)⇒u=6 gives v=((15)/2)∉N So,consider s=3;we get (1/t)+(1/u)+(1/v)=(2/3) (3/t)>(2/3)⇒t<4.5⇒t=4⇒(1/u)+(1/v)=(5/(12)) (2/u)>(5/(12))⇒u<4.8⇒no solution since u>t=4 s=2⇒t=3⇒(u,v)=(10,15),(9,18),(8,24),(7,42) s=2⇒t=4⇒(u,v)=(5,20);(6,12)
$${s}<{t}\Rightarrow\frac{\mathrm{1}}{{s}}>\frac{\mathrm{1}}{{t}}\Rightarrow\frac{\mathrm{4}}{{s}}>\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1}\Rightarrow{s}<\mathrm{4} \\ $$$${s}=\mathrm{1}\:{gives}\:{absurd}\:{result},{so}\:{s}=\mathrm{2}\:{or}\:\mathrm{3} \\ $$$${when}\:{s}=\mathrm{2},\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{3}}{{t}}>\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{t}<\mathrm{6} \\ $$$${t}=\mathrm{1},\mathrm{2}\:{gives}\:{absurd}\:{results}..\:{So}\:{consider}\:\mathrm{3},\mathrm{4}\:{or}\:\mathrm{5} \\ $$$${t}=\mathrm{3}\Rightarrow\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{\mathrm{6}}\Rightarrow{u}=\frac{\mathrm{6}\left({v}−\mathrm{6}\right)+\mathrm{36}}{{v}−\mathrm{6}}=\mathrm{6}+\frac{\mathrm{36}}{{v}−\mathrm{6}}<{v} \\ $$$$\Rightarrow\mathrm{6}{v}<{v}^{\mathrm{2}} −\mathrm{6}{v}\Rightarrow{v}^{\mathrm{2}} −\mathrm{12}{v}>\mathrm{0}\Rightarrow{v}>\mathrm{12} \\ $$$$\Rightarrow{v}−\mathrm{6}\mid\mathrm{36}\Rightarrow{v}=\mathrm{15},\mathrm{18},\mathrm{24},\mathrm{42}\left({since}\:{v}>\mathrm{12}\right) \\ $$$${For}\:{each}\:{of}\:{the}\:{values},{we}\:{get}\:{an}\:{integer}\:{value} \\ $$$${for}\:{u}=\mathrm{10},\mathrm{9},\mathrm{8},\mathrm{7}\:{consecutively} \\ $$$${Similarly},{we}\:{can}\:{get}\:{values}\:{for}\:{t}=\mathrm{4}\:{or}\:\mathrm{5} \\ $$$${t}=\mathrm{4}\Rightarrow\frac{\mathrm{2}}{{u}}>\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{u}<\mathrm{8}..{Checking}\:{for}\:\mathrm{5},\mathrm{6},\mathrm{7} \\ $$$${we}\:{get}\:{u},{v}=\left(\mathrm{5},\mathrm{20}\right);\left(\mathrm{6},\mathrm{12}\right) \\ $$$${t}=\mathrm{5}\Rightarrow\frac{\mathrm{2}}{{u}}>\frac{\mathrm{3}}{\mathrm{10}}\Rightarrow{u}<\frac{\mathrm{20}}{\mathrm{3}}\Rightarrow{u}=\mathrm{6}\:{gives}\:{v}=\frac{\mathrm{15}}{\mathrm{2}}\notin\mathbb{N} \\ $$$${So},{consider}\:{s}=\mathrm{3};{we}\:{get}\:\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\mathrm{3}}{{t}}>\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow{t}<\mathrm{4}.\mathrm{5}\Rightarrow{t}=\mathrm{4}\Rightarrow\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\frac{\mathrm{2}}{{u}}>\frac{\mathrm{5}}{\mathrm{12}}\Rightarrow{u}<\mathrm{4}.\mathrm{8}\Rightarrow{no}\:{solution}\:{since}\:{u}>{t}=\mathrm{4} \\ $$$${s}=\mathrm{2}\Rightarrow{t}=\mathrm{3}\Rightarrow\left({u},{v}\right)=\left(\mathrm{10},\mathrm{15}\right),\left(\mathrm{9},\mathrm{18}\right),\left(\mathrm{8},\mathrm{24}\right),\left(\mathrm{7},\mathrm{42}\right) \\ $$$${s}=\mathrm{2}\Rightarrow{t}=\mathrm{4}\Rightarrow\left({u},{v}\right)=\left(\mathrm{5},\mathrm{20}\right);\left(\mathrm{6},\mathrm{12}\right) \\ $$
Commented by Frix last updated on 29/Jun/23
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