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Question Number 128636 by john_santu last updated on 09/Jan/21

Solve diopthantine equation (1/a)+(1/b) = (2/(17)).

$$\mathrm{Solve}\:\mathrm{diopthantine}\:\mathrm{equation}\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}\:=\:\frac{\mathrm{2}}{\mathrm{17}}. \\ $$

Answered by liberty last updated on 09/Jan/21

⇒ (1/a)+(1/b)=(2/(17)) ; 2ab = 17(a+b) consider (2a−17)(2b−17)=4ab−34(a+b)+17^2 it follows that 4ab−34(a+b)=0 so we get (2a−17)(2b−17)=17^2 since a and b are positive integer we can take (i) 2a−17=1 and 2a−17=17^2 which yields (a,b)=(9,103) (ii) 2a−17=17 and 2a−17=17which yields (a,b)=(17,17) (iii) 2a−17=17^2 and 2a−17=1 which yields (a,b)=(103, 9) ∴ ≡ solution {(9,103),(17,17),(103,9)}

$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{2}}{\mathrm{17}}\:;\:\mathrm{2ab}\:=\:\mathrm{17}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{consider}\:\left(\mathrm{2a}−\mathrm{17}\right)\left(\mathrm{2b}−\mathrm{17}\right)=\mathrm{4ab}−\mathrm{34}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{17}^{\mathrm{2}} \\ $$$$\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{4ab}−\mathrm{34}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{2a}−\mathrm{17}\right)\left(\mathrm{2b}−\mathrm{17}\right)=\mathrm{17}^{\mathrm{2}} \\ $$$$\mathrm{since}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{we}\:\mathrm{can}\: \\ $$$$\mathrm{take}\:\left(\mathrm{i}\right)\:\mathrm{2a}−\mathrm{17}=\mathrm{1}\:\mathrm{and}\:\mathrm{2a}−\mathrm{17}=\mathrm{17}^{\mathrm{2}} \\ $$$$\:\mathrm{which}\:\mathrm{yields}\:\left(\mathrm{a},\mathrm{b}\right)=\left(\mathrm{9},\mathrm{103}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{2a}−\mathrm{17}=\mathrm{17}\:\mathrm{and}\:\mathrm{2a}−\mathrm{17}=\mathrm{17which}\:\mathrm{yields} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)=\left(\mathrm{17},\mathrm{17}\right) \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{2a}−\mathrm{17}=\mathrm{17}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{2a}−\mathrm{17}=\mathrm{1}\:\mathrm{which}\:\mathrm{yields} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)=\left(\mathrm{103},\:\mathrm{9}\right) \\ $$$$\therefore\:\equiv\:\mathrm{solution}\:\left\{\left(\mathrm{9},\mathrm{103}\right),\left(\mathrm{17},\mathrm{17}\right),\left(\mathrm{103},\mathrm{9}\right)\right\}\: \\ $$$$ \\ $$$$ \\ $$

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