Question-96508

Question Number 96508 by bemath last updated on 02/Jun/20

Answered by 1549442205 last updated on 02/Jun/20

From the hypothesis we have b=a^(3/2) =(√a^3 )=a(√a) d=c^(5/4) =^4 (√c^5 )=c^4 (√c)⇒b−d=a(√a)−c^4 (√c) Since b−d∈N^∗ ,so we need must have a=m^2 and c=n^4 .Then b−d=m^3 −n^5 (for m,n∈N^∗ ).From the original condition a−c=9 we have m^2 −n^4 =9⇔(m+n^2 )(m−n^2 )=9(1).Because (m+n^2 )+(m−n^2 )=2m is an even number and their product is odd,both are odd. Also,m+n^2 >m−n^2 .Hence,from (1) we get { ((m+n^2 =9)),((m−n^2 =1)) :}⇔ { ((m=5)),((n=2)) :} Which implies that b−d=m^3 −n^5 = =5^3 −2^5 =125−32=93 is unique root

$$\boldsymbol{\mathrm{F}}\mathrm{rom}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have}\:\mathrm{b}=\mathrm{a}^{\frac{\mathrm{3}}{\mathrm{2}}} =\sqrt{\mathrm{a}^{\mathrm{3}} }=\mathrm{a}\sqrt{\mathrm{a}} \\ $$$$\mathrm{d}=\mathrm{c}^{\frac{\mathrm{5}}{\mathrm{4}}} =^{\mathrm{4}} \sqrt{\mathrm{c}^{\mathrm{5}} }=\mathrm{c}^{\mathrm{4}} \sqrt{\mathrm{c}}\Rightarrow\mathrm{b}−\mathrm{d}=\mathrm{a}\sqrt{\mathrm{a}}−\mathrm{c}^{\mathrm{4}} \sqrt{\mathrm{c}} \\ $$$$\mathrm{Since}\:\mathrm{b}−\mathrm{d}\in\mathbb{N}^{\ast} ,\mathrm{so}\:\mathrm{we}\:\mathrm{need}\:\mathrm{must}\:\mathrm{have}\:\mathrm{a}=\mathrm{m}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{c}=\mathrm{n}^{\mathrm{4}} .\mathrm{Then} \\ $$$$\mathrm{b}−\mathrm{d}=\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{5}} \left(\mathrm{for}\:\mathrm{m},\mathrm{n}\in\mathbb{N}^{\ast} \right).\mathrm{From}\:\mathrm{the} \\ $$$$\mathrm{original}\:\mathrm{condition}\:\:\mathrm{a}−\mathrm{c}=\mathrm{9}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{4}} =\mathrm{9}\Leftrightarrow\left(\mathrm{m}+\mathrm{n}^{\mathrm{2}} \right)\left(\mathrm{m}−\mathrm{n}^{\mathrm{2}} \right)=\mathrm{9}\left(\mathrm{1}\right).\mathrm{Because} \\ $$$$\left(\mathrm{m}+\mathrm{n}^{\mathrm{2}} \right)+\left(\mathrm{m}−\mathrm{n}^{\mathrm{2}} \right)=\mathrm{2m}\:\mathrm{is}\:\:\mathrm{an}\:\mathrm{even}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{their}\:\mathrm{product}\:\mathrm{is}\:\mathrm{odd},\mathrm{both}\:\mathrm{are}\:\mathrm{odd}. \\ $$$$\mathrm{Also},\mathrm{m}+\mathrm{n}^{\mathrm{2}} >\mathrm{m}−\mathrm{n}^{\mathrm{2}} .\mathrm{Hence},\mathrm{from}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{\mathrm{m}+\mathrm{n}^{\mathrm{2}} =\mathrm{9}}\\{\mathrm{m}−\mathrm{n}^{\mathrm{2}} =\mathrm{1}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{m}=\mathrm{5}}\\{\mathrm{n}=\mathrm{2}}\end{cases} \\ $$$$\mathrm{Which}\:\mathrm{implies}\:\mathrm{that}\:\mathrm{b}−\mathrm{d}=\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{5}} = \\ $$$$=\mathrm{5}^{\mathrm{3}} −\mathrm{2}^{\mathrm{5}} =\mathrm{125}−\mathrm{32}=\mathrm{93}\:\mathrm{is}\:\mathrm{unique}\:\mathrm{root} \\ $$

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