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Question Number 180877 by Sheshdevsahu last updated on 18/Nov/22

Q. find the largest value of such that the positive integers a, b > 1 satisfy. a^b .b^a + a^b + b^a = 5329

$$\boldsymbol{\mathrm{Q}}.\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{largest}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}} \\ $$ $$\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integers}}\:\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}}\:>\:\mathrm{1}\:\boldsymbol{\mathrm{satisfy}}. \\ $$ $$\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{b}}} .\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{a}}} \:+\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{b}}} \:+\:\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{a}}} \:=\:\mathrm{5329} \\ $$

Commented byMJS_new last updated on 18/Nov/22

the largest value of what? anyway if a, b ∈Z^+ \{1} we have these solutions: a=3∧b=4∨a=4∧b=3

$$\mathrm{the}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:{what}? \\ $$ $$\mathrm{anyway}\:\mathrm{if}\:{a},\:{b}\:\in\mathbb{Z}^{+} \backslash\left\{\mathrm{1}\right\}\:\mathrm{we}\:\mathrm{have}\:\mathrm{these}\:\mathrm{solutions}: \\ $$ $${a}=\mathrm{3}\wedge{b}=\mathrm{4}\vee{a}=\mathrm{4}\wedge{b}=\mathrm{3} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Nov/22

a^b .b^a + a^b + b^a = 5329 a^b (b^a +1)+b^a +1=5330 (a^b +1)(b^a +1)=5330=2∙5∙13∙41 k ∣ 5330 : a^b +1=k ∧ b^a +1=5330/k ⇒k=65 , 82 { ((a^b +1=65 ∧ b^a +1=82)),((a^b +1=82 ∧ b^a +1=65)) :} { ((a^b =64 ∧ b^a =81⇒4^3 =64∧3^4 =81)),((a^b =81 ∧ b^a =64⇒3^4 =81∧4^3 =64)) :} {a,b}={3,4}

$$\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{b}}} .\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{a}}} \:+\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{b}}} \:+\:\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{a}}} \:=\:\mathrm{5329} \\ $$ $$\mathrm{a}^{\mathrm{b}} \left(\mathrm{b}^{\mathrm{a}} +\mathrm{1}\right)+\mathrm{b}^{\mathrm{a}} +\mathrm{1}=\mathrm{5330} \\ $$ $$\left(\mathrm{a}^{\mathrm{b}} +\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{a}} +\mathrm{1}\right)=\mathrm{5330}=\mathrm{2}\centerdot\mathrm{5}\centerdot\mathrm{13}\centerdot\mathrm{41} \\ $$ $$\:\:{k}\:\mid\:\mathrm{5330}\:: \\ $$ $$\:\mathrm{a}^{\mathrm{b}} +\mathrm{1}={k}\:\wedge\:\mathrm{b}^{\mathrm{a}} +\mathrm{1}=\mathrm{5330}/{k} \\ $$ $$\Rightarrow{k}=\mathrm{65}\:,\:\mathrm{82} \\ $$ $$\begin{cases}{\mathrm{a}^{\mathrm{b}} +\mathrm{1}=\mathrm{65}\:\wedge\:\mathrm{b}^{\mathrm{a}} +\mathrm{1}=\mathrm{82}}\\{\mathrm{a}^{\mathrm{b}} +\mathrm{1}=\mathrm{82}\:\wedge\:\mathrm{b}^{\mathrm{a}} +\mathrm{1}=\mathrm{65}}\end{cases}\: \\ $$ $$\begin{cases}{\mathrm{a}^{\mathrm{b}} =\mathrm{64}\:\wedge\:\mathrm{b}^{\mathrm{a}} =\mathrm{81}\Rightarrow\mathrm{4}^{\mathrm{3}} =\mathrm{64}\wedge\mathrm{3}^{\mathrm{4}} =\mathrm{81}}\\{\mathrm{a}^{\mathrm{b}} =\mathrm{81}\:\wedge\:\mathrm{b}^{\mathrm{a}} =\mathrm{64}\Rightarrow\mathrm{3}^{\mathrm{4}} =\mathrm{81}\wedge\mathrm{4}^{\mathrm{3}} =\mathrm{64}}\end{cases}\: \\ $$ $$\left\{\mathrm{a},\mathrm{b}\right\}=\left\{\mathrm{3},\mathrm{4}\right\} \\ $$

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