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Question Number 34429 by $@ty@m last updated on 06/May/18

$$Provethat$$$$3^m+3^n+1isnotaperfectsquare.$$$$wheremandnarepositiveintegers.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18

$$letassumem=2k$$$$={\left(3^k\right)}^2+2.3^k.1+1^2+3^n-2.3^k$$$$={(3^k+1)}^2+3^k\times (3^{n-k}-2)$$$$forperfectsquare{3}^k(3^{n-k}-2)=0$$$$so{3}^{n-k}=2foranyvalueofnandk$$$$thethevalueof{3}^{n-k}cannotbe2.$$$$so{3}^m+3^n+1csnnotbeperfectsquae$$

Commented by $@ty@m last updated on 06/May/18

$$Niceattempt.$$$$butincompletebecausem=2k$$$$meansmiseven.$$$$Whatifbothmandareodd?$$

Commented by Rasheed.Sindhi last updated on 07/May/18

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{necessary}\mathrm{to}\mathrm{become}$$$$3^k(3^{n-k}-2)=0\mathrm{for}\mathrm{being}{(3^k+1)}^2+3^k\times (3^{n-k}-2)$$$$\mathrm{square}.$$$$\mathrm{because}\mathrm{sum}\mathrm{of}\mathrm{a}\mathrm{square}\mathrm{\&}\mathrm{a}\mathrm{number}$$$$\left(\mathrm{although}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}0\right)\mathrm{may}\mathrm{be}\mathrm{a}\mathrm{square}.$$$$\mathrm{For}\mathrm{example}$$$$\mathrm{if}{(3^k+1)}^2=25\mathrm{\&}{3}^k\times (3^{n-k}-2)=11$$$${(3^k+1)}^2+3^k\times (3^{n-k}-2)=36$$$$$$

Answered by Rasheed.Sindhi last updated on 08/May/18

$$3^{\mathrm{m}},3^{\mathrm{n}}\mathrm{\&}1\mathrm{are}\mathrm{odd}\mathrm{numbers}.$$$$\because \mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{odd}\mathrm{numbers}\mathrm{is}\mathrm{odd}$$$$\therefore 3^{\mathrm{m}}+3^{\mathrm{n}}+1\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}.$$$$\mathrm{And}\mathrm{if}\mathrm{it}\mathbf{were}\mathrm{perfect}\mathrm{square},\mathrm{it}\mathrm{must}\mathrm{have}\mathrm{been}$$$$\mathrm{square}\mathrm{of}\mathrm{an}\mathrm{odd}\mathrm{number}.[\mathrm{Square}\mathrm{of}\mathrm{an}$$$$\mathrm{even}\mathrm{number}\mathrm{is}\mathrm{even}\mathrm{of}\mathrm{course}.]$$$$\mathrm{Let}{3}^{\mathrm{m}}+3^{\mathrm{n}}+1={(2\mathrm{k}+1)}^2$$$$3^{\mathrm{m}}+3^{\mathrm{n}}+1={4\mathrm{k}}^2+4\mathrm{k}+1$$$$3^{\mathrm{m}}+3^{\mathrm{n}}={4\mathrm{k}}^2+4\mathrm{k}$$$$3^{\mathrm{m}}+3^{\mathrm{n}}=4({\mathrm{k}}^2+\mathrm{k})$$$${\mathrm{k}}^2+\mathrm{k}\in \mathbb{E},\mathrm{so}\mathrm{right}\mathrm{side}\mathbf{is}\mathbf{divisible}\mathrm{by}8$$$$\mathrm{whereas}\mathrm{left}\mathrm{side}\mathbf{is}\mathbf{not}\mathbf{divisible}\mathrm{by}8$$$${\left(\mathrm{See}\mathrm{proof}\mathrm{at}\mathrm{the}\mathrm{end}\right)}^{\ast }$$$$\because 3^{\mathrm{m}}+3^{\mathrm{n}}+1\mathrm{cannot}\mathrm{be}\mathrm{square}\mathrm{of}\mathrm{even}\mathrm{nunmber}.$$$$\mathrm{because}\mathrm{it}\mathrm{an}\mathrm{odd}\mathrm{number}.$$$$\because 3^{\mathrm{m}}+3^{\mathrm{n}}+1\mathrm{cannot}\mathrm{be}\mathrm{square}\mathrm{of}\mathrm{odd}\mathrm{nunmber}$$$$\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{above}\mathrm{contrdiction}.$$$$\therefore 3^{\mathrm{m}}+3^{\mathrm{n}}+1{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{perfect}\mathrm{square}.$$$$-----------$$$$\ast 3^{\mathrm{m}}\left(\mathrm{mod}8\right)=1\mathrm{or}3$$$$3^{\mathrm{m}}+3^{\mathrm{n}}\left(\mathrm{mod}8\right)=2,4,6\mathrm{only}$$$$\therefore 3^{\mathrm{m}}+3^{\mathrm{n}}\left(\mathrm{mod}8\right)\ne 0$$$$\mathrm{I}-\mathrm{e}{3}^{\mathrm{m}}+3^{\mathrm{n}}\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}8$$

Commented by MJS last updated on 09/May/18

$$\mathrm{great}!$$

Commented by MrW3 last updated on 14/Jun/18

$$wonderfulproof!$$

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