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Question Number 21795 by hi147 last updated on 04/Oct/17

help x∈N determine x where 7 divise 2^x +3^x

$${help} \\ $$$${x}\in{N} \\ $$$${determine}\:{x}\:{where}\:\mathrm{7}\:{divise}\:\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 04/Oct/17

x=3,9

$$\mathrm{x}=\mathrm{3},\mathrm{9} \\ $$

Commented by prakash jain last updated on 04/Oct/17

How did you get x=3?

$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:{x}=\mathrm{3}? \\ $$

Commented by math solver last updated on 05/Oct/17

how did you get x=9

$$ \\ $$$${how}\:{did}\:{you}\:{get}\:{x}=\mathrm{9} \\ $$

Answered by mrW1 last updated on 04/Oct/17

x=3^n , n∈N prove: for n=1, 2^3^1 +3^3^1 =8+27=35 mod 7=0 it′s true for n=1. let′s assume it′s true for n, i.e. 2^3^n +3^3^n mod 7=0 for n+1: (2^3^(n+1) +3^3^(n+1) ) mod 7 =(2^(3×3^n ) +3^(3×3^n ) ) mod 7 =[(2^3^n )^3 +(3^3^n )^3 ] mod 7 =[(2^3^n )+(3^3^n )][(2^3^n )^2 −(2^3^n )(3^3^n )+(3^3^n )^2 ] mod 7 =[(2^3^n )+(3^3^n )] mod 7 =0 ⇒x=3^n is a solution

$$\mathrm{x}=\mathrm{3}^{\mathrm{n}} ,\:\mathrm{n}\in\mathrm{N} \\ $$$$ \\ $$$$\mathrm{prove}: \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{1}, \\ $$$$\mathrm{2}^{\mathrm{3}^{\mathrm{1}} } +\mathrm{3}^{\mathrm{3}^{\mathrm{1}} } =\mathrm{8}+\mathrm{27}=\mathrm{35}\:\mathrm{mod}\:\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{1}. \\ $$$$ \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{assume}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n},\:\mathrm{i}.\mathrm{e}. \\ $$$$\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } +\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \:\mathrm{mod}\:\mathrm{7}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}+\mathrm{1}: \\ $$$$\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}+\mathrm{1}} } +\mathrm{3}^{\mathrm{3}^{\mathrm{n}+\mathrm{1}} } \:\right)\:\mathrm{mod}\:\mathrm{7} \\ $$$$=\left(\mathrm{2}^{\mathrm{3}×\mathrm{3}^{\mathrm{n}} } +\mathrm{3}^{\mathrm{3}×\mathrm{3}^{\mathrm{n}} } \:\right)\:\mathrm{mod}\:\mathrm{7} \\ $$$$=\left[\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } \right)^{\mathrm{3}} +\left(\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \right)^{\mathrm{3}} \:\right]\:\mathrm{mod}\:\mathrm{7} \\ $$$$=\left[\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } \right)+\left(\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \right)\right]\left[\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } \right)^{\mathrm{2}} −\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } \right)\left(\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \right)+\left(\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \right)^{\mathrm{2}} \right]\:\:\mathrm{mod}\:\mathrm{7} \\ $$$$=\left[\left(\mathrm{2}^{\mathrm{3}^{\mathrm{n}} } \right)+\left(\mathrm{3}^{\mathrm{3}^{\mathrm{n}} } \right)\right]\:\mathrm{mod}\:\mathrm{7} \\ $$$$=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\mathrm{x}=\mathrm{3}^{\mathrm{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$

Commented by hi147 last updated on 04/Oct/17

thx but how u think in 3^(n ) exactly? the prove is not complete... it seems that we must not use congruence methode because this exercise is under the title(divisibility) in secondary book.so we must just use (n divise x ⇔x=kn/k∈Z)... i wait for an other answer ...

$${thx} \\ $$$${but}\:{how}\:{u}\:{think}\:{in}\:\mathrm{3}^{{n}\:} {exactly}?\: \\ $$$${the}\:{prove}\:{is}\:{not}\:{complete}... \\ $$$${it}\:{seems}\:{that}\:{we}\:{must}\:{not}\:{use}\: \\ $$$${congruence}\:{methode}\:{because}\:{this}\: \\ $$$${exercise}\:{is}\:{under}\:{the}\:{title}\left({divisibility}\right) \\ $$$${in}\:{secondary}\:{book}.{so}\:{we}\:{must}\:{just} \\ $$$${use}\:\left({n}\:{divise}\:{x}\:\Leftrightarrow{x}={kn}/{k}\in{Z}\right)... \\ $$$$\:{i}\:{wait}\:{for}\:{an}\:{other}\:{answer}\:... \\ $$

Commented by hi147 last updated on 04/Oct/17

the question is from where come 3^n ? i mean before using the demonstration with concurrence....if we say from imagination.it is not a prove.and as you say.maybe we have other x.so we still need a complet solution...thx

$${the}\:{question}\:{is}\:{from}\:{where}\:{come}\:\mathrm{3}^{{n}} ? \\ $$$${i}\:{mean}\:{before}\:{using}\:{the}\:{demonstration} \\ $$$${with}\:{concurrence}....{if}\:{we}\:{say}\:{from} \\ $$$${imagination}.{it}\:{is}\:{not}\:{a}\:{prove}.{and}\:{as}\: \\ $$$${you}\:{say}.{maybe}\:{we}\:{have}\:{other}\:{x}.{so} \\ $$$${we}\:{still}\:{need}\:{a}\:{complet}\:{solution}...{thx}\: \\ $$$$ \\ $$

Commented by mrW1 last updated on 04/Oct/17

with x=3^n we have proved that 7 divides 2^x +3^x . this prove is complete. but indeed it is not proved that no other solution than 3^n exists.

$$\mathrm{with}\:\mathrm{x}=\mathrm{3}^{\mathrm{n}} \:\mathrm{we}\:\mathrm{have}\:\mathrm{proved}\:\mathrm{that} \\ $$$$\mathrm{7}\:\mathrm{divides}\:\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} . \\ $$$$\mathrm{this}\:\mathrm{prove}\:\mathrm{is}\:\mathrm{complete}. \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{indeed}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{proved}\:\mathrm{that}\:\mathrm{no} \\ $$$$\mathrm{other}\:\mathrm{solution}\:\mathrm{than}\:\mathrm{3}^{\mathrm{n}} \:\mathrm{exists}. \\ $$

Commented by mrW1 last updated on 05/Oct/17

by try and error we get a solution x=3. since a^3 +b^3 =(a+b)(a^2 −ab+b^2 ) we know that x=3×3 is then also a solution, and x=3×3×3 is also a solution, etc. that means x=3^n is a solution.

$$\mathrm{by}\:\mathrm{try}\:\mathrm{and}\:\mathrm{error}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{x}=\mathrm{3}. \\ $$$$\mathrm{since}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right)\:\mathrm{we} \\ $$$$\mathrm{know}\:\mathrm{that}\:\mathrm{x}=\mathrm{3}×\mathrm{3}\:\mathrm{is}\:\mathrm{then}\:\mathrm{also}\:\mathrm{a} \\ $$$$\mathrm{solution},\:\mathrm{and}\:\mathrm{x}=\mathrm{3}×\mathrm{3}×\mathrm{3}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a} \\ $$$$\mathrm{solution},\:\mathrm{etc}.\:\mathrm{that}\:\mathrm{means}\:\mathrm{x}=\mathrm{3}^{\mathrm{n}} \:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{solution}. \\ $$

Commented by mrW1 last updated on 05/Oct/17

generally 7 divides (2^3 )^(2k+1) +(3^3 )^(2k+1) so x=3(2k+1) is a solution, k≥0. i.e. x=3,9,15,21,27......

$$\mathrm{generally}\:\mathrm{7}\:\mathrm{divides}\:\left(\mathrm{2}^{\mathrm{3}} \right)^{\mathrm{2k}+\mathrm{1}} +\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2k}+\mathrm{1}} \\ $$$$\mathrm{so}\:\mathrm{x}=\mathrm{3}\left(\mathrm{2k}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{k}\geqslant\mathrm{0}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{x}=\mathrm{3},\mathrm{9},\mathrm{15},\mathrm{21},\mathrm{27}...... \\ $$

Commented by hi147 last updated on 06/Oct/17

good.but how did u get (2^3 )^(2k+1) +(3^3 )^(2k+1) and how we can prove that 3(2k+1) is the only solution thx.

$${good}.{but}\:{how}\:{did}\:{u}\:{get}\:\left(\mathrm{2}^{\mathrm{3}} \right)^{\mathrm{2}{k}+\mathrm{1}} +\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2}{k}+\mathrm{1}} \\ $$$${and}\:{how}\:{we}\:{can}\:{prove}\:{that}\:\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)\:{is} \\ $$$${the}\:{only}\:{solution} \\ $$$${thx}. \\ $$

Commented by mrW1 last updated on 06/Oct/17

2^(3(2k+1)) +3^(3(2k+1)) =(2^3 )^(2k+1) +(3^3 )^(2k+1) =(2^3 +3^3 )(.......) =35×(......) =7×5×(......) ⇒x=3(2k+1) is a solution. I don′t know if this is the only solution.

$$\mathrm{2}^{\mathrm{3}\left(\mathrm{2k}+\mathrm{1}\right)} +\mathrm{3}^{\mathrm{3}\left(\mathrm{2k}+\mathrm{1}\right)} \\ $$$$=\left(\mathrm{2}^{\mathrm{3}} \right)^{\mathrm{2k}+\mathrm{1}} +\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2k}+\mathrm{1}} =\left(\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} \right)\left(.......\right) \\ $$$$=\mathrm{35}×\left(......\right) \\ $$$$=\mathrm{7}×\mathrm{5}×\left(......\right) \\ $$$$\Rightarrow\mathrm{x}=\mathrm{3}\left(\mathrm{2k}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}. \\ $$

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