Solve for $m$: $m^5 + 7m + 8 equiv 0 pmod 3$












0












$begingroup$


I got about this far:



$$m^5 + 7m equiv -8 equiv 1 pmod 3$$



I'd have a much easier time solving a linear equation, but I have no clue about this one.



$$m(m^4 + 7) equiv 1 pmod 3$$



m itself is not divisible by three, but we already knew that... Any thoughts?






modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    $large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
    $endgroup$
    – Bill Dubuque
    Dec 17 '18 at 18:54












  • $begingroup$
    The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 10:30


















0












$begingroup$


I got about this far:



$$m^5 + 7m equiv -8 equiv 1 pmod 3$$



I'd have a much easier time solving a linear equation, but I have no clue about this one.



$$m(m^4 + 7) equiv 1 pmod 3$$



m itself is not divisible by three, but we already knew that... Any thoughts?






modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    $large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
    $endgroup$
    – Bill Dubuque
    Dec 17 '18 at 18:54












  • $begingroup$
    The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 10:30
















0












0








0





$begingroup$


I got about this far:



$$m^5 + 7m equiv -8 equiv 1 pmod 3$$



I'd have a much easier time solving a linear equation, but I have no clue about this one.



$$m(m^4 + 7) equiv 1 pmod 3$$



m itself is not divisible by three, but we already knew that... Any thoughts?






modular-arithmetic







share|cite|improve this question











$endgroup$




I got about this far:



$$m^5 + 7m equiv -8 equiv 1 pmod 3$$



I'd have a much easier time solving a linear equation, but I have no clue about this one.



$$m(m^4 + 7) equiv 1 pmod 3$$



m itself is not divisible by three, but we already knew that... Any thoughts?






modular-arithmetic




modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 18:02









Tianlalu

3,09421138




3,09421138










asked Dec 17 '18 at 17:58









MinahMinah

183




183












  • $begingroup$
    $large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
    $endgroup$
    – Bill Dubuque
    Dec 17 '18 at 18:54












  • $begingroup$
    The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 10:30




















  • $begingroup$
    $large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
    $endgroup$
    – Bill Dubuque
    Dec 17 '18 at 18:54












  • $begingroup$
    The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 10:30


















$begingroup$
$large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
$endgroup$
– Bill Dubuque
Dec 17 '18 at 18:54






$begingroup$
$large bmod 3!: , mnotequiv 0,Rightarrow, mequiv pm1,Rightarrow,color{#C00}{m^4equiv 1}, $ so $large 1equiv (color{#c00}{m^4}!+!7)mequiv -m $
$endgroup$
– Bill Dubuque
Dec 17 '18 at 18:54














$begingroup$
The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:30






$begingroup$
The fastest method here would be to plug in $m=0$, $m=1$ and $m=2$, and see which (if any) of those work. Every integer is congruent to one of those three, so you are done at this point. More power to you, if you are asking for techniques handling more difficult cases. Those are not really for paper & pencil work though.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:30












4 Answers
4






active

oldest

votes


















5












$begingroup$

You are working in the space $mathbb{Z}/3mathbb{Z}={0,1,2}$ so there are only three elements to check.



For $mequiv 0$, $m^{5}+7m+8=8equiv 2mod 3$, so $0$ is not a solution.



For $mequiv 1$, $m^{5}+7m+8=16equiv 1mod 3$, so $1$ is not a solution.



For $mequiv 2$, $m^{5}+7m+8=54equiv 0mod 3$, so $2$ is a solution.



So $2$ is the only solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
    $endgroup$
    – Minah
    Dec 17 '18 at 18:18










  • $begingroup$
    For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
    $endgroup$
    – pwerth
    Dec 17 '18 at 18:23



















2












$begingroup$

$$m^3equiv mpmod3$$



$$m^5equiv m^3equiv mpmod3$$



So, we need $3|8(m+1)iff mequiv-1equiv2pmod3$ as $(8,3)=1$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Take the coefficients mod $3$:
    $$m^5+7m+8equiv m^5 + m+2equiv 0pmod 3$$
    Apply the Little Theoreom of Fermat, which is $m^pequiv m pmod{p}$ for a prime $p$ and any $m$:
    $$mcdot m^2 + m+2equiv m + m+2 equiv 2m+2 equiv 0pmod 3$$
    Therefore:
    $$m equiv -1 equiv 2pmod 3$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is essentially a dupe of lab's answer.
      $endgroup$
      – Bill Dubuque
      Dec 17 '18 at 18:40





















    1












    $begingroup$

    Alternatively:
    $$m^5+7m+8equiv m^5-m+8m+8equiv m(m-1)(m+1)(m^2+1)+8(m+1)equiv \8(m+1)equiv 0pmod{3}.$$
    Here is the linear congruence equation for you.






    share|cite|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      You are working in the space $mathbb{Z}/3mathbb{Z}={0,1,2}$ so there are only three elements to check.



      For $mequiv 0$, $m^{5}+7m+8=8equiv 2mod 3$, so $0$ is not a solution.



      For $mequiv 1$, $m^{5}+7m+8=16equiv 1mod 3$, so $1$ is not a solution.



      For $mequiv 2$, $m^{5}+7m+8=54equiv 0mod 3$, so $2$ is a solution.



      So $2$ is the only solution.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
        $endgroup$
        – Minah
        Dec 17 '18 at 18:18










      • $begingroup$
        For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
        $endgroup$
        – pwerth
        Dec 17 '18 at 18:23
















      5












      $begingroup$

      You are working in the space $mathbb{Z}/3mathbb{Z}={0,1,2}$ so there are only three elements to check.



      For $mequiv 0$, $m^{5}+7m+8=8equiv 2mod 3$, so $0$ is not a solution.



      For $mequiv 1$, $m^{5}+7m+8=16equiv 1mod 3$, so $1$ is not a solution.



      For $mequiv 2$, $m^{5}+7m+8=54equiv 0mod 3$, so $2$ is a solution.



      So $2$ is the only solution.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
        $endgroup$
        – Minah
        Dec 17 '18 at 18:18










      • $begingroup$
        For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
        $endgroup$
        – pwerth
        Dec 17 '18 at 18:23














      5












      5








      5





      $begingroup$

      You are working in the space $mathbb{Z}/3mathbb{Z}={0,1,2}$ so there are only three elements to check.



      For $mequiv 0$, $m^{5}+7m+8=8equiv 2mod 3$, so $0$ is not a solution.



      For $mequiv 1$, $m^{5}+7m+8=16equiv 1mod 3$, so $1$ is not a solution.



      For $mequiv 2$, $m^{5}+7m+8=54equiv 0mod 3$, so $2$ is a solution.



      So $2$ is the only solution.






      share|cite|improve this answer









      $endgroup$



      You are working in the space $mathbb{Z}/3mathbb{Z}={0,1,2}$ so there are only three elements to check.



      For $mequiv 0$, $m^{5}+7m+8=8equiv 2mod 3$, so $0$ is not a solution.



      For $mequiv 1$, $m^{5}+7m+8=16equiv 1mod 3$, so $1$ is not a solution.



      For $mequiv 2$, $m^{5}+7m+8=54equiv 0mod 3$, so $2$ is a solution.



      So $2$ is the only solution.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 17 '18 at 18:00









      pwerthpwerth

      3,300417




      3,300417












      • $begingroup$
        Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
        $endgroup$
        – Minah
        Dec 17 '18 at 18:18










      • $begingroup$
        For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
        $endgroup$
        – pwerth
        Dec 17 '18 at 18:23


















      • $begingroup$
        Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
        $endgroup$
        – Minah
        Dec 17 '18 at 18:18










      • $begingroup$
        For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
        $endgroup$
        – pwerth
        Dec 17 '18 at 18:23
















      $begingroup$
      Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
      $endgroup$
      – Minah
      Dec 17 '18 at 18:18




      $begingroup$
      Delightful! May I ask, how would we proceed with a larger modulo where checking possible solutions would consume too much time?
      $endgroup$
      – Minah
      Dec 17 '18 at 18:18












      $begingroup$
      For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
      $endgroup$
      – pwerth
      Dec 17 '18 at 18:23




      $begingroup$
      For larger moduli, it would depend on the problem. It's impossible to give a general method for solving any congruence, but some useful tools are factoring (like you did), Fermat's little theorem (for prime moduli), or using quadratic residues.
      $endgroup$
      – pwerth
      Dec 17 '18 at 18:23











      2












      $begingroup$

      $$m^3equiv mpmod3$$



      $$m^5equiv m^3equiv mpmod3$$



      So, we need $3|8(m+1)iff mequiv-1equiv2pmod3$ as $(8,3)=1$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$m^3equiv mpmod3$$



        $$m^5equiv m^3equiv mpmod3$$



        So, we need $3|8(m+1)iff mequiv-1equiv2pmod3$ as $(8,3)=1$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$m^3equiv mpmod3$$



          $$m^5equiv m^3equiv mpmod3$$



          So, we need $3|8(m+1)iff mequiv-1equiv2pmod3$ as $(8,3)=1$






          share|cite|improve this answer









          $endgroup$



          $$m^3equiv mpmod3$$



          $$m^5equiv m^3equiv mpmod3$$



          So, we need $3|8(m+1)iff mequiv-1equiv2pmod3$ as $(8,3)=1$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 18:02









          lab bhattacharjeelab bhattacharjee

          227k15158276




          227k15158276























              1












              $begingroup$

              Take the coefficients mod $3$:
              $$m^5+7m+8equiv m^5 + m+2equiv 0pmod 3$$
              Apply the Little Theoreom of Fermat, which is $m^pequiv m pmod{p}$ for a prime $p$ and any $m$:
              $$mcdot m^2 + m+2equiv m + m+2 equiv 2m+2 equiv 0pmod 3$$
              Therefore:
              $$m equiv -1 equiv 2pmod 3$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is essentially a dupe of lab's answer.
                $endgroup$
                – Bill Dubuque
                Dec 17 '18 at 18:40


















              1












              $begingroup$

              Take the coefficients mod $3$:
              $$m^5+7m+8equiv m^5 + m+2equiv 0pmod 3$$
              Apply the Little Theoreom of Fermat, which is $m^pequiv m pmod{p}$ for a prime $p$ and any $m$:
              $$mcdot m^2 + m+2equiv m + m+2 equiv 2m+2 equiv 0pmod 3$$
              Therefore:
              $$m equiv -1 equiv 2pmod 3$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is essentially a dupe of lab's answer.
                $endgroup$
                – Bill Dubuque
                Dec 17 '18 at 18:40
















              1












              1








              1





              $begingroup$

              Take the coefficients mod $3$:
              $$m^5+7m+8equiv m^5 + m+2equiv 0pmod 3$$
              Apply the Little Theoreom of Fermat, which is $m^pequiv m pmod{p}$ for a prime $p$ and any $m$:
              $$mcdot m^2 + m+2equiv m + m+2 equiv 2m+2 equiv 0pmod 3$$
              Therefore:
              $$m equiv -1 equiv 2pmod 3$$






              share|cite|improve this answer









              $endgroup$



              Take the coefficients mod $3$:
              $$m^5+7m+8equiv m^5 + m+2equiv 0pmod 3$$
              Apply the Little Theoreom of Fermat, which is $m^pequiv m pmod{p}$ for a prime $p$ and any $m$:
              $$mcdot m^2 + m+2equiv m + m+2 equiv 2m+2 equiv 0pmod 3$$
              Therefore:
              $$m equiv -1 equiv 2pmod 3$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 17 '18 at 18:26









              I like SerenaI like Serena

              4,3321822




              4,3321822












              • $begingroup$
                This is essentially a dupe of lab's answer.
                $endgroup$
                – Bill Dubuque
                Dec 17 '18 at 18:40




















              • $begingroup$
                This is essentially a dupe of lab's answer.
                $endgroup$
                – Bill Dubuque
                Dec 17 '18 at 18:40


















              $begingroup$
              This is essentially a dupe of lab's answer.
              $endgroup$
              – Bill Dubuque
              Dec 17 '18 at 18:40






              $begingroup$
              This is essentially a dupe of lab's answer.
              $endgroup$
              – Bill Dubuque
              Dec 17 '18 at 18:40













              1












              $begingroup$

              Alternatively:
              $$m^5+7m+8equiv m^5-m+8m+8equiv m(m-1)(m+1)(m^2+1)+8(m+1)equiv \8(m+1)equiv 0pmod{3}.$$
              Here is the linear congruence equation for you.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Alternatively:
                $$m^5+7m+8equiv m^5-m+8m+8equiv m(m-1)(m+1)(m^2+1)+8(m+1)equiv \8(m+1)equiv 0pmod{3}.$$
                Here is the linear congruence equation for you.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Alternatively:
                  $$m^5+7m+8equiv m^5-m+8m+8equiv m(m-1)(m+1)(m^2+1)+8(m+1)equiv \8(m+1)equiv 0pmod{3}.$$
                  Here is the linear congruence equation for you.






                  share|cite|improve this answer









                  $endgroup$



                  Alternatively:
                  $$m^5+7m+8equiv m^5-m+8m+8equiv m(m-1)(m+1)(m^2+1)+8(m+1)equiv \8(m+1)equiv 0pmod{3}.$$
                  Here is the linear congruence equation for you.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 18:29









                  farruhotafarruhota

                  21.3k2841




                  21.3k2841






























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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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