Prove that $3^{2n} +7$ is divisible by 8












1












$begingroup$



Prove by induction that $3^{2n} +7$ is divisible by 8 for $n in Bbb N$



So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?




If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$



Assume true for $n=k$ so $$ 8vert 3^{2k} +7$$



If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A in Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B in Bbb Z$
$$3^2 times 3^{2k}+7$$
$$3^2 times (8A)=72A$$
$$72A =8(9A)=9B$$



So by induction $3^{2n} +7$ is divisibe by 8 $forall n in Bbb N$






elementary-number-theory induction divisibility







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$endgroup$












  • $begingroup$
    Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
    $endgroup$
    – lulu
    Jan 9 at 18:57












  • $begingroup$
    Do you know modular arithmetic / congruences?
    $endgroup$
    – Bill Dubuque
    Jan 9 at 18:59










  • $begingroup$
    You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
    $endgroup$
    – KM101
    Jan 9 at 18:59












  • $begingroup$
    Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
    $endgroup$
    – AlephNull
    Jan 9 at 19:03
















1












$begingroup$



Prove by induction that $3^{2n} +7$ is divisible by 8 for $n in Bbb N$



So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?




If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$



Assume true for $n=k$ so $$ 8vert 3^{2k} +7$$



If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A in Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B in Bbb Z$
$$3^2 times 3^{2k}+7$$
$$3^2 times (8A)=72A$$
$$72A =8(9A)=9B$$



So by induction $3^{2n} +7$ is divisibe by 8 $forall n in Bbb N$






elementary-number-theory induction divisibility







share|cite|improve this question











$endgroup$












  • $begingroup$
    Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
    $endgroup$
    – lulu
    Jan 9 at 18:57












  • $begingroup$
    Do you know modular arithmetic / congruences?
    $endgroup$
    – Bill Dubuque
    Jan 9 at 18:59










  • $begingroup$
    You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
    $endgroup$
    – KM101
    Jan 9 at 18:59












  • $begingroup$
    Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
    $endgroup$
    – AlephNull
    Jan 9 at 19:03














1












1








1





$begingroup$



Prove by induction that $3^{2n} +7$ is divisible by 8 for $n in Bbb N$



So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?




If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$



Assume true for $n=k$ so $$ 8vert 3^{2k} +7$$



If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A in Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B in Bbb Z$
$$3^2 times 3^{2k}+7$$
$$3^2 times (8A)=72A$$
$$72A =8(9A)=9B$$



So by induction $3^{2n} +7$ is divisibe by 8 $forall n in Bbb N$






elementary-number-theory induction divisibility







share|cite|improve this question











$endgroup$





Prove by induction that $3^{2n} +7$ is divisible by 8 for $n in Bbb N$



So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?




If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$



Assume true for $n=k$ so $$ 8vert 3^{2k} +7$$



If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A in Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B in Bbb Z$
$$3^2 times 3^{2k}+7$$
$$3^2 times (8A)=72A$$
$$72A =8(9A)=9B$$



So by induction $3^{2n} +7$ is divisibe by 8 $forall n in Bbb N$






elementary-number-theory induction divisibility




elementary-number-theory induction divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 21:35









Maria Mazur

49.7k1361124




49.7k1361124










asked Jan 9 at 18:54









H.LinkhornH.Linkhorn

485213




485213












  • $begingroup$
    Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
    $endgroup$
    – lulu
    Jan 9 at 18:57












  • $begingroup$
    Do you know modular arithmetic / congruences?
    $endgroup$
    – Bill Dubuque
    Jan 9 at 18:59










  • $begingroup$
    You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
    $endgroup$
    – KM101
    Jan 9 at 18:59












  • $begingroup$
    Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
    $endgroup$
    – AlephNull
    Jan 9 at 19:03


















  • $begingroup$
    Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
    $endgroup$
    – lulu
    Jan 9 at 18:57












  • $begingroup$
    Do you know modular arithmetic / congruences?
    $endgroup$
    – Bill Dubuque
    Jan 9 at 18:59










  • $begingroup$
    You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
    $endgroup$
    – KM101
    Jan 9 at 18:59












  • $begingroup$
    Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
    $endgroup$
    – AlephNull
    Jan 9 at 19:03
















$begingroup$
Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
$endgroup$
– lulu
Jan 9 at 18:57






$begingroup$
Not following. Yes, you may assume (inductively) that $3^{2n}+7=8A$ and yes, you want to argue that this implies that $3^{3k+2}+7=8B$, but where do you argue that? It is simply not true that $3^2times 3^{2k}+7=3^2times (8A)$.
$endgroup$
– lulu
Jan 9 at 18:57














$begingroup$
Do you know modular arithmetic / congruences?
$endgroup$
– Bill Dubuque
Jan 9 at 18:59




$begingroup$
Do you know modular arithmetic / congruences?
$endgroup$
– Bill Dubuque
Jan 9 at 18:59












$begingroup$
You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
$endgroup$
– KM101
Jan 9 at 18:59






$begingroup$
You made an error. Note that $3^2cdot 3^{2k}+7 color{red}{neq 3^2cdot left(3^{2k}+7right)}$.
$endgroup$
– KM101
Jan 9 at 18:59














$begingroup$
Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
$endgroup$
– AlephNull
Jan 9 at 19:03




$begingroup$
Not relevant to the proof by induction, but it is worth noting that there is a very simple proof of this fact using basic binomial expansion: $3^{2n}+7=9^n+7=(1+8)^n+7$ and after expansion all your terms are divisible by $8$.
$endgroup$
– AlephNull
Jan 9 at 19:03










4 Answers
4






active

oldest

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3












$begingroup$

No, it is not correct. It sholud be done something like this:
$$3^{2k}= 8A-7$$ sobegin{eqnarray} 3^{2k+2}+7 &=& 9cdot 3^{2k}+7\ &=& 9cdot (8A-7)+7 \ &=& 72A-56\ &=& 8(9A-7)end{eqnarray}






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    An option:



    Step $n+1$:



    $3^{2n+2}+7 =9 cdot 3^{2n} +7=$



    $(8+1)cdot 3^{2n} +7=$



    $(3^{2n}+7) +8;$



    The first term is divisible by $8$ by hypothesis, so is the second term.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Posting this community wiki answer as a protest - instructors should not ask for proofs by induction when there are much more straightforward ones like
      $$
      3^{2n} + 7 = 9^n + 7 equiv 1^n + 7 equiv 0 pmod{8} .
      $$



      Save induction for questions where it's really important.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        You've got the right idea. Here's how to view the inductive step intuitively as multiplication



        $$begin{align}
        3^{large 2k} &= color{#c00}{-7} + 8aqquad {rm i.e.} P(k)\
        times 3^{large 2} &= color{#c00} 1 + 8phantom{I_{I_{I_I}}}\
        hline
        Rightarrow 3^{large 2(k+1)} &= {-7} + 8(cdots) {rm i.e.} P(k!+!1)
        end{align}qquad $$



        Renark $ $ Note $, -7! +! 8a = 1! +! 8(a!-!1) $ so viewed this way, instead of $, color{#c00}{-7 times 1},$ we get $,1times 1,,$ so the above boils down to $, (1+8j)(1+8k) = 1+8n,,$ i.e. $bmod 8!: 1^2equiv 1, $ so the result boils down to $,9equiv 1,Rightarrow, 9^nequiv 1^nequiv 1,$ by the Congruence Power Rule.






        share|cite|improve this answer











        $endgroup$














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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          No, it is not correct. It sholud be done something like this:
          $$3^{2k}= 8A-7$$ sobegin{eqnarray} 3^{2k+2}+7 &=& 9cdot 3^{2k}+7\ &=& 9cdot (8A-7)+7 \ &=& 72A-56\ &=& 8(9A-7)end{eqnarray}






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            No, it is not correct. It sholud be done something like this:
            $$3^{2k}= 8A-7$$ sobegin{eqnarray} 3^{2k+2}+7 &=& 9cdot 3^{2k}+7\ &=& 9cdot (8A-7)+7 \ &=& 72A-56\ &=& 8(9A-7)end{eqnarray}






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              No, it is not correct. It sholud be done something like this:
              $$3^{2k}= 8A-7$$ sobegin{eqnarray} 3^{2k+2}+7 &=& 9cdot 3^{2k}+7\ &=& 9cdot (8A-7)+7 \ &=& 72A-56\ &=& 8(9A-7)end{eqnarray}






              share|cite|improve this answer









              $endgroup$



              No, it is not correct. It sholud be done something like this:
              $$3^{2k}= 8A-7$$ sobegin{eqnarray} 3^{2k+2}+7 &=& 9cdot 3^{2k}+7\ &=& 9cdot (8A-7)+7 \ &=& 72A-56\ &=& 8(9A-7)end{eqnarray}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 9 at 18:57









              Maria MazurMaria Mazur

              49.7k1361124




              49.7k1361124























                  1












                  $begingroup$

                  An option:



                  Step $n+1$:



                  $3^{2n+2}+7 =9 cdot 3^{2n} +7=$



                  $(8+1)cdot 3^{2n} +7=$



                  $(3^{2n}+7) +8;$



                  The first term is divisible by $8$ by hypothesis, so is the second term.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    An option:



                    Step $n+1$:



                    $3^{2n+2}+7 =9 cdot 3^{2n} +7=$



                    $(8+1)cdot 3^{2n} +7=$



                    $(3^{2n}+7) +8;$



                    The first term is divisible by $8$ by hypothesis, so is the second term.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      An option:



                      Step $n+1$:



                      $3^{2n+2}+7 =9 cdot 3^{2n} +7=$



                      $(8+1)cdot 3^{2n} +7=$



                      $(3^{2n}+7) +8;$



                      The first term is divisible by $8$ by hypothesis, so is the second term.






                      share|cite|improve this answer









                      $endgroup$



                      An option:



                      Step $n+1$:



                      $3^{2n+2}+7 =9 cdot 3^{2n} +7=$



                      $(8+1)cdot 3^{2n} +7=$



                      $(3^{2n}+7) +8;$



                      The first term is divisible by $8$ by hypothesis, so is the second term.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 9 at 19:08









                      Peter SzilasPeter Szilas

                      11.7k2822




                      11.7k2822























                          0












                          $begingroup$

                          Posting this community wiki answer as a protest - instructors should not ask for proofs by induction when there are much more straightforward ones like
                          $$
                          3^{2n} + 7 = 9^n + 7 equiv 1^n + 7 equiv 0 pmod{8} .
                          $$



                          Save induction for questions where it's really important.






                          share|cite|improve this answer











                          $endgroup$


















                            0












                            $begingroup$

                            Posting this community wiki answer as a protest - instructors should not ask for proofs by induction when there are much more straightforward ones like
                            $$
                            3^{2n} + 7 = 9^n + 7 equiv 1^n + 7 equiv 0 pmod{8} .
                            $$



                            Save induction for questions where it's really important.






                            share|cite|improve this answer











                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Posting this community wiki answer as a protest - instructors should not ask for proofs by induction when there are much more straightforward ones like
                              $$
                              3^{2n} + 7 = 9^n + 7 equiv 1^n + 7 equiv 0 pmod{8} .
                              $$



                              Save induction for questions where it's really important.






                              share|cite|improve this answer











                              $endgroup$



                              Posting this community wiki answer as a protest - instructors should not ask for proofs by induction when there are much more straightforward ones like
                              $$
                              3^{2n} + 7 = 9^n + 7 equiv 1^n + 7 equiv 0 pmod{8} .
                              $$



                              Save induction for questions where it's really important.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              answered Jan 9 at 19:12


























                              community wiki





                              Ethan Bolker
























                                  0












                                  $begingroup$

                                  You've got the right idea. Here's how to view the inductive step intuitively as multiplication



                                  $$begin{align}
                                  3^{large 2k} &= color{#c00}{-7} + 8aqquad {rm i.e.} P(k)\
                                  times 3^{large 2} &= color{#c00} 1 + 8phantom{I_{I_{I_I}}}\
                                  hline
                                  Rightarrow 3^{large 2(k+1)} &= {-7} + 8(cdots) {rm i.e.} P(k!+!1)
                                  end{align}qquad $$



                                  Renark $ $ Note $, -7! +! 8a = 1! +! 8(a!-!1) $ so viewed this way, instead of $, color{#c00}{-7 times 1},$ we get $,1times 1,,$ so the above boils down to $, (1+8j)(1+8k) = 1+8n,,$ i.e. $bmod 8!: 1^2equiv 1, $ so the result boils down to $,9equiv 1,Rightarrow, 9^nequiv 1^nequiv 1,$ by the Congruence Power Rule.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You've got the right idea. Here's how to view the inductive step intuitively as multiplication



                                    $$begin{align}
                                    3^{large 2k} &= color{#c00}{-7} + 8aqquad {rm i.e.} P(k)\
                                    times 3^{large 2} &= color{#c00} 1 + 8phantom{I_{I_{I_I}}}\
                                    hline
                                    Rightarrow 3^{large 2(k+1)} &= {-7} + 8(cdots) {rm i.e.} P(k!+!1)
                                    end{align}qquad $$



                                    Renark $ $ Note $, -7! +! 8a = 1! +! 8(a!-!1) $ so viewed this way, instead of $, color{#c00}{-7 times 1},$ we get $,1times 1,,$ so the above boils down to $, (1+8j)(1+8k) = 1+8n,,$ i.e. $bmod 8!: 1^2equiv 1, $ so the result boils down to $,9equiv 1,Rightarrow, 9^nequiv 1^nequiv 1,$ by the Congruence Power Rule.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You've got the right idea. Here's how to view the inductive step intuitively as multiplication



                                      $$begin{align}
                                      3^{large 2k} &= color{#c00}{-7} + 8aqquad {rm i.e.} P(k)\
                                      times 3^{large 2} &= color{#c00} 1 + 8phantom{I_{I_{I_I}}}\
                                      hline
                                      Rightarrow 3^{large 2(k+1)} &= {-7} + 8(cdots) {rm i.e.} P(k!+!1)
                                      end{align}qquad $$



                                      Renark $ $ Note $, -7! +! 8a = 1! +! 8(a!-!1) $ so viewed this way, instead of $, color{#c00}{-7 times 1},$ we get $,1times 1,,$ so the above boils down to $, (1+8j)(1+8k) = 1+8n,,$ i.e. $bmod 8!: 1^2equiv 1, $ so the result boils down to $,9equiv 1,Rightarrow, 9^nequiv 1^nequiv 1,$ by the Congruence Power Rule.






                                      share|cite|improve this answer











                                      $endgroup$



                                      You've got the right idea. Here's how to view the inductive step intuitively as multiplication



                                      $$begin{align}
                                      3^{large 2k} &= color{#c00}{-7} + 8aqquad {rm i.e.} P(k)\
                                      times 3^{large 2} &= color{#c00} 1 + 8phantom{I_{I_{I_I}}}\
                                      hline
                                      Rightarrow 3^{large 2(k+1)} &= {-7} + 8(cdots) {rm i.e.} P(k!+!1)
                                      end{align}qquad $$



                                      Renark $ $ Note $, -7! +! 8a = 1! +! 8(a!-!1) $ so viewed this way, instead of $, color{#c00}{-7 times 1},$ we get $,1times 1,,$ so the above boils down to $, (1+8j)(1+8k) = 1+8n,,$ i.e. $bmod 8!: 1^2equiv 1, $ so the result boils down to $,9equiv 1,Rightarrow, 9^nequiv 1^nequiv 1,$ by the Congruence Power Rule.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Jan 9 at 19:21

























                                      answered Jan 9 at 19:05









                                      Bill DubuqueBill Dubuque

                                      213k29196654




                                      213k29196654






























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                                          Bressuire

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                                          Cabo Verde

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                                            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                                          Gyllenstierna

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