find the value $2^nequiv ?pmod {12}$












0












$begingroup$


if $n>1$ odd number,find $$2^nequiv ?pmod {12}$$



it seem the answer is $8$,because
$$2^3=8equiv 8pmod{12}$$
$$2^5=32equiv 8pmod {12}$$
$$2^7=128equiv 8pmod {12}$$
$$2^9=512equiv 8pmod {12}$$
$$cdots $$
But How to prove it for all postive integers $n$?






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
    $endgroup$
    – Will Jagy
    Jan 19 at 3:04
















0












$begingroup$


if $n>1$ odd number,find $$2^nequiv ?pmod {12}$$



it seem the answer is $8$,because
$$2^3=8equiv 8pmod{12}$$
$$2^5=32equiv 8pmod {12}$$
$$2^7=128equiv 8pmod {12}$$
$$2^9=512equiv 8pmod {12}$$
$$cdots $$
But How to prove it for all postive integers $n$?






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
    $endgroup$
    – Will Jagy
    Jan 19 at 3:04














0












0








0





$begingroup$


if $n>1$ odd number,find $$2^nequiv ?pmod {12}$$



it seem the answer is $8$,because
$$2^3=8equiv 8pmod{12}$$
$$2^5=32equiv 8pmod {12}$$
$$2^7=128equiv 8pmod {12}$$
$$2^9=512equiv 8pmod {12}$$
$$cdots $$
But How to prove it for all postive integers $n$?






number-theory







share|cite|improve this question











$endgroup$




if $n>1$ odd number,find $$2^nequiv ?pmod {12}$$



it seem the answer is $8$,because
$$2^3=8equiv 8pmod{12}$$
$$2^5=32equiv 8pmod {12}$$
$$2^7=128equiv 8pmod {12}$$
$$2^9=512equiv 8pmod {12}$$
$$cdots $$
But How to prove it for all postive integers $n$?






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 3:07







inequality

















asked Jan 19 at 3:02









inequalityinequality

734520




734520








  • 1




    $begingroup$
    $4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
    $endgroup$
    – Will Jagy
    Jan 19 at 3:04














  • 1




    $begingroup$
    $4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
    $endgroup$
    – Will Jagy
    Jan 19 at 3:04








1




1




$begingroup$
$4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
$endgroup$
– Will Jagy
Jan 19 at 3:04




$begingroup$
$4 cdot 8 = 32 = 24 + 8 equiv 8 pmod{12}$
$endgroup$
– Will Jagy
Jan 19 at 3:04










5 Answers
5






active

oldest

votes


















4












$begingroup$

To be precise, we want to prove that if $n$ is an odd number $geq 3$, then $$2^nequiv 8pmod{12}.$$
Since you've verify the initial case $n=3$, we assume if $kgeq3$ is an odd number and $2^kequiv 8pmod{12}$ holds, then $$2^{k+2}equiv 8times 4equiv 32equiv 8pmod{12}$$
holds as well. Hence we completed the proof by induction.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
    $endgroup$
    – Randall
    Jan 19 at 4:01










  • $begingroup$
    @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
    $endgroup$
    – Bill Dubuque
    Jan 19 at 14:09





















2












$begingroup$

$8cdot 4^{large n}!bmod 12, =, 4(2cdot 4^{large n}! bmod 3), =, 4(2)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
    $endgroup$
    – Bill Dubuque
    Jan 19 at 5:44












  • $begingroup$
    Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
    $endgroup$
    – Bill Dubuque
    Jan 19 at 14:12





















0












$begingroup$

We first calculate $$2^n mod{3}, 2^n mod{4}$$
and then combine the results with Chinese Remainder Theorem.



Both should be easy to calculate.
Writing n=2k+1, $$2^{2k+1} equiv {-1}^{2k+1} equiv ({-1}^{2})^{k} times {-1}^1 equiv 1 times -1 equiv -1 equiv 2 mod{3}$$
$$2^{2k+1} equiv 4^ktimes2equiv0mod{4}$$
Therefore by CRT, $$2^{2k+1} equiv 8 mod{12}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can prove it with the Chinese remainder theorem: $12=2^2cdot3$.



    We have $2^ncong0pmod{2^2}$, and$2^ncong2pmod3$ ( since by Fermat's little theorem, $2^2cong1pmod3$).



    Using Bezout's identity, $1cdot2^2-1cdot 3=1$, we get $2cdot1cdot2^2+0cdot1cdot 3=8$, as our solution.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      A slicker way to apply CRT here is via mod distributivity - see my answer.
      $endgroup$
      – Bill Dubuque
      Jan 19 at 5:49










    • $begingroup$
      Nice @BillDubuque. I will try to use it some time.
      $endgroup$
      – Chris Custer
      Jan 19 at 6:34



















    0












    $begingroup$

    Notice that:



    $$2^{2m+1}-8=2(2^m+2)(2^m-2)$$



    We wish to show the RHS is a multiple of $12$. The outside $2$ means we need one of the brackets to be a multiple of $6$. Can you show that if two even numbers are four apart, one of them must be a multiple of $6$?






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      To be precise, we want to prove that if $n$ is an odd number $geq 3$, then $$2^nequiv 8pmod{12}.$$
      Since you've verify the initial case $n=3$, we assume if $kgeq3$ is an odd number and $2^kequiv 8pmod{12}$ holds, then $$2^{k+2}equiv 8times 4equiv 32equiv 8pmod{12}$$
      holds as well. Hence we completed the proof by induction.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
        $endgroup$
        – Randall
        Jan 19 at 4:01










      • $begingroup$
        @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:09


















      4












      $begingroup$

      To be precise, we want to prove that if $n$ is an odd number $geq 3$, then $$2^nequiv 8pmod{12}.$$
      Since you've verify the initial case $n=3$, we assume if $kgeq3$ is an odd number and $2^kequiv 8pmod{12}$ holds, then $$2^{k+2}equiv 8times 4equiv 32equiv 8pmod{12}$$
      holds as well. Hence we completed the proof by induction.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
        $endgroup$
        – Randall
        Jan 19 at 4:01










      • $begingroup$
        @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:09
















      4












      4








      4





      $begingroup$

      To be precise, we want to prove that if $n$ is an odd number $geq 3$, then $$2^nequiv 8pmod{12}.$$
      Since you've verify the initial case $n=3$, we assume if $kgeq3$ is an odd number and $2^kequiv 8pmod{12}$ holds, then $$2^{k+2}equiv 8times 4equiv 32equiv 8pmod{12}$$
      holds as well. Hence we completed the proof by induction.






      share|cite|improve this answer









      $endgroup$



      To be precise, we want to prove that if $n$ is an odd number $geq 3$, then $$2^nequiv 8pmod{12}.$$
      Since you've verify the initial case $n=3$, we assume if $kgeq3$ is an odd number and $2^kequiv 8pmod{12}$ holds, then $$2^{k+2}equiv 8times 4equiv 32equiv 8pmod{12}$$
      holds as well. Hence we completed the proof by induction.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 19 at 3:10









      kelvin hong 方kelvin hong 方

      62018




      62018








      • 1




        $begingroup$
        For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
        $endgroup$
        – Randall
        Jan 19 at 4:01










      • $begingroup$
        @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:09
















      • 1




        $begingroup$
        For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
        $endgroup$
        – Randall
        Jan 19 at 4:01










      • $begingroup$
        @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:09










      1




      1




      $begingroup$
      For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
      $endgroup$
      – Randall
      Jan 19 at 4:01




      $begingroup$
      For quite a while I was trying to do this without induction, but this really is nice, short, and sweet.
      $endgroup$
      – Randall
      Jan 19 at 4:01












      $begingroup$
      @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
      $endgroup$
      – Bill Dubuque
      Jan 19 at 14:09






      $begingroup$
      @Randall We can reduce the induction to the triviality $,1^nequiv 1,$ by factoring out $4$ (or $8)$ as I explain in my answer and its comments. Using this mod distributivity often simplifies modular arithmetic (it can be viewed as an operational form of CRT more amenable to computation).
      $endgroup$
      – Bill Dubuque
      Jan 19 at 14:09













      2












      $begingroup$

      $8cdot 4^{large n}!bmod 12, =, 4(2cdot 4^{large n}! bmod 3), =, 4(2)$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 5:44












      • $begingroup$
        Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:12


















      2












      $begingroup$

      $8cdot 4^{large n}!bmod 12, =, 4(2cdot 4^{large n}! bmod 3), =, 4(2)$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 5:44












      • $begingroup$
        Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:12
















      2












      2








      2





      $begingroup$

      $8cdot 4^{large n}!bmod 12, =, 4(2cdot 4^{large n}! bmod 3), =, 4(2)$






      share|cite|improve this answer











      $endgroup$



      $8cdot 4^{large n}!bmod 12, =, 4(2cdot 4^{large n}! bmod 3), =, 4(2)$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 19 at 5:38

























      answered Jan 19 at 5:20









      Bill DubuqueBill Dubuque

      210k29192645




      210k29192645












      • $begingroup$
        $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 5:44












      • $begingroup$
        Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:12




















      • $begingroup$
        $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 5:44












      • $begingroup$
        Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
        $endgroup$
        – Bill Dubuque
        Jan 19 at 14:12


















      $begingroup$
      $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
      $endgroup$
      – Bill Dubuque
      Jan 19 at 5:44






      $begingroup$
      $large {rm By} abbmod ac = a(bbmod c) = $ mod Distributive Law, and 4 = 1 (mod 3) $ $
      $endgroup$
      – Bill Dubuque
      Jan 19 at 5:44














      $begingroup$
      Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
      $endgroup$
      – Bill Dubuque
      Jan 19 at 14:12






      $begingroup$
      Or $,large 8cdot 4^{large n}!bmod 24, =, 8(color{#c00}4^{large n}! bmod 3), =, 8(1),$ more generally, by $,large color{#c00}4^nequiv color{#c00}1^nequiv 1pmod{!3}. $ Notice that factoring out the $,8,$ simplifies the induction to a completely trivial instance: $,,large 1^nequiv 1 $
      $endgroup$
      – Bill Dubuque
      Jan 19 at 14:12













      0












      $begingroup$

      We first calculate $$2^n mod{3}, 2^n mod{4}$$
      and then combine the results with Chinese Remainder Theorem.



      Both should be easy to calculate.
      Writing n=2k+1, $$2^{2k+1} equiv {-1}^{2k+1} equiv ({-1}^{2})^{k} times {-1}^1 equiv 1 times -1 equiv -1 equiv 2 mod{3}$$
      $$2^{2k+1} equiv 4^ktimes2equiv0mod{4}$$
      Therefore by CRT, $$2^{2k+1} equiv 8 mod{12}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We first calculate $$2^n mod{3}, 2^n mod{4}$$
        and then combine the results with Chinese Remainder Theorem.



        Both should be easy to calculate.
        Writing n=2k+1, $$2^{2k+1} equiv {-1}^{2k+1} equiv ({-1}^{2})^{k} times {-1}^1 equiv 1 times -1 equiv -1 equiv 2 mod{3}$$
        $$2^{2k+1} equiv 4^ktimes2equiv0mod{4}$$
        Therefore by CRT, $$2^{2k+1} equiv 8 mod{12}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We first calculate $$2^n mod{3}, 2^n mod{4}$$
          and then combine the results with Chinese Remainder Theorem.



          Both should be easy to calculate.
          Writing n=2k+1, $$2^{2k+1} equiv {-1}^{2k+1} equiv ({-1}^{2})^{k} times {-1}^1 equiv 1 times -1 equiv -1 equiv 2 mod{3}$$
          $$2^{2k+1} equiv 4^ktimes2equiv0mod{4}$$
          Therefore by CRT, $$2^{2k+1} equiv 8 mod{12}$$






          share|cite|improve this answer









          $endgroup$



          We first calculate $$2^n mod{3}, 2^n mod{4}$$
          and then combine the results with Chinese Remainder Theorem.



          Both should be easy to calculate.
          Writing n=2k+1, $$2^{2k+1} equiv {-1}^{2k+1} equiv ({-1}^{2})^{k} times {-1}^1 equiv 1 times -1 equiv -1 equiv 2 mod{3}$$
          $$2^{2k+1} equiv 4^ktimes2equiv0mod{4}$$
          Therefore by CRT, $$2^{2k+1} equiv 8 mod{12}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 3:20









          Gareth MaGareth Ma

          321113




          321113























              0












              $begingroup$

              You can prove it with the Chinese remainder theorem: $12=2^2cdot3$.



              We have $2^ncong0pmod{2^2}$, and$2^ncong2pmod3$ ( since by Fermat's little theorem, $2^2cong1pmod3$).



              Using Bezout's identity, $1cdot2^2-1cdot 3=1$, we get $2cdot1cdot2^2+0cdot1cdot 3=8$, as our solution.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                A slicker way to apply CRT here is via mod distributivity - see my answer.
                $endgroup$
                – Bill Dubuque
                Jan 19 at 5:49










              • $begingroup$
                Nice @BillDubuque. I will try to use it some time.
                $endgroup$
                – Chris Custer
                Jan 19 at 6:34
















              0












              $begingroup$

              You can prove it with the Chinese remainder theorem: $12=2^2cdot3$.



              We have $2^ncong0pmod{2^2}$, and$2^ncong2pmod3$ ( since by Fermat's little theorem, $2^2cong1pmod3$).



              Using Bezout's identity, $1cdot2^2-1cdot 3=1$, we get $2cdot1cdot2^2+0cdot1cdot 3=8$, as our solution.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                A slicker way to apply CRT here is via mod distributivity - see my answer.
                $endgroup$
                – Bill Dubuque
                Jan 19 at 5:49










              • $begingroup$
                Nice @BillDubuque. I will try to use it some time.
                $endgroup$
                – Chris Custer
                Jan 19 at 6:34














              0












              0








              0





              $begingroup$

              You can prove it with the Chinese remainder theorem: $12=2^2cdot3$.



              We have $2^ncong0pmod{2^2}$, and$2^ncong2pmod3$ ( since by Fermat's little theorem, $2^2cong1pmod3$).



              Using Bezout's identity, $1cdot2^2-1cdot 3=1$, we get $2cdot1cdot2^2+0cdot1cdot 3=8$, as our solution.






              share|cite|improve this answer











              $endgroup$



              You can prove it with the Chinese remainder theorem: $12=2^2cdot3$.



              We have $2^ncong0pmod{2^2}$, and$2^ncong2pmod3$ ( since by Fermat's little theorem, $2^2cong1pmod3$).



              Using Bezout's identity, $1cdot2^2-1cdot 3=1$, we get $2cdot1cdot2^2+0cdot1cdot 3=8$, as our solution.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 19 at 4:49

























              answered Jan 19 at 3:19









              Chris CusterChris Custer

              13.2k3827




              13.2k3827












              • $begingroup$
                A slicker way to apply CRT here is via mod distributivity - see my answer.
                $endgroup$
                – Bill Dubuque
                Jan 19 at 5:49










              • $begingroup$
                Nice @BillDubuque. I will try to use it some time.
                $endgroup$
                – Chris Custer
                Jan 19 at 6:34


















              • $begingroup$
                A slicker way to apply CRT here is via mod distributivity - see my answer.
                $endgroup$
                – Bill Dubuque
                Jan 19 at 5:49










              • $begingroup$
                Nice @BillDubuque. I will try to use it some time.
                $endgroup$
                – Chris Custer
                Jan 19 at 6:34
















              $begingroup$
              A slicker way to apply CRT here is via mod distributivity - see my answer.
              $endgroup$
              – Bill Dubuque
              Jan 19 at 5:49




              $begingroup$
              A slicker way to apply CRT here is via mod distributivity - see my answer.
              $endgroup$
              – Bill Dubuque
              Jan 19 at 5:49












              $begingroup$
              Nice @BillDubuque. I will try to use it some time.
              $endgroup$
              – Chris Custer
              Jan 19 at 6:34




              $begingroup$
              Nice @BillDubuque. I will try to use it some time.
              $endgroup$
              – Chris Custer
              Jan 19 at 6:34











              0












              $begingroup$

              Notice that:



              $$2^{2m+1}-8=2(2^m+2)(2^m-2)$$



              We wish to show the RHS is a multiple of $12$. The outside $2$ means we need one of the brackets to be a multiple of $6$. Can you show that if two even numbers are four apart, one of them must be a multiple of $6$?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Notice that:



                $$2^{2m+1}-8=2(2^m+2)(2^m-2)$$



                We wish to show the RHS is a multiple of $12$. The outside $2$ means we need one of the brackets to be a multiple of $6$. Can you show that if two even numbers are four apart, one of them must be a multiple of $6$?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Notice that:



                  $$2^{2m+1}-8=2(2^m+2)(2^m-2)$$



                  We wish to show the RHS is a multiple of $12$. The outside $2$ means we need one of the brackets to be a multiple of $6$. Can you show that if two even numbers are four apart, one of them must be a multiple of $6$?






                  share|cite|improve this answer









                  $endgroup$



                  Notice that:



                  $$2^{2m+1}-8=2(2^m+2)(2^m-2)$$



                  We wish to show the RHS is a multiple of $12$. The outside $2$ means we need one of the brackets to be a multiple of $6$. Can you show that if two even numbers are four apart, one of them must be a multiple of $6$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 19 at 5:02









                  Rhys HughesRhys Hughes

                  6,7071530




                  6,7071530






























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