Unable to solve this exponential equation - Diffie-Hellman key exchange












1












$begingroup$


By looking at it, I can deduce that $a = 6$, and $b = 5$, but how do I can solve for $a$ and $b$ without guessing?



$$2^a = 11b + 9$$






elementary-number-theory cryptography discrete-logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider taking $mod 11$ perhaps.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 3:03












  • $begingroup$
    You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:04










  • $begingroup$
    @ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:08












  • $begingroup$
    @YadatiKiran If would be great if you could be more elaborate
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:10










  • $begingroup$
    If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:17


















1












$begingroup$


By looking at it, I can deduce that $a = 6$, and $b = 5$, but how do I can solve for $a$ and $b$ without guessing?



$$2^a = 11b + 9$$






elementary-number-theory cryptography discrete-logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider taking $mod 11$ perhaps.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 3:03












  • $begingroup$
    You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:04










  • $begingroup$
    @ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:08












  • $begingroup$
    @YadatiKiran If would be great if you could be more elaborate
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:10










  • $begingroup$
    If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:17
















1












1








1





$begingroup$


By looking at it, I can deduce that $a = 6$, and $b = 5$, but how do I can solve for $a$ and $b$ without guessing?



$$2^a = 11b + 9$$






elementary-number-theory cryptography discrete-logarithms







share|cite|improve this question











$endgroup$




By looking at it, I can deduce that $a = 6$, and $b = 5$, but how do I can solve for $a$ and $b$ without guessing?



$$2^a = 11b + 9$$






elementary-number-theory cryptography discrete-logarithms




elementary-number-theory cryptography discrete-logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 3:26







Almost Handsome

















asked Dec 8 '18 at 2:57









Almost HandsomeAlmost Handsome

1085




1085












  • $begingroup$
    Consider taking $mod 11$ perhaps.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 3:03












  • $begingroup$
    You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:04










  • $begingroup$
    @ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:08












  • $begingroup$
    @YadatiKiran If would be great if you could be more elaborate
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:10










  • $begingroup$
    If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:17




















  • $begingroup$
    Consider taking $mod 11$ perhaps.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 3:03












  • $begingroup$
    You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:04










  • $begingroup$
    @ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:08












  • $begingroup$
    @YadatiKiran If would be great if you could be more elaborate
    $endgroup$
    – Almost Handsome
    Dec 8 '18 at 3:10










  • $begingroup$
    If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 3:17


















$begingroup$
Consider taking $mod 11$ perhaps.
$endgroup$
– Yadati Kiran
Dec 8 '18 at 3:03






$begingroup$
Consider taking $mod 11$ perhaps.
$endgroup$
– Yadati Kiran
Dec 8 '18 at 3:03














$begingroup$
You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:04




$begingroup$
You tagged this “calculus”. What context did you encounter this problem in? I suspect it wasn’t calculus...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:04












$begingroup$
@ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
$endgroup$
– Almost Handsome
Dec 8 '18 at 3:08






$begingroup$
@ArturoMagidin I am not sure if it is correctly calculus, it is a Diffie-Hellman Key exchange problem, I am trying to find the private key of Alice which is $a$, the equation in the question is the one I derived from the data provided in the problem, so the common prime $q = 11$, primitive root $alpha = 2$, and Alice's public generated key is 9, find Alice's private key
$endgroup$
– Almost Handsome
Dec 8 '18 at 3:08














$begingroup$
@YadatiKiran If would be great if you could be more elaborate
$endgroup$
– Almost Handsome
Dec 8 '18 at 3:10




$begingroup$
@YadatiKiran If would be great if you could be more elaborate
$endgroup$
– Almost Handsome
Dec 8 '18 at 3:10












$begingroup$
If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:17






$begingroup$
If it is Diffie-Hellman, then it is most definitely not calculus, as it involves a discrete problem. This is the discrete logarithm problem, cryptography, perhaps elementary number theory or number theory. I’ve fixed your tags.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:17












2 Answers
2






active

oldest

votes


















5












$begingroup$

Write $$2^aequiv9pmod{11}.$$ Rewrite it as $$2^aequiv-2pmod{11},$$or
$$2^{a-1}equiv-1pmod{11}.$$ Squaring both sides,
$$2^{2a-2}equiv1pmod{11}.$$ By Fermat's little theorem, we know that one solution is $$2a-2=10iff a=6$$ Any smaller exponent must be must be a factor of $10$, and the only even factor of $10$ is $2,$ but $2a-2=2$ gives $a=2,$ which doesn't give a solution.



Since $2^{10}equiv 1pmod{11},$ all values of the form $a=10k+6$ actually lead to solutions.



As has been pointed out in the comments, this is a discrete logarithm problem, and is very difficult in general. Usually the problem is to find the smallest exponent that would work, which involves find the factors of the exponent that you get from little Fermat. Since factoring is very difficult, so is this problem.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This equation in a way can be related to Collatz conjecture. There is a set of infinite odd numbers which satisfy the condition of Collatz conjecture; this set is:



    $K={1,5,21,85,341,1365,5461,21845 . . .(4k_{i-1}+1)}$; $i ∈ N$



    So that:



    $3times k_i+1=2^n$



    Which due to Collatz conjecture leads to 1 by cintinious dividing by 2.



    Hence the equation can also be written as:



    $11a+9=3k+1$; $k ∈ K$



    The general solutions of this equation are:



    $a=4t+5$



    $k=11t+21$



    Where 5 and 21 are a solution of equation. We can see that is set K, in addition to $k_3=21$ which gives $a=5$ and $n=6$ there is another numbers such as $k_8=21845$ which gives $n=16$ and $a=5957$:



    $3times21845+1=65536=2^{16}$



    $65536=5957times 11+9$



    which sre resulted with $t=1488$.



    It is not known the number of solutions of this equation is infinite, however using general solution and try more numbers for t we may find even more solutions by Brut force.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Write $$2^aequiv9pmod{11}.$$ Rewrite it as $$2^aequiv-2pmod{11},$$or
      $$2^{a-1}equiv-1pmod{11}.$$ Squaring both sides,
      $$2^{2a-2}equiv1pmod{11}.$$ By Fermat's little theorem, we know that one solution is $$2a-2=10iff a=6$$ Any smaller exponent must be must be a factor of $10$, and the only even factor of $10$ is $2,$ but $2a-2=2$ gives $a=2,$ which doesn't give a solution.



      Since $2^{10}equiv 1pmod{11},$ all values of the form $a=10k+6$ actually lead to solutions.



      As has been pointed out in the comments, this is a discrete logarithm problem, and is very difficult in general. Usually the problem is to find the smallest exponent that would work, which involves find the factors of the exponent that you get from little Fermat. Since factoring is very difficult, so is this problem.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Write $$2^aequiv9pmod{11}.$$ Rewrite it as $$2^aequiv-2pmod{11},$$or
        $$2^{a-1}equiv-1pmod{11}.$$ Squaring both sides,
        $$2^{2a-2}equiv1pmod{11}.$$ By Fermat's little theorem, we know that one solution is $$2a-2=10iff a=6$$ Any smaller exponent must be must be a factor of $10$, and the only even factor of $10$ is $2,$ but $2a-2=2$ gives $a=2,$ which doesn't give a solution.



        Since $2^{10}equiv 1pmod{11},$ all values of the form $a=10k+6$ actually lead to solutions.



        As has been pointed out in the comments, this is a discrete logarithm problem, and is very difficult in general. Usually the problem is to find the smallest exponent that would work, which involves find the factors of the exponent that you get from little Fermat. Since factoring is very difficult, so is this problem.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Write $$2^aequiv9pmod{11}.$$ Rewrite it as $$2^aequiv-2pmod{11},$$or
          $$2^{a-1}equiv-1pmod{11}.$$ Squaring both sides,
          $$2^{2a-2}equiv1pmod{11}.$$ By Fermat's little theorem, we know that one solution is $$2a-2=10iff a=6$$ Any smaller exponent must be must be a factor of $10$, and the only even factor of $10$ is $2,$ but $2a-2=2$ gives $a=2,$ which doesn't give a solution.



          Since $2^{10}equiv 1pmod{11},$ all values of the form $a=10k+6$ actually lead to solutions.



          As has been pointed out in the comments, this is a discrete logarithm problem, and is very difficult in general. Usually the problem is to find the smallest exponent that would work, which involves find the factors of the exponent that you get from little Fermat. Since factoring is very difficult, so is this problem.






          share|cite|improve this answer









          $endgroup$



          Write $$2^aequiv9pmod{11}.$$ Rewrite it as $$2^aequiv-2pmod{11},$$or
          $$2^{a-1}equiv-1pmod{11}.$$ Squaring both sides,
          $$2^{2a-2}equiv1pmod{11}.$$ By Fermat's little theorem, we know that one solution is $$2a-2=10iff a=6$$ Any smaller exponent must be must be a factor of $10$, and the only even factor of $10$ is $2,$ but $2a-2=2$ gives $a=2,$ which doesn't give a solution.



          Since $2^{10}equiv 1pmod{11},$ all values of the form $a=10k+6$ actually lead to solutions.



          As has been pointed out in the comments, this is a discrete logarithm problem, and is very difficult in general. Usually the problem is to find the smallest exponent that would work, which involves find the factors of the exponent that you get from little Fermat. Since factoring is very difficult, so is this problem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 3:31









          saulspatzsaulspatz

          14.3k21329




          14.3k21329























              0












              $begingroup$

              This equation in a way can be related to Collatz conjecture. There is a set of infinite odd numbers which satisfy the condition of Collatz conjecture; this set is:



              $K={1,5,21,85,341,1365,5461,21845 . . .(4k_{i-1}+1)}$; $i ∈ N$



              So that:



              $3times k_i+1=2^n$



              Which due to Collatz conjecture leads to 1 by cintinious dividing by 2.



              Hence the equation can also be written as:



              $11a+9=3k+1$; $k ∈ K$



              The general solutions of this equation are:



              $a=4t+5$



              $k=11t+21$



              Where 5 and 21 are a solution of equation. We can see that is set K, in addition to $k_3=21$ which gives $a=5$ and $n=6$ there is another numbers such as $k_8=21845$ which gives $n=16$ and $a=5957$:



              $3times21845+1=65536=2^{16}$



              $65536=5957times 11+9$



              which sre resulted with $t=1488$.



              It is not known the number of solutions of this equation is infinite, however using general solution and try more numbers for t we may find even more solutions by Brut force.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This equation in a way can be related to Collatz conjecture. There is a set of infinite odd numbers which satisfy the condition of Collatz conjecture; this set is:



                $K={1,5,21,85,341,1365,5461,21845 . . .(4k_{i-1}+1)}$; $i ∈ N$



                So that:



                $3times k_i+1=2^n$



                Which due to Collatz conjecture leads to 1 by cintinious dividing by 2.



                Hence the equation can also be written as:



                $11a+9=3k+1$; $k ∈ K$



                The general solutions of this equation are:



                $a=4t+5$



                $k=11t+21$



                Where 5 and 21 are a solution of equation. We can see that is set K, in addition to $k_3=21$ which gives $a=5$ and $n=6$ there is another numbers such as $k_8=21845$ which gives $n=16$ and $a=5957$:



                $3times21845+1=65536=2^{16}$



                $65536=5957times 11+9$



                which sre resulted with $t=1488$.



                It is not known the number of solutions of this equation is infinite, however using general solution and try more numbers for t we may find even more solutions by Brut force.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This equation in a way can be related to Collatz conjecture. There is a set of infinite odd numbers which satisfy the condition of Collatz conjecture; this set is:



                  $K={1,5,21,85,341,1365,5461,21845 . . .(4k_{i-1}+1)}$; $i ∈ N$



                  So that:



                  $3times k_i+1=2^n$



                  Which due to Collatz conjecture leads to 1 by cintinious dividing by 2.



                  Hence the equation can also be written as:



                  $11a+9=3k+1$; $k ∈ K$



                  The general solutions of this equation are:



                  $a=4t+5$



                  $k=11t+21$



                  Where 5 and 21 are a solution of equation. We can see that is set K, in addition to $k_3=21$ which gives $a=5$ and $n=6$ there is another numbers such as $k_8=21845$ which gives $n=16$ and $a=5957$:



                  $3times21845+1=65536=2^{16}$



                  $65536=5957times 11+9$



                  which sre resulted with $t=1488$.



                  It is not known the number of solutions of this equation is infinite, however using general solution and try more numbers for t we may find even more solutions by Brut force.






                  share|cite|improve this answer









                  $endgroup$



                  This equation in a way can be related to Collatz conjecture. There is a set of infinite odd numbers which satisfy the condition of Collatz conjecture; this set is:



                  $K={1,5,21,85,341,1365,5461,21845 . . .(4k_{i-1}+1)}$; $i ∈ N$



                  So that:



                  $3times k_i+1=2^n$



                  Which due to Collatz conjecture leads to 1 by cintinious dividing by 2.



                  Hence the equation can also be written as:



                  $11a+9=3k+1$; $k ∈ K$



                  The general solutions of this equation are:



                  $a=4t+5$



                  $k=11t+21$



                  Where 5 and 21 are a solution of equation. We can see that is set K, in addition to $k_3=21$ which gives $a=5$ and $n=6$ there is another numbers such as $k_8=21845$ which gives $n=16$ and $a=5957$:



                  $3times21845+1=65536=2^{16}$



                  $65536=5957times 11+9$



                  which sre resulted with $t=1488$.



                  It is not known the number of solutions of this equation is infinite, however using general solution and try more numbers for t we may find even more solutions by Brut force.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 13:30









                  siroussirous

                  1,6291513




                  1,6291513






























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                      Dieringhausen

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                      Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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