How to use modular arithmetic for $a^b-1 $












1












$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







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$endgroup$








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40
















1












$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40














1












1








1





$begingroup$


The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




The problem is: If $a,b ge 2$ are natural numbers s.t $a^b - 1$ is prime, then $a = 2$.



My ideas so far:



Since the we want to show that $a = 2$, we can take the the expression modulo $a$.



$a^b-1 equiv -1 pmod{a}$, so we know that there's a prime $p>2$ such that $p equiv -1 pmod{a}$ However I am not really sure where to go from here.






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 21:05









Namaste

1




1










asked Dec 29 '18 at 21:03









xAlyxAly

404




404








  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40














  • 1




    $begingroup$
    I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:40








1




1




$begingroup$
I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
$endgroup$
– Wojowu
Dec 29 '18 at 21:40




$begingroup$
I think the intent might have been to look at this modulo $a-1$ - since $aequiv 1pmod{a-1}$, $a^b-1equiv 1^b-1=0pmod{a-1}$, so $a-1$ is a factor of $a^b-1$. This says nothing more than the answer below, but it's still worth pointing out.
$endgroup$
– Wojowu
Dec 29 '18 at 21:40










2 Answers
2






active

oldest

votes


















3












$begingroup$

Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10










  • $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17



















1












$begingroup$

Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



These are the Mersenne primes. Which are well researched.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    3












    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17
















    3












    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17














    3












    3








    3





    $begingroup$

    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.






    share|cite|improve this answer











    $endgroup$



    Suppose, there is an $a>2$, and a $b$ such that $a^b-1$ is prime. Observe that $a-1mid a^b-1$, $2leq a-1<a^b-1$ establishes, $a^b-1$ is not a prime.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 29 '18 at 21:18

























    answered Dec 29 '18 at 21:07









    AaronAaron

    1,922415




    1,922415








    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17














    • 1




      $begingroup$
      What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
      $endgroup$
      – Wojowu
      Dec 29 '18 at 21:10










    • $begingroup$
      No need to. I was thinking as I was writing. Condensed it.
      $endgroup$
      – Aaron
      Dec 29 '18 at 21:17








    1




    1




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10




    $begingroup$
    What's the point in taking $q$? $a-1mid a^b-1$ already proves $a^b-1$ is not prime.
    $endgroup$
    – Wojowu
    Dec 29 '18 at 21:10












    $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17




    $begingroup$
    No need to. I was thinking as I was writing. Condensed it.
    $endgroup$
    – Aaron
    Dec 29 '18 at 21:17











    1












    $begingroup$

    Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



    The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



    These are the Mersenne primes. Which are well researched.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



      The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



      These are the Mersenne primes. Which are well researched.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



        The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



        These are the Mersenne primes. Which are well researched.






        share|cite|improve this answer









        $endgroup$



        Hmm, If $b ge 2$ then $a^b -1 = (a-1)(a^{b-1}+ .... + 1)$ and $a ge 2$ then $a^{b-1}+ .... + 1> 1$ so the only way $a^b - 1$ is prime is if $a-1 = 1$ or $a = 2$.



        The real question is under what conditions of $b$ can we assume $2^b - 1$ is prime. If $b= nk$ is not prime then you have $2^b - 1 = 2^n)^k - 1 = (2^n-1)((2^n)^{k-1} + ..... +1)$ so $b$ must be prime if $2^b -1$ is to be prime.



        These are the Mersenne primes. Which are well researched.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 21:43









        fleabloodfleablood

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