Prove that if $a,b ge2$ such that $a^b-1$ is prime, then $a=2$












1












$begingroup$


I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?






proof-verification







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  • 1




    $begingroup$
    Hint: the polynomial $x^n-1$ is divisible by $x-1$.
    $endgroup$
    – lulu
    Jan 9 at 18:06










  • $begingroup$
    Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
    $endgroup$
    – fleablood
    Jan 9 at 18:22
















1












$begingroup$


I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?






proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: the polynomial $x^n-1$ is divisible by $x-1$.
    $endgroup$
    – lulu
    Jan 9 at 18:06










  • $begingroup$
    Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
    $endgroup$
    – fleablood
    Jan 9 at 18:22














1












1








1





$begingroup$


I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?






proof-verification







share|cite|improve this question











$endgroup$




I want to start with a contradiction assuming $a$ isn't $2$ but no idea how to do that so I'm assuming it's wrong. Any help?






proof-verification




proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 18:17









egreg

185k1486207




185k1486207










asked Jan 9 at 18:04









AnoUser1AnoUser1

836




836








  • 1




    $begingroup$
    Hint: the polynomial $x^n-1$ is divisible by $x-1$.
    $endgroup$
    – lulu
    Jan 9 at 18:06










  • $begingroup$
    Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
    $endgroup$
    – fleablood
    Jan 9 at 18:22














  • 1




    $begingroup$
    Hint: the polynomial $x^n-1$ is divisible by $x-1$.
    $endgroup$
    – lulu
    Jan 9 at 18:06










  • $begingroup$
    Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
    $endgroup$
    – fleablood
    Jan 9 at 18:22








1




1




$begingroup$
Hint: the polynomial $x^n-1$ is divisible by $x-1$.
$endgroup$
– lulu
Jan 9 at 18:06




$begingroup$
Hint: the polynomial $x^n-1$ is divisible by $x-1$.
$endgroup$
– lulu
Jan 9 at 18:06












$begingroup$
Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
$endgroup$
– fleablood
Jan 9 at 18:22




$begingroup$
Try to prove $a^b -1$ is not prime. In other words that it can be factored. Hint: $(a -1)(a^c + a^{c-1} + ..... + 1) = a^{c+1} - 1$.
$endgroup$
– fleablood
Jan 9 at 18:22










1 Answer
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$begingroup$

Hint: Use the difference of $n^{th}$ powers:



$$a^n-b^n = (a-b)left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}right)$$



Apply this to $a^b-1$:



$$implies a^b-1 = color{blue}{(a-1)}left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1right)$$



What happens if $color{blue}{a > 2}$?






share|cite|improve this answer











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    $begingroup$

    Hint: Use the difference of $n^{th}$ powers:



    $$a^n-b^n = (a-b)left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}right)$$



    Apply this to $a^b-1$:



    $$implies a^b-1 = color{blue}{(a-1)}left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1right)$$



    What happens if $color{blue}{a > 2}$?






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Hint: Use the difference of $n^{th}$ powers:



      $$a^n-b^n = (a-b)left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}right)$$



      Apply this to $a^b-1$:



      $$implies a^b-1 = color{blue}{(a-1)}left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1right)$$



      What happens if $color{blue}{a > 2}$?






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: Use the difference of $n^{th}$ powers:



        $$a^n-b^n = (a-b)left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}right)$$



        Apply this to $a^b-1$:



        $$implies a^b-1 = color{blue}{(a-1)}left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1right)$$



        What happens if $color{blue}{a > 2}$?






        share|cite|improve this answer











        $endgroup$



        Hint: Use the difference of $n^{th}$ powers:



        $$a^n-b^n = (a-b)left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}ab^{n-2}+b^{n-1}right)$$



        Apply this to $a^b-1$:



        $$implies a^b-1 = color{blue}{(a-1)}left(a^{b-1}+a^{b-2}+a^{b-3}+…+a^2+a+1right)$$



        What happens if $color{blue}{a > 2}$?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 at 18:17

























        answered Jan 9 at 18:11









        KM101KM101

        6,0901525




        6,0901525






























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