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Find prime p s.t. a^4-b^4=p

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Find all primes p such that [tex]\exists a,b \in \mathbf{Z}[/tex] such that [tex]a^4-b^4=p[/tex].


    2. Relevant equations



    3. The attempt at a solution
    For simplicity, we can limit a and b to the positive integers.

    Factoring, we have [tex]p=(a^2+b^2)(a-b)(a+b)[/tex]. By the unique factorization theorem, we are limited to three cases:

    (1) [tex]a+b=1[/tex] and [tex]a-b=1[/tex], which gives [tex]a=1[/tex] and [tex]b=0[/tex], so p must be 1. But since 1 is not a prime, case 1 is eliminated.

    (2) [tex]a^2+b^2=1[/tex] and [tex]a-b=1[/tex], which gives [tex]a^2+b^2-2ab=1[/tex] and then [tex]-2ab=0[/tex]. Again, we are left with [tex]a=1[/tex] and [tex]b=0[/tex], so case 2 is eliminated.

    (3) [tex]a^2+b^2=1[/tex] and [tex]a+b=1[/tex], which gives [tex]a^2+b^2+2ab=1[/tex] and then [tex]2ab=0[/tex]. Again, we are left with [tex]a=1[/tex] and [tex]b=0[/tex], so case 3 is eliminated.

    Therefore, no primes satisfy the equation. Q.E.D.



    Is my proof valid? If it is, is there a "more elegant" proof?


    edit: I accidentally put the question as [tex]a^4+b^4=p[/tex] instead of what I currently have up there. >_< Oops!
     
  2. jcsd
  3. Jul 19, 2008 #2
    Do a and b have to be distinct? Because a=1=b gives p=2, which is prime.
     
  4. Jul 19, 2008 #3
    Hi,
    Jeffreydk, if a=1=b then p=0 from the origional equation.

    foxjwill,
    on part 2,
    I don't see how your algebra got you [tex] a^2+b^2-2ab=1[/tex]
    I agree with your result up to a point, though.
    Here's what I did:
    [tex]a^2+b^2=1[/tex] and [tex]a-b=1 \Longrightarrow a=b+1 [/tex]
    then plug that into the first one:
    [tex](b+1)^2+b^2=1[/tex]
    [tex]b^2+2b+1+b^2=1[/tex]
    which eventually gives 2 solutions:
    [tex]b=0[/tex] and [tex] b=-1[/tex]
    both of which still don't make p prime when you solve for a...you just have to make sure that you cover all possibilities.

    I didn't work out the 3rd part, but I bet you get 2 results out of it as well.

    CC
     
    Last edited: Jul 19, 2008
  5. Jul 19, 2008 #4
    yes, but he gave this remark before I had corrected my typo. So, when he saw it, a=b=1 did give p=2

    I got that by squaring [tex]a-b=1[/tex]
     
  6. Jul 19, 2008 #5
    I did not see the edit....sorry.

    I took it as a system of equations. Your method is also valid, but not quite complete.
    -1 has to be included to be completely rigorous...no matter what, no primes will satisfy the thing :)
    CC
     
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