1. Joined
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    05 Nov '08 17:25
    Consider this equation:
    n^4 + 4*m^4 = p
    where x and y is an positive integer and p is a prime.

    Does this equation have
    (a) no solutions,
    (b) one solution,
    (c) many solutions, or
    (d) infinitely many solutions?

    If possible, show the a, b and p in the solution(s).
  2. Joined
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    05 Nov '08 17:48
    Originally posted by FabianFnas
    Consider this equation:
    n^4 + 4*m^4 = p
    where x and y is an positive integer and p is a prime.

    Does this equation have
    (a) no solutions,
    (b) one solution,
    (c) many solutions, or
    (d) infinitely many solutions?

    If possible, show the a, b and p in the solution(s).
    perhaps a silly question... but 3 pairs of variables? n-m x-y or a-b? i assume you mean to have the question be consistent in choosing one of these pairings (i.e. n^3 + 4*m^4 ... where n and m are pos int.... find n and m)?

    in terms of finding the solution, i have to think for a bit 🙂
  3. Standard memberPalynka
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    05 Nov '08 18:221 edit
    One solution: (m,n,p) = (1,1,5)

    There aren't any other ones. Now prove me wrong. 😛
  4. Joined
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    05 Nov '08 20:253 edits
    Couldn't (-1, -1, 5) be correct?

    -1^4=1, 1+(4*1)=5

    Oh, and 😛
  5. Joined
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    05 Nov '08 22:08
    Originally posted by ketch90
    Couldn't (-1, -1, 5) be correct?

    -1^4=1, 1+(4*1)=5

    Oh, and 😛
    No, -1 is negative integer, not positive integer.

    And right - misprint. Should be a^4 + 4*b^4 = p.

    Okay, we have one solution. So the answer of (a) is no, and proven. One solution is found.
    But are there others? Yes? Give examples! No? Prove it!
    Are there infinitely many? Prove it!
  6. Joined
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    06 Nov '08 00:51
    Any other solution will require either a or b be divisible by 5, and a be odd.

    The reason being any number not divisible by 5 will have a 4th power of the form 5k+1, so if you add 5 of these together (4 b^4s and a a^4a), the answer will be divisible by 5.

    So here's what I have so far as being necessary for any solution.

    * a is odd, otherwise p is divisible by 4.
    * a and b are relatively prime, obviously..
    * either a or b is divisible by 5, save for the singular counterexample of a=1, b=1
  7. Subscribersonhouse
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    06 Nov '08 10:221 edit
    Originally posted by Palynka
    One solution: (m,n,p) = (1,1,5)

    There aren't any other ones. Now prove me wrong. 😛
    Why isn't P=257 if m and n is 1? 1^4+ 4*1^4=257
    Oh, I think the equation should have been written
    X^4+ 4*(Y^4)
    That makes it =5. Otherwise its 257
    I took it to be 1^4 + (4*1)^4 =P
  8. Standard memberPalynka
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    06 Nov '08 12:23
    Originally posted by sonhouse
    Why isn't P=257 if m and n is 1? 1^4+ 4*1^4=257
    Oh, I think the equation should have been written
    X^4+ 4*(Y^4)
    That makes it =5. Otherwise its 257
    I took it to be 1^4 + (4*1)^4 =P
    PEMDAS.
  9. Standard memberChronicLeaky
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    06 Nov '08 21:081 edit
    I'd keep the things geepamoogle mentioned in mind and then try factorising b^4 over the ring of integers in Q[sqrt(p)], ie the ring D = Z[(1+sqrt(p))/2]. This gives, where d = (1+sqrt(p))/2:

    4b^4 = (2d -a^2 -1)(2d + a^2 - 1)

    Now let (a,b) be an integer solution for some given prime p and consider the ideals H = (2d - a^2 -1) and I = (2d + a^2 -1) in the ring where we live. Then there's some business about checking the ideal class group of Q[sqrt(p)]. You ultimately get some other equation whose integral solutions give rise to the integral solutions of the equation you want. I forget the details right now, but I can refresh my memory later perhaps.

    EDIT On second thought, it isn't at all easy to see what the ideal class group will look like in general, so this probably isn't the approach to take.
  10. Joined
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    07 Nov '08 10:221 edit
    Hint:
    See if a factorization of a^4 + 4*b^4 gives something...

    Did 4b^4 = (2d -a^2 -1)(2d + a^2 - 1), where d = (1+sqrt(p))/2, give something? I'd rather try a^4 + 4*b^4 and care about p later...

    There can be, of course, several lines to the solution. I have one of them, others I cannot perhaps even understand...
  11. Standard memberChronicLeaky
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    07 Nov '08 13:071 edit
    Oh. Now I feel silly for talking about class groups, because:

    (a^4 + 4b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

    If this is to be prime, then the smaller of the two factors must be 1. Now either a = 1 = b or:

    a^2 - 2ab + 2b^2 = (a-b)^2 + b^2 = 1, so the above is the only solution.
  12. Standard memberChronicLeaky
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    07 Nov '08 13:11
    Originally posted by Palynka
    PEMDAS.
    Palynka Expresses Mathematical Doctrine Acronym-Style?
  13. Standard memberPalynka
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    07 Nov '08 14:11
    Originally posted by ChronicLeaky
    Palynka Expresses Mathematical Doctrine Acronym-Style?
    Palynka Exhibits Mortified Dermis in Avatar Snip
  14. Joined
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    07 Nov '08 14:31
    Originally posted by ChronicLeaky
    Oh. Now I feel silly for talking about class groups, because:

    (a^4 + 4b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

    If this is to be prime, then the smaller of the two factors must be 1. Now either a = 1 = b or:

    a^2 - 2ab + 2b^2 = (a-b)^2 + b^2 = 1, so the above is the only solution.
    Nice done!
  15. Joined
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    07 Nov '08 15:00
    I didn't think an expression like a^4 + 4*b^4 would factor into 2 parts so nicely.

    Nice job..

    Just as a small note to an above post, standard order of operations dictates that all powers operations (squares, cubes, etc) are evaluated before multiplication and division.

    http://en.wikipedia.org/wiki/Order_of_operations

    The order, as I understand it.

    1) Expressions inside parentheses
    2) Exponents/Roots
    3) Multiplcation/Division
    4) Addition/Subtraction
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