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X^p + y^p = z^p

  1. Apr 24, 2004 #1
    X^2 :

    3^2 = 2*(1+2)+3

    4^2 = 2*(1+2+3)+4

    5^2 = 2*(1+2+3+4)+5

    6^2 = 2*(1+2+3+4+5)+6

    X^3 :

    3^3 = 3*(1*2 + 2*3)+3

    4^3 = 3*(1*2+2*3+3*4)+4

    5^3 = 3*(1*2+2*3+3*4+4*5)+5

    6^3 = 3*(1*2+2*3+3*4+4*5+5*6)+6
     
  2. jcsd
  3. Apr 25, 2004 #2

    matt grime

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    you keep writing these as if there is something deep going on. A little elementary algebra should show you it's not.
     
  4. Apr 25, 2004 #3

    honestrosewater

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    Russell,

    for x and n in N, you can write the sequence x^n using partial sums and n!s. (n factorials)

    For n=1, think sigma[y+1], y is nonegative integer
    For n=2, think sigma[2y+1]
    For n=3, think sigma[6([y^2+y]/2)+1]
    ...

    You can see the connection by expanding (x+1)^n.
    Now look at n! and the partial sums of (1, 1, 1...), i.e. (1, 2, 3...) (1, 3, 6...) (1, 4, 10...)...

    I didn't see the connection at first either :)

    Happy thougths
    Rachel
     
  5. Apr 26, 2004 #4
    :eek: :eek: :eek:


    It's called brainstorming. Everything does not necessarily start out with ...deep thoughts.


    3^2 = (1+1)*(1+2) + 3

    4^2 = (1+1)*(1+2+3) + 4

    5^2 = (1+1)*(1+2+3+4) + 5

    [...]

    3^2 = (1+1)*(1+2) +3

    3^3 = (1+2+3)*(1+3) + 3

    3^5 = (1+2+3)*(1+2+3+4)*(1+3) + 3

    3^7 = (1+2+3+4+5+6+7+8+9+10+11+12+13)*(1+2+3)*(1+3) +3
     
  6. Apr 26, 2004 #5
    5^2 = 4*5/2 + 5*6/2

    13^2 = 12*13/2 + 13*14/2

    17^2 = 16*17/2 + 17*18/2

    25^2 = 24*25/2 + 25*26/2

    5^2 = 10+15

    13^2 = 78+91

    17^2 = 136+153

    25^2 = 300+325

    1+2+3+...+N = N*[N+1]/2

    X^2 = X*[X-1]/2 + X*[X+1]/2

    X^3 = X*[X^2-1]/2 + X*[X^2+1]/2

    X^4 = X*[X^3-1]/2 + X*[X^3+1]/2

    X^n = X*[X^(n-1) -1]/2 + X*[X^(n-1) + 1]/2
     
  7. Apr 26, 2004 #6

    matt grime

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    But why do you feel the need to just write them out repeatedly without explanation?
     
  8. May 12, 2004 #7
    5*2-1 = 3^2

    5*3+1 = 4^2

    [5*2-1] + [5*3+1] = 5*2 + 5*3 = 5*[2+3] = 5^2


    [13*2-1] + [13*11+1] = 5^2 + 12^2 = 13^2

    [17*4 - 4] + [17*13 + 4] = 17^2


    Z*U + K = X^2

    Z*V - K = Y^2



    [Z*U + K] + [Z*V - K] = Z*U + Z*V = Z*[U + V] = Z*Z = Z^2

    = X^2 + Y^2


    Z = U + V
     
  9. May 16, 2004 #8
    The General Equation?

    3*(3+4) + 4*|4-3| = 5^2

    3*(3+4) + 4*1 = 5^2

    3*(3+4) + 4*(1 + 5^2) = 5^3

    3*(3+4) + 4*(1 + 5^2 + 5^3) = 5^4

    3*(3+4) + 4*(1 + 5^2 + 5^3 + 5^4) = 5^5

    etc...


    5*(5+12) + 12*|12 - 5| = 13^2

    5*(5+12) + 12*(7 + 13^2) = 13^3

    5*(5+12) + 12*(7 + 13^2 + 13^3) = 13^4

    etc...

    The equation? :

    p is a prime number > 2.

    z^p = x*(x+y) + y*( |y-x| +...+ z^(p-1) )
     
  10. Jun 5, 2004 #9
    More random thoughts:


    a^n + b^n = c^n

    [a+b] > c

    [a+b] - d = c


    [a+b] = [c+d]


    [a+b]^2 = [c+d]^2


    a^2 + 2ab + b^2 = c^2 + 2cd + d^2

    iff

    a^2 + b^2 = c^2

    then

    2cd + d^2 = 2ab

    [...]

    [a+b]^3 = [c+d]^3

    a^3 + 3ba^2 + 3ab^2 + b^3 = c^3 + 3dc^2 + 3cd^2 + d^3


    iff

    a^3 + b^3 = c^3


    then

    3ba^2 + 3ab^2 = 3dc^2 + 3cd^2 + d^3

    ba^2 + ab^2 = dc^2 + cd^2 + [d^3]/3
     
  11. Jun 27, 2004 #10
    Fermat Algebra

    a^3 + b^3 = (a+c)^3


    a^3 + b^3 = (b+d)^3


    (a+c) = (b+d)


    a^3 = (b+d)^3 - b^3


    b^3 = (a+c)^3 - a^3


    a^3 + b^3 = (b+d)^3 - b^3 + (a+c)^3 - a^3


    a^3 + b^3 = 3ca^2 + 3ac^2 + 3db^2 + 3bd^2


    a^3 + b^3 = 3ac(a+c) + 3bd(b+d)


    a+c = b+d


    a^3 + b^3 = 3ac(a+c) + 3bd(a+c)


    a^3 + b^3 = 3*(a+c)*(ac+bd)


    (a+c)*(ac+bd) must be a certain multiple of 3 in order for a^3 + b^3 to be a "cube"


    3*9 = 3^3


    3*72 = 6^3


    3*243 = 9^3


    3*576 = 12^3


    Interesting...
     
    Last edited: Jun 27, 2004
  12. Jun 27, 2004 #11

    Hurkyl

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    Sheesh, the arithmetic's even wrong. (what happened to d^3 and c^3?)
     
  13. Jun 27, 2004 #12

    Thanks for your excellent help Hurkyl

    a^3 + b^3 = (a+c)^3

    a^3 + b^3 = (b+d)^3

    a^3 + b^3 = a^3 + 3ac^2 + 3ca^2 + c^3

    a^3 + b^3 = b^3 + 3bd^2 + 3db^2 + d^3


    b^3 = 3ac^2 + 3ca^2 + c^3

    a^3 = 3db^2 + 3bd^2 + d^3

    a^3 - d^3 = 3db^2 + 3bd^2

    b^3 - c^3 = 3ca^2 + 3ac^2


    same principle:


    3*9 = 3^3


    3*72 = 6^3


    3*243 = 9^3


    3*576 = 12^3


    Interesting...
     
  14. Jun 27, 2004 #13
    The abc conjecture:

    http://www.math.unicaen.fr/~nitaj/abc.html#Consequences

    Asymptotic Fermat using L'Hopital's rule:

    B > A

    A^x + B^x

    [A^x + B^x]^[1/x]

    L'Hopital's Rule:

    Limit f(x)/g(x) = Limit f'(x)/g'(x)

    Take the natural log

    Ln[A^x + B^x]^[1/x] = Ln[A^x + B^x]/x


    = f(x)/g(x)


    L'Hopital's Rule...

    Limit Ln[A^x + B^x]/x =

    [(A^x)*Ln[A] + (B^x)*Ln]/[A^x + B^x] / 1

    = LnA/[1 + [B/A]^x] + LnB/[1+[A/B]^x]

    Take the limit

    = 0 + Ln

    e^LnB = B

    Limit

    for B > A

    [A^x + B^x]^[1/x] = B
     
  15. Aug 20, 2004 #14
    Interesting...

    x+y = A

    x-y = B

    [A+B]/2 = x

    [A-B]/2 = y


    x^2 + y^2 = [A^2 + B^2]/2

    x^3 + y^3 = [A^3 + 3AB^2]/4

    x^5 + y^5 = [A^5 + 10A^3 B^2 +5AB^4]/16



    So [A^3 + 3AB^2]/4 cannot be a cube...

    It is a cube if A = B but x and y are integers greater than zero so A cannot equal B...
     
  16. Aug 21, 2004 #15
    [itex]x+y = A[/itex]

    [itex]x-y = B[/itex]

    [itex][A+B]/2 = x[/itex]

    [itex][A-B]/2 = y[/itex]


    [itex]x^2+y^2 = [A^2 + B^2]/2[/itex]

    [itex]x^3+y^3 = [A^3 + 3AB^2]/4[/itex]

    [itex]x^4+y^4 = [A^4 + 6A^2 B^2 + B^4]/8[/itex]

    [itex]x^5+y^5 = [A^5 + 10A^3 B^2 + 5AB^4]/16[/itex]

    [itex]x^6+y^6 = [A^6 + 15A^4 B^2 + 15A^2 B^4 + B^6]/32[/itex]

    [itex]x^7+y^7 = [A^7 + 21A^5 B^2 + 35A^3 B^4 + 7AB^6]/64[/itex]

    etc...
     
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