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  1. 1

    Tình sầu vạn ý

    Chẳng biết nói sao với các công thức của bác. Ví dụ cho bác thấy nhé
    CÔNG THỨC 1

    ζ(s) = α^2 + β^2 = 1

    ζ(s) = 1 & ζ(1) = 0

    Nghĩa là ζ(s) = 1 = 0 ?!!!

    Rồi ở đoạn dưới
    ζ(s) = r^n + s^n = 1
    ζ(s) = c & ζ(c) = 0
    Nghĩa là ζ(s) = 1 = 0 ?!!!

  2. 2

    trantancuong

    z^3=[z(z+1)/2]^2-[(z-m-a-1)(z-m-a)/2]^2 – [z(z-1)/2]^3+[(z-m-a)(z-m-a+1)/2]^2 – (z-m-a)^3.
    (a ) from 1 to 2,3,4….endless
    z^n =
    z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 – [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 – (z-m-1)^3}
    =
    z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 – [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 – (z-m-2)^3}
    ….endless
    Impossible (z,x,y) are integer in (z^n=x^n+y^n)

  3. 3

    trantancuong

    z^n=z^(n-3)*{[z(z+1)/2]^2-[(z-m-a-1)(z-m-a)/2]^2 – [z(z-1)/2]^3+[(z-m-a)(z-m-a+1)/2]^2 – (z-m-a)^3}
    The number (m ) is integer and the number (a ) move from 1 to 2,3,4….endless
    Therefore
    z is the integer.
    However the values of x,y and z are not the integers when (a) move from 1 to 2,3,4….endless in the equation (z^n=x^n+y^n)

  4. 4

    bichngoc

    Special case:
    z^4=/x^4+y^4.
    Named
    Z=z^2 X=x^2 and Y=y^2.
    So
    z^2=a^2+b^2.
    x^2=a^2-b^2.
    y^2=2ab.
    Because
    z^2=a^2+b^2.
    So
    z=c^2+d^2.
    a=c^2-d^2.
    b=2cd.
    Because
    x^2=a^2-b^2
    Because a is the hypotenuse.
    x and b are not the hypotenuses.
    So
    a=c^2-d^2
    And
    b=2cd
    So
    x^2=(c^2-d^2)^2-(2cd)^2
    Named
    e=c^2,f=d^2
    So
    x^2=e^2+f^2-6ef.
    Because
    [n(3^1/2)e-(3^1/2)/nf]^2=3n^2e^2+3/n^2f^2-6ef.
    =3n^2e^2-e^2+3/n^2f^2-f^2+(e^2+f^2-6ef)
    =3n^2e^2-e^2+3/n^2f^2-f^2+x^2.
    So
    x^2=[n(3^1/2e-(3^1/2)/nf]^2-[3n^2e^2-e^2+3/n^2f^2-f^2].
    Because
    [n(3^1/2e-(3^1/2)/nf]^2 is an irrational number.
    And
    [3n^2e^2-e^2+3/n^2f^2-f^2] is a rational number.
    So
    x is an irrational number.
    So
    z^4=/x^4+y^4

  5. 5

    trantancuong

    Dear friends.

    Object and strategy.
    Exist an equation in which is any integer d. If Z ^ n = X ^ n + Y ^ n This will break the properties of the equation.

    Define function F.
    F(Z)=Z(Z+1)/2.
    F^2(Z)=1+2^3+..+(Z-1)^3+Z^3.
    Z=4.
    (4*5/2)^2=10^2=1+2^3+3^3+4^3=1+8+27+64=100.

    F[ Z^(2^a) ]= 2 F^2 { Z^[ 2^(a-1) ] } – Z^{ 3[ 2^(a-1) ] } .
    Z=2 and a=3.
    F[ Z^(2^a) ].=32896.
    2 F^2 { Z^[ 2^(a-1) ] } – Z^{ 3[ 2^(a-1) ] }=36992. – 4096=32896.

    Define
    F^2[ Z^(2^a) ] is the function G of F^2 { Z^[ 2^(a-1) ] } and Z^{ 3[ 2^(a-1) ]
    Then
    F{ Z^[ 2^(a-1) ] } =2 F^2 { Z^[ 2^(a-2) ] } – Z^{ 3[ 2^(a-2) ] } .
    F^2{ Z^[ 2^(a-1) ] } is the function G of F^2 { Z^[ 2^(a-2) ] } and Z^{ 3[ 2^(a-2) ]
    Continue same
    First equation.
    F^2{ Z^[2^(a-b)] }=1+2^3+..+{ Z^[2^(a-b)] -1 }^3+{ Z^[2^(a-b)] }^3=the function G of F^2 { Z^[ 2^(a-b-1) ] } and Z^{ 3[ 2^(a-b-1) ] }.= the function G of F^2 { Z^[ 2^(a-b-2) ] } and Z^{ 3[ 2^(a-b-2) ] }. and Z^{ 3[ 2^(a-b-1) ] }.= ……..=the function G of F^2 { Z^[ 2^(a-b-c) ] } and Z^{ 3[ 2^(a-b-c) ] }.and Z^{ 3[ 2^(a-b-c+1) ] }.and Z^{ 3[ 2^(a-b-c+2) ] } and ……..and Z^{ 3[ 2^(a–b-2) ] }.and Z^{ 3[ 2^(a-b-1) ] }.
    This system consists of (c +1) polynomials for each equation
    and there are (b +1) equations.

    Because;
    F{ Z^[ 2^(a+1) ] }= 2F^2 [ Z^(2^a) ] – Z^[3(2^a)]
    So
    F^2 [ Z^(2^a) ] = the function of F{ Z^[ 2^(a+1) ] } and Z^[3(2^a)]
    So
    F^(2^d) [ Z^(2^a) ] =the function H of F^[2^(d-1){ Z^[ 2^(a+1) ] } and Z^[3(2^a)]
    Continue same
    Second equation:
    F^(2^d) [ Z^(2^a) ]= the function of F^[2^(d-d+1) { Z^[2^(a+d-1) ] } and ….and …. and Z^[3(2^a)].

    Combined first & second equations.
    F^(2^d) [ Z^(2^a) ]= the function of F^[2^(d-d+1) { Z^[2^(a+d-1) ] } and ….and …. and Z^[3(2^a)]
    And
    F^[2^(d-d+1) { Z^[2^(a+d-1) ] }=the function G of F^2 { Z^[ 2^(a+d-c) ] } and Z^{ 3[ 2^(a+d-c) ] }.and Z^{ 3[ 2^(a+d-c+1) ] }.and Z^{ 3[ 2^(a+d-c+2) ] } and ……..and Z^{ 3[ 2^(a+d-3) ] }.and Z^{ 3[ 2^(a+d-2) ] }.
    Creating third equation.

    Special properties of third equation: is always contain the sum of consecutive cube and each polynomial in every equation contain Z^n
    The properties of third equation is always absolute with only integer Z ^ n. However, if Z ^ n = X ^ n + Y ^ n . We substitute (X ^ n + y ^ n) instead of Z ^ n into the third equation.:,Two integers X and Y can not keeping this properties with d is an integer of any, so there exists at least an integer d which make third equation: is wrong. So ,X or Y are not integers.

    So
    Z^n No=X^n+Y^n.

    Attention! Pythagore.
    3^2+4^2=5^2.
    So
    (3^2+4^2)^n=(5^2)^n.
    Because the number n is an arbitrary number should have infinity equations. However, they have no special principle such as always exist the sum of consecutive integers cube in each polynomial of third equation in my system.

    Easily.
    Third equation contain:
    (F^2)^(d-1) [ Z^(2^a) ] and F^2 { Z^[2^(a+d-1) ] } which are the sum of consecutive integers cube
    a^3=F^2(a)-F^2(a-1)
    So changing (F^2)^(d-1) [ Z^(2^a) ] and F^2 { Z^[2^(a+d-1) ] } become the sum of integers square.
    Because all the rest are multiples of Z ^ 3.which is the sum of integers square.
    So
    Creating the series of equal sum of integers square
    Replace X^n+Y^n for Z^n .
    There exists at least an integer d for X or Y are not the integers which is the cause Z^n No=X^n+Y^n.
    Imagin:
    1+2^2+4^2+X^n+Y^n=3^2+5^2+X^n+Y^n=5^2+4*7^2+X^2+Y^2=9^2+7*4^2+X^n+Y^n+..+..=…limited polynomials of degree n depends on unspecified integer d
    Unacceptable X and Y are integers.

    ISHTAR.

  6. 6

    trantancuong

    Anh Ran than men .
    Bay gio toi la ban cua Ong Bill Gates nho vao bai giai nay.

    Object and strategy.
    Exist an equation in which is any integer d. If Z ^ n = X ^ n + Y ^ n This will break the properties of the equation.

    Define function F.
    F(Z)=Z(Z+1)/2.
    F^2(Z)=1+2^3+..+(Z-1)^3+Z^3.
    Z=4.
    (4*5/2)^2=10^2=1+2^3+3^3+4^3=1+8+27+64=100.

    F[ Z^(2^a) ]= 2 F^2 { Z^[ 2^(a-1) ] } – Z^{ 3[ 2^(a-1) ] } .
    Z=2 and a=3.
    F[ Z^(2^a) ].=32896.
    2 F^2 { Z^[ 2^(a-1) ] } – Z^{ 3[ 2^(a-1) ] }=36992. – 4096=32896.

    Define
    F^2[ Z^(2^a) ] is the function G of F^2 { Z^[ 2^(a-1) ] } and Z^{ 3[ 2^(a-1) ]
    Then
    F{ Z^[ 2^(a-1) ] } =2 F^2 { Z^[ 2^(a-2) ] } – Z^{ 3[ 2^(a-2) ] } .
    F^2{ Z^[ 2^(a-1) ] } is the function G of F^2 { Z^[ 2^(a-2) ] } and Z^{ 3[ 2^(a-2) ]
    Continue same
    First equation.
    F^2{ Z^[2^(a-b)] }=1+2^3+..+{ Z^[2^(a-b)] -1 }^3+{ Z^[2^(a-b)] }^3=the function G of F^2 { Z^[ 2^(a-b-1) ] } and Z^{ 3[ 2^(a-b-1) ] }.= the function G of F^2 { Z^[ 2^(a-b-2) ] } and Z^{ 3[ 2^(a-b-2) ] }. and Z^{ 3[ 2^(a-b-1) ] }.= ……..=the function G of F^2 { Z^[ 2^(a-b-c) ] } and Z^{ 3[ 2^(a-b-c) ] }.and Z^{ 3[ 2^(a-b-c+1) ] }.and Z^{ 3[ 2^(a-b-c+2) ] } and ……..and Z^{ 3[ 2^(a–b-2) ] }.and Z^{ 3[ 2^(a-b-1) ] }.
    This system consists of (c +1) polynomials for each equation
    and there are (b +1) equations.

    Because;
    F{ Z^[ 2^(a+1) ] }= 2F^2 [ Z^(2^a) ] – Z^[3(2^a)]
    So
    F^2 [ Z^(2^a) ] = the function of F{ Z^[ 2^(a+1) ] } and Z^[3(2^a)]
    So
    F^(2^d) [ Z^(2^a) ] =the function H of F^[2^(d-1){ Z^[ 2^(a+1) ] } and Z^[3(2^a)]
    Continue same
    Second equation:
    F^(2^d) [ Z^(2^a) ]= the function of F^[2^(d-d+1) { Z^[2^(a+d-1) ] } and ….and …. and Z^[3(2^a)].

    Combined first & second equations.
    F^(2^d) [ Z^(2^a) ]= the function of F^[2^(d-d+1) { Z^[2^(a+d-1) ] } and ….and …. and Z^[3(2^a)]
    And
    F^[2^(d-d+1) { Z^[2^(a+d-1) ] }=the function G of F^2 { Z^[ 2^(a+d-c) ] } and Z^{ 3[ 2^(a+d-c) ] }.and Z^{ 3[ 2^(a+d-c+1) ] }.and Z^{ 3[ 2^(a+d-c+2) ] } and ……..and Z^{ 3[ 2^(a+d-3) ] }.and Z^{ 3[ 2^(a+d-2) ] }.
    Creating third equation.

    Special properties of third equation: is always contain the sum of consecutive cube and each polynomial in every equation contain Z^n
    The properties of third equation is always absolute with only integer Z ^ n. However, if Z ^ n = X ^ n + Y ^ n . We substitute (X ^ n + y ^ n) instead of Z ^ n into the third equation.:,Two integers X and Y can not keeping this properties with d is an integer of any, so there exists at least an integer d which make third equation: is wrong. So ,X or Y are not integers.

    So
    Z^n No=X^n+Y^n.

    Attention! Pythagore.
    3^2+4^2=5^2.
    So
    (3^2+4^2)^n=(5^2)^n.
    Because the number n is an arbitrary number should have infinity equations. However, they have no special principle such as always exist the sum of consecutive integers cube in each polynomial of third equation in my system.

    Easily.
    Third equation contain:
    (F^2)^(d-1) [ Z^(2^a) ] and F^2 { Z^[2^(a+d-1) ] } which are the sum of consecutive integers cube
    a^3=F^2(a)-F^2(a-1)
    So changing (F^2)^(d-1) [ Z^(2^a) ] and F^2 { Z^[2^(a+d-1) ] } become the sum of integers square.
    Because all the rest are multiples of Z ^ 3.which is the sum of integers square.
    So
    Creating the series of equal sum of integers square
    Replace X^n+Y^n for Z^n .
    There exists at least an integer d for X or Y are not the integers which is the cause Z^n No=X^n+Y^n.
    Imagin:
    1+2^2+4^2+X^n+Y^n=3^2+5^2+X^n+Y^n=5^2+4*7^2+X^2+Y^2=9^2+7*4^2+X^n+Y^n+..+..=…limited polynomials of degree n depends on unspecified integer d
    Unacceptable X and Y are integers.

    ISHTAR.

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