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December 7th, 2014, 11:43 PM   #1
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Beal Conjecture Special Proof for x^m+y^m=z^n

Let a,b,n out of n , n>3

x:= (a^2-b^2)

y:= (b^2-a^2)

Z:= ((a^2-b^2)/4)



(a^2-b^2)^(2*n-1) - (b^2-a^2)^(2*n-1)= ((a^2-b^2)/4)^(2*n)

<=>

((a^2-b^2)^(2 n)+(b^2-a^2)^(2 n))/((a-b) (a+b)) = 4^(-2 n) (a^2-b^2)^(2 n)


=>
If the system have a integer solution

(a+b) = (b+a) is the common factor of x,y,z

The number of Integer solutions is 8n



M_B_S
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Last edited by M_B_S; December 8th, 2014 at 12:03 AM.
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December 8th, 2014, 05:06 AM   #2
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Quote:
Originally Posted by M_B_S View Post
x:= (a^2-b^2)

y:= (b^2-a^2)

Z:= ((a^2-b^2)/4)
Your proof is flawed from the start because of these assumptions.
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December 8th, 2014, 10:35 PM   #3
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x:= (a^2-b^2)

y:= (b^2-a^2)

Z:= ((a^2-b^2)/(2^n))



(a^2-b^2)^(2*n-c) - (b^2-a^2)^(2*n-c) = ((a^2-b^2)/(2^n))^(2*n)
****************

This is a proof that Beal is correct in this special case
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December 9th, 2014, 05:05 AM   #4
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Originally Posted by M_B_S View Post
This is a proof that Beal is correct in this special case
Oh. In that case let me offer a yet more general proof: if x and y have a common factor > 1, and $x^n+y^n=z^n,$ then Beal's conjecture holds for (x, y, z, n).
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December 9th, 2014, 09:02 PM   #5
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<=>

So if Beal is correct

X^a+Y^b=Z^c can be found in N when X;Y;Z have a common factor or a,b,c <= 2 in N

Use Goldbach Conjecture

X = (n+a) prime number odd

Y = (n-a) prime number odd

Z = (n²-a²) no prime (n+a)(n-a) odd

=>

X^a+ Y^b = Z^c

X odd +Y odd is even = Z is odd => no solution for primes only 2

=> 2 is the common factor or X and Y are not both prime

Last edited by M_B_S; December 9th, 2014 at 09:41 PM.
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December 9th, 2014, 11:42 PM   #6
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Oh. In that case let me offer a yet more general proof: if x and y have a common factor > 1, and $x^n+y^n=z^n,$ then Beal's conjecture holds for (x, y, z, n).
(x+y)^(2*n-1)+(x^2-y^2)^(2*n-1) = ((x^2-y^2)/(4))^(2*n)

or

(x+y)^(2*n)+(x^2-y^2)^(2*n) = ((x^2-y^2)/(4))^(2*n+1)

The system has only one solution for x,y in N for all n

(x+y) is the common factor
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December 10th, 2014, 07:12 AM   #7
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Quote:
Originally Posted by M_B_S View Post
So if Beal is correct

X^a+Y^b=Z^c can be found in N when X;Y;Z have a common factor or a,b,c <= 2 in N

Use Goldbach Conjecture

X = (n+a) prime number odd

Y = (n-a) prime number odd

Z = (n²-a²) no prime (n+a)(n-a) odd

=>

X^a+ Y^b = Z^c

X odd +Y odd is even = Z is odd => no solution for primes only 2

=> 2 is the common factor or X and Y are not both prime
You don't understand how proofs work, do you.
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December 10th, 2014, 10:57 PM   #8
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You don't understand how proofs work, do you.
You mean elegant proofs like this one:

Every 4ab number is the the difference of two perfect square numbers


4ab = (a+b)²-(a-b)²

=> Euclids Pythagorean Triple Formula

(2ab)²=(a²+b²)²-(a²-b²)²

=> Einsteins E=mC²

E²=p²c²+m²C²C²


More elegant X^n+Y^n = Z^n has no anti Beal solution with X Y Z primes


(b/a+a)^n (+)or( -) (b/a-a)^n = 2^n

Last edited by M_B_S; December 10th, 2014 at 11:26 PM.
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December 11th, 2014, 06:45 AM   #9
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Quote:
Originally Posted by M_B_S View Post
Every 4ab number is the the difference of two perfect square numbers


4ab = (a+b)²-(a-b)²

=> Euclids Pythagorean Triple Formula

(2ab)²=(a²+b²)²-(a²-b²)²

=> Einsteins E=mC²

E²=p²c²+m²C²C²
Those are great examples of non-proofs, thanks. Unless you meant
4ab = (a+b)²-(a-b)²
as a proof (rather than an example) -- that one is actually correct.* No actual proofs are given, just claimed implications, and no attempt is made in the last one of connecting the variables to the quantities to which they relate.

* It's incomplete -- you can represent any number not = 2 mod 4 as a difference of two squares.

Quote:
Originally Posted by M_B_S View Post
More elegant X^n+Y^n = Z^n has no anti Beal solution with X Y Z primes


(b/a+a)^n (+)or( -) (b/a-a)^n = 2^n
Another good example of a non-proof.

Last edited by CRGreathouse; December 11th, 2014 at 06:47 AM.
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December 12th, 2014, 12:19 AM   #10
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Beal

You may like to publish your proof at the Unsolved Problems web site.

Unsolved Problems Home
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