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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
Attached Images Beal_x^m-y^m=z^n.gif (1.6 KB, 2 views) BealProof.gif (1.5 KB, 0 views)

Last edited by M_B_S; December 8th, 2014 at 12:03 AM. December 8th, 2014, 05:06 AM   #2
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Quote:
 Originally Posted by M_B_S 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. December 8th, 2014, 10:35 PM #3 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory 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  December 9th, 2014, 05:05 AM   #4
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Quote:
 Originally Posted by M_B_S 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). December 9th, 2014, 09:02 PM #5 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory <=> 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 = (nÂ˛-aÂ˛) 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. December 9th, 2014, 11:42 PM   #6
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Quote:
 Originally Posted by CRGreathouse 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 December 10th, 2014, 07:12 AM   #7
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Quote:
 Originally Posted by M_B_S 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 = (nÂ˛-aÂ˛) 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. December 10th, 2014, 10:57 PM   #8
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Quote:
 Originally Posted by CRGreathouse 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)Â˛-(a-b)Â˛

=> Euclids Pythagorean Triple Formula

(2ab)Â˛=(aÂ˛+bÂ˛)Â˛-(aÂ˛-bÂ˛)Â˛

=> Einsteins E=mCÂ˛

EÂ˛=pÂ˛cÂ˛+mÂ˛CÂ˛CÂ˛

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. December 11th, 2014, 06:45 AM   #9
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Quote:
 Originally Posted by M_B_S Every 4ab number is the the difference of two perfect square numbers 4ab = (a+b)Â˛-(a-b)Â˛ => Euclids Pythagorean Triple Formula (2ab)Â˛=(aÂ˛+bÂ˛)Â˛-(aÂ˛-bÂ˛)Â˛ => Einsteins E=mCÂ˛ EÂ˛=pÂ˛cÂ˛+mÂ˛CÂ˛CÂ˛
Those are great examples of non-proofs, thanks. Unless you meant
4ab = (a+b)Â˛-(a-b)Â˛
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 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. December 12th, 2014, 12:19 AM #10 Newbie   Joined: Aug 2007 Posts: 3 Thanks: 0 Beal You may like to publish your proof at the Unsolved Problems web site. Unsolved Problems Home Tags beal, conjecture, proof, special, ymzn ,

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