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December 7th, 2014, 11:43 PM  #1 
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Beal Conjecture Special Proof for x^m+y^m=z^n
Let a,b,n out of n , n>3 x:= (a^2b^2) y:= (b^2a^2) Z:= ((a^2b^2)/4) (a^2b^2)^(2*n1)  (b^2a^2)^(2*n1)= ((a^2b^2)/4)^(2*n) <=> ((a^2b^2)^(2 n)+(b^2a^2)^(2 n))/((ab) (a+b)) = 4^(2 n) (a^2b^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 Last edited by M_B_S; December 8th, 2014 at 12:03 AM. 
December 8th, 2014, 05:06 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
December 8th, 2014, 10:35 PM  #3 
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  x:= (a^2b^2) y:= (b^2a^2) Z:= ((a^2b^2)/(2^n)) (a^2b^2)^(2*nc)  (b^2a^2)^(2*nc) = ((a^2b^2)/(2^n))^(2*n) **************** This is a proof that Beal is correct in this special case 
December 9th, 2014, 05:05 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
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 = (na) prime number odd Z = (nÂ²aÂ²) no prime (n+a)(na) 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  
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  Quote:
or (x+y)^(2*n)+(x^2y^2)^(2*n) = ((x^2y^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  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 
December 10th, 2014, 10:57 PM  #8 
Senior Member Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory  You mean elegant proofs like this one: Every 4ab number is the the difference of two perfect square numbers 4ab = (a+b)Â²(ab)Â² => Euclids Pythagorean Triple Formula (2ab)Â²=(aÂ²+bÂ²)Â²(aÂ²bÂ²)Â² => Einsteins E=mCÂ² EÂ²=pÂ²cÂ²+mÂ²CÂ²CÂ² More elegant X^n+Y^n = Z^n has no anti Beal solution with X Y Z primes (b/a+a)^n (+)or( ) (b/aa)^n = 2^n Last edited by M_B_S; December 10th, 2014 at 11:26 PM. 
December 11th, 2014, 06:45 AM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
4ab = (a+b)Â²(ab)Â² 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. Another good example of a nonproof. 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 

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