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MrAwojobi
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SIMPLE PROOF OF BEAL’S CONJECTURE
(THE $100 000 PRIZE ANSWER)
Beal’s Conjecture
Beal’s conjecture states that if A^x + B^y = C^z where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
Examples
......Common Prime Factor
2^3 + 2^3 = 2^4 => 2
2^9 + 8^3 = 4^5 => 2
3^3 + 6^3 = 3^5 => 3
3^9 + 54^3 = 3^11 => 3
27^4 + 162^3 = 9^7 => 3
7^6 + 7^7 = 98^3 => 7
33^5 + 66^5 = 33^6 => 11
34^5 + 51^4 = 85^4 => 17
19^4 + 38^3 = 57^3 => 19
Primitive Pythagorean Triples
A primitive Pythagorean triple is one in which the integer lengths of the right angled triangle do not have a common prime factor. Examples are….
( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
Hence the reason why x, y and z in Beal’s conjecture equation have to be greater than 2. For the purpose of the proof of Beal’s conjecture, the author wants to stress right from the onset that the equation of Pythagoras theorem is an entirely different type of equation to Beal’s equation even though they look similar. It should therefore not be surprising that it has to be exempt from Beal’s equation. The difference between these 2 equations is simply due to the fact that A^2 + B^2 = C^2 can be rewritten as A^2 = (C+B)(C-B), i.e. A^2 is a product of 2 numbers i.e. it can be factorised. A^x + B^y = C^z cannot be factorised hence the big difference between the 2 equations.
Simple Proof
It should be clear that each term in the equation A^x + B^y = C^z can be broken down into the product of its prime factors. Thus the equation could be rewritten as
abcde + fghij = klmno for instance, where a,b,c,d,e,f,g,h,i,j,k,l,m,n,o are prime . It isn’t difficult to see that the only way for
the 1st product + the 2nd product = the 3rd product
is if and only if the left hand side of the equation can be factorised. This will therefore guarantee that A,B and C share a common prime factor.
(THE $100 000 PRIZE ANSWER)
Beal’s Conjecture
Beal’s conjecture states that if A^x + B^y = C^z where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
Examples
......Common Prime Factor
2^3 + 2^3 = 2^4 => 2
2^9 + 8^3 = 4^5 => 2
3^3 + 6^3 = 3^5 => 3
3^9 + 54^3 = 3^11 => 3
27^4 + 162^3 = 9^7 => 3
7^6 + 7^7 = 98^3 => 7
33^5 + 66^5 = 33^6 => 11
34^5 + 51^4 = 85^4 => 17
19^4 + 38^3 = 57^3 => 19
Primitive Pythagorean Triples
A primitive Pythagorean triple is one in which the integer lengths of the right angled triangle do not have a common prime factor. Examples are….
( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
Hence the reason why x, y and z in Beal’s conjecture equation have to be greater than 2. For the purpose of the proof of Beal’s conjecture, the author wants to stress right from the onset that the equation of Pythagoras theorem is an entirely different type of equation to Beal’s equation even though they look similar. It should therefore not be surprising that it has to be exempt from Beal’s equation. The difference between these 2 equations is simply due to the fact that A^2 + B^2 = C^2 can be rewritten as A^2 = (C+B)(C-B), i.e. A^2 is a product of 2 numbers i.e. it can be factorised. A^x + B^y = C^z cannot be factorised hence the big difference between the 2 equations.
Simple Proof
It should be clear that each term in the equation A^x + B^y = C^z can be broken down into the product of its prime factors. Thus the equation could be rewritten as
abcde + fghij = klmno for instance, where a,b,c,d,e,f,g,h,i,j,k,l,m,n,o are prime . It isn’t difficult to see that the only way for
the 1st product + the 2nd product = the 3rd product
is if and only if the left hand side of the equation can be factorised. This will therefore guarantee that A,B and C share a common prime factor.