SIMPLE PROOF OF BEALíS CONJECTURE
(THE $100 000 PRIZE ANSWER)
Bealís 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.
..............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 areÖ.
( 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 Bealís conjecture equation have to be greater than 2.
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 after numbers have been substituted into it. The equation can be rewritten as
the 1st product + the 2nd product = the 3rd product .
This holds, for x, y and z greater than 2, if and only if the left hand side of the equation can be factorized, i.e. rewritten in the form P(Q + R) where P,Q and R are positive integers. This will therefore guarantee that A, B and C share a common prime factor. The reason why this statement doesnít always hold for the Pythagorean equation A^2 + B^2 = C^2 is due to the fact that A^2 + B^2, i.e. the left hand side of the equation, canít be written in the form P(Q+R) where P,Q and R are positive integers. An example of an expression that could be written in the form P(Q+R), even without actual numbers substituted into the algebraic expression, is A^3 + B^3 = (A+B)(A^2-AB+B^2). One could wonder what argument stops A^3 + B^3 = (A+B)(A^2-AB+B^2) = C^z when A and B have no common prime factors. The simple argument is that even though (A+B)(A^2-AB+B^2) and C^z are products, one is a positive integer raised to a positive power i.e. C^z and the other is something else other than that. The only way that (A+B)(A^2-AB+B^2) stands a chance of being equal to C^z is if and only if A and B share a common prime factor which will give (A+B)(A^2-AB+B^2) a chance to become a positive integer raised to a positive integer power.
One could also raise the argument as to why equations of the form
10^5 + 41^3 = 411^2 or 25^2 + 6^3= 29^2
do not conform with the Beal equation. The answer lies in the fact that these equations are simply derived from the right hand side of the Pythagorean triples equations 105^2 + 88^2 = 137^2 and 21^2 + 20^2 = 29^2 respectively which cannot be factorized.
Note that 411^2 = (3x137)^2.