MrAwojobi

Hi all, I have added an extra equation to this revised proof to show why the Pythagorean triples do not obey Beal's equation

SIMPLE PROOF OF BEAL’S CONJECTURE

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. These triples do not obey Beal's equation because of

the unique factorisation of a^2+b^2=c^2 to a^2 = (c+b)(c-b)

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 after numbers have been substituted into it. 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 1st product + the 2nd product = the 3rd product

if and only if the left hand side of the equation can be factorised, 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.