# Simple Proof of Beal's Conjecture

MrAwojobi
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. 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.

JCVD
Your statements about factorization are vague and mostly untrue. For instance, for any positive a,b,c and any integer n>1, a^n+b^n=c^n can be written as a^n=c^n-b^n=(c-b)(c^(n-1)+c^(n-2)b+...+cb^(n-1)+b^(n-1))

The logic you employ in the "proof" doesn't seem to use anything specific about exponents being greater than 2; what I'm trying to say is, if your argument were valid, it would say "since neither 2 nor 3 is divisible by 5, 2+3 cannot be divisible by 5."

MrAwojobi
Hi viwers

I think I have to concede that my proof is not water tight based on an issue.
Since a^3 + b^3 and a^3 - b^3 also factorise in a unique way just like Pythagoras theorem equation factorises in its own unique way, then counter examples to Beal's conjecture could be searched for based on these. However, if one can show that the above equations will still yield to the Beal's conjecture then I think my proof should still stand unless maybe there are other unique factorisations of A^x + B^y or C^z - B^y.

MrAwojobi
Hi all, I have slightly altered my proof and I actually believe it is correct. Please before criticising it, read it carefully first.

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.

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.

JCVD
If you are given the equation A^x+B^y=C^z, and P is any prime dividing C, then yes, P divides A^x+B^y, as you state, but that does not automatically imply that P divides both A and B; once again, consider the case x=y=z=1 and A=2,B=3,C=5. Nowhere in your proof do you use the fact that x,y,z>2; this should raise a red flag since the statement is not true without that assumption.

Another red flag: note that Beal's Conjecture generalizes Fermat's Last Theorem, which was proved in over 100 pages 358 years after it was first conjectured. Odds are any proof of Beal's Conjecture not using Fermat's Last Theorem will be longer than a paragraph.

MrAwojobi
Hi all, I have added an equation to the revised proof to show why the Pythagorean triples
don't 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.
The triples don't obey Beal's equation because they factorise in a unique way i.e.
A^2+B^2=C^2 can be rewritten as 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.

JCVD
This is not a proof. It contains several errors/gaps in logic, which when removed/corrected, leave it, at best, a restatement of the conjecture.

MrAwojobi
JCVD

This is not a criticism. It contains no highlights of the so called several errors/gaps in logic, which when removed/corrected, leave it, at best, a restatement of the conjecture.