# Simple Proof of Beal's Conjecture

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• Aug 9th 2010, 12:46 PM
MrAwojobi
Simple Proof of Beal's Conjecture
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.
• Aug 9th 2010, 01:29 PM
Bacterius
You haven't proved anything, you have merely stated the conjecture in a less mathematical way.
• Aug 9th 2010, 01:38 PM
MrAwojobi
Bacterius

I bet you just can't believe how simple the proof of this \$100 000 problem is. Instead of flippantly dismissing the proof, look for valid 'holes' in my proof. If you can't find any then simply accept that I have proved the conjecture.
• Aug 9th 2010, 04:12 PM
wonderboy1953
Look at it this way. It wouldn't be a conjecture anymore if it's proven.
• Aug 9th 2010, 05:29 PM
Vlasev
Mr. Awojobi, take

2 + 3 = 5.

They don't have prime factors in common, yet the equation holds.

Also, 3^2+4^2 = 5^2, this one doesn't have prime factors in common, either.
• Aug 9th 2010, 08:11 PM
chiph588@
Quote:

Originally Posted by Vlasev
Mr. Awojobi, take

2 + 3 = 5.

They don't have prime factors in common, yet the equation holds.

Also, 3^2+4^2 = 5^2, this one doesn't have prime factors in common, either.

$\displaystyle x,y,z>2$

Quote:

Originally Posted by MrAwojobi
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.

This is not a proof. The flaw is that the if and only if part of your statement does not follow. Also notice you didn't use the fact that $\displaystyle x,y,z>2$, so one could apply your proof to the example of $\displaystyle 3^2+4^2=5^2$, which is absurd.
• Aug 10th 2010, 06:17 AM
wonderboy1953
Posted before?
I have the feeling that this thread was posted before by someone.
• Aug 10th 2010, 09:20 AM
MrAwojobi
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 Queen of Hearts
http://www.mymathforum.com/images/ranks/clover_2.gif Posts: 30 Joined: Thu Aug 05, 2010 8:12 am
• Aug 10th 2010, 09:24 AM
chiph588@
Quote:

Originally Posted by MrAwojobi
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 Queen of Hearts
http://www.mymathforum.com/images/ranks/clover_2.gif Posts: 30 Joined: Thu Aug 05, 2010 8:12 am

Your proof is no where near complete. Have you addressed the issues I pointed out?
• Aug 10th 2010, 07:39 PM
Bacterius
Quote:

Instead of flippantly dismissing the proof, look for valid 'holes' in my proof. If you can't find any then simply accept that I have proved the conjecture.
I don't think so, no ... what you seem to say is that a mathematical work is valid until pronounced invalid. Wrong ... mathematical work is invalid until pronounced valid. And reading what you have written, no, this is not a proof, this is a flawed proof at best. I'm not trying to be mean or anything, I'm just pointing out that the contents of the topic do not reach the expectations of its title.

By the way, there is perhaps a reason why a \$100.000 prize is attached to this conjecture ... maybe it is not "simple", and mind that your line of reasoning has probably been followed hundreds of times by other people who unsuccessfully tried to prove the conjecture. Anyway, my point is that a prized conjecture should not be taken lightly, and that most of the time, no simple proof is ever found for these.

But, yeah, you can still keep to your flawed proof and try to make it pass as a valid proof, but it just won't work. It's up to you.
• Aug 10th 2010, 07:45 PM
Vlasev
• Aug 11th 2010, 10:32 AM
elemental
Quote:

Originally Posted by MrAwojobi
SIMPLE PROOF OF BEAL’S CONJECTURE
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.

Your if and only if statement makes sense (not really, but it's not completely wrong), but your immediate conclusion afterwards does not. The left side can indeed be factorized (as the prime factorization of the right side), but this does not guarantee that A, B, and C share a common prime factor. It only guarantees that A^X + B^Y shares a common factor with C^Z, which is inherently obvious. Realize that the prime factorization of a sum is not necessarily inclusive of a prime factor of each term.
• Aug 11th 2010, 01:21 PM
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.
• Aug 11th 2010, 02:28 PM
Defunkt
Quote:

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.
Your 'proof' essentially comes down to this line.
Now tell me, if I apply the same logic with x,y,z=2, is there any reason why it wouldn't work?

Also, what elemental said. And you didn't change anything if I'm not mistaken - you just added a few letters...
• Aug 11th 2010, 03:04 PM
MrAwojobi
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.
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