Revised Simple Proof of Beal's Conjecture

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MrAwojobi
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Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Wed Aug 11, 2010 9:39 pm UTC

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.

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Re: Revised Simple Proof of Beal's Conjecture

Postby heyitsguay » Wed Aug 11, 2010 10:09 pm UTC

Very good, you fixed the flaw, that $100,000 should be yours in no time. The next step you should take is to email math professors at several universities. Explain to them what you've done exactly as you explained it to us. Pick your favorite university, go to its website, and find its math faculty. You should probably email all of them just to be sure you get a response.

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Re: Revised Simple Proof of Beal's Conjecture

Postby Tirian » Wed Aug 11, 2010 10:18 pm UTC

MrAwojobi wrote:Please before criticising it, read it carefully first.


Why on earth should we be expected to do that? You clearly post this crap before reading it carefully first.

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Re: Revised Simple Proof of Beal's Conjecture

Postby jestingrabbit » Wed Aug 11, 2010 10:39 pm UTC

MrAwojobi wrote: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.


I can't see that at all. Can you make it clearer? That is, can you explain why you think this is true.
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Re: Revised Simple Proof of Beal's Conjecture

Postby jaap » Wed Aug 11, 2010 10:41 pm UTC

MrAwojobi wrote: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.


Where in this simple proof have you used the fact that x,y,z need to be greater than 2?
If it is not difficult to see there must be a common factor, how come 32 + 24 = 52 even though 3,2,5 have no common factor?

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Re: Revised Simple Proof of Beal's Conjecture

Postby Talith » Wed Aug 11, 2010 10:49 pm UTC

Not even going to bother this time.

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Re: Revised Simple Proof of Beal's Conjecture

Postby Syrin » Wed Aug 11, 2010 11:20 pm UTC

MrAwojobi wrote:I actually believe it is correct.


Not the best of things with which to preface your proof.

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Re: Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Wed Aug 11, 2010 11:52 pm UTC

Hi again my critics. I have added an equation to my revised proof just to clarify why the Pythagorean triples are the exception to the rule. I hope gmalivuk doesn't close me down again because it seems you guys and girls, if any, are loving this. The way the hits keep coming is amazing.


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 are the exception to the rule because of their unique factorisation 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.
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.

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Re: Revised Simple Proof of Beal's Conjecture

Postby squareroot » Wed Aug 11, 2010 11:56 pm UTC

Syrin wrote:
MrAwojobi wrote:I actually believe it is correct.


Not the best of things with which to preface your proof.


I lol'ed at that. X-D

MrAwojobi, please... this proof is barely any different from the first one you posted. Although I'm pretty sure this thread will eventually degrade in to people flaming you and your logic, please just accept that there are dramatic flaws in your proof. Believe me, if a proof was this short, it would have been found. In the days of Euler, or Diophantus, it's true that many people could have sat down and - with a basic knowledge of math, and some time - discover something new. Today, that is sadly not true. Go take some more math classes, or grab yourself a textbook on number theory. You'll do soo much more that way, than if you try to prove something this hard. It's like trying to learn how to walk by doing barrel rolls as fast as you can at a marathon.

So, believe us when we tell you that you're wrong. It's not an insult in any way, and you'll come out looking more graceful if you just tried to understand the flaws in your logic.

Thanks, in advance.
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Re: Revised Simple Proof of Beal's Conjecture

Postby letterX » Thu Aug 12, 2010 12:18 am UTC

MrAwojobi wrote:I hope gmalivuk doesn't close me down again because it seems you guys and girls, if any, are loving this. The way the hits keep coming is amazing.


Troll, much? Anyways, yes. I watch with the same fascination as a twenty car pileup.

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Re: Revised Simple Proof of Beal's Conjecture

Postby scarecrovv » Thu Aug 12, 2010 12:52 am UTC

MrAwojobi wrote: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.

Wonderful, now all you need to do is show that the left hand side of the equation can, in fact, be factorised.

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Re: Revised Simple Proof of Beal's Conjecture

Postby mathboy » Thu Aug 12, 2010 2:33 am UTC

MrAwojobi wrote:The triples are the exception to the rule because of their unique factorisation i.e. A^2+B^2=C^2 can be rewritten as A^2=(C+B)(C-B)

What makes that factorisation unique? Compared with that A^x+B^y = C^y can be rewritten as A^x = (C-B)(C^(y-1) + B*C^(y-2) + B^2*C^(y-3) + ... + B^(y-3)*C^2 + B^(y-2)*C + B^(y-1)).

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Re: Revised Simple Proof of Beal's Conjecture

Postby Penitent87 » Thu Aug 12, 2010 9:00 pm UTC

Talith wrote:Not even going to bother this time.



Ditto.

Clearly the OP took no notice of the advice given.

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Re: Revised Simple Proof of Beal's Conjecture

Postby njperrone » Thu Aug 12, 2010 9:38 pm UTC

MrAwojobi wrote:"It should be clear" "It isn’t difficult to see"


If you choose to revise this more, try to avoid using these phrases and instead show the connections that you imply with them. It may work to your advantage when communicating it.

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Re: Revised Simple Proof of Beal's Conjecture

Postby achan1058 » Fri Aug 13, 2010 3:16 am UTC

MrAwojobi wrote:I hope gmalivuk doesn't close me down again because it seems you guys and girls, if any, are loving this. The way the hits keep coming is amazing.
So are the people who showed up for Florence Jenkins' opera singing.

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Re: Revised Simple Proof of Beal's Conjecture

Postby supremum » Fri Aug 13, 2010 5:20 am UTC

MrAwojobi wrote: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 statement is wrong. And if it isn't wrong, please be more clear about why not. I have read this thread and the previous one containing this proof and you haven't made any effort to justify this statement at all.

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Re: Revised Simple Proof of Beal's Conjecture

Postby Luonnos » Fri Aug 13, 2010 6:27 am UTC

MrAwojobi wrote: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.

It's only wrong in one direction. If a+b=c, then the left side can always be factored into the form P(Q+R). This is true. The problem comes from the fact that if the factorization is P=1, Q=a, and R=b, then this gets you nothing, because all we've established is that 1 divides a, b, and c. If MrAwojobi tried to make this stronger by saying that P!=1, then his "proof" would just be a restatement of the conjecture (protip: this is why proofs are usually more than one sentence).

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Re: Revised Simple Proof of Beal's Conjecture

Postby DavCrav » Fri Aug 13, 2010 11:44 am UTC

Just thought I'd say, in before the lock! :twisted:

Also, this:
Luonnos wrote:It's only wrong in one direction. If a+b=c, then the left side can always be factored into the form P(Q+R). This is true. The problem comes from the fact that if the factorization is P=1, Q=a, and R=b, then this gets you nothing, because all we've established is that 1 divides a, b, and c. If MrAwojobi tried to make this stronger by saying that P!=1, then his "proof" would just be a restatement of the conjecture (protip: this is why proofs are usually more than one sentence).


Quoted because it's the point.

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Re: Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Fri Aug 13, 2010 5:16 pm UTC

Hi all, I believe I have addressed most of the issues you raised in my proof. Herein lies the latest updated version.

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.

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. 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 doesn’t 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, can’t 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.

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Re: Revised Simple Proof of Beal's Conjecture

Postby jaap » Fri Aug 13, 2010 6:20 pm UTC

MrAwojobi wrote: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.


And 29 = 132 + 73 is derived from the right hand side of which Pythagorean triple?

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Re: Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Fri Aug 13, 2010 8:42 pm UTC

Yes jaap

because 12^2 + 5^2 = 13^2

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Re: Revised Simple Proof of Beal's Conjecture

Postby Doraki » Fri Aug 13, 2010 8:43 pm UTC

It's amazing, proving this conjecture (of which the Fermat-Wiles Theorem is a simple consequence), only 2 weeks after proving the Twin Primes Conjecture !!
And in only one page.
Clearly, the thousands of bright people during all those centuries who worked on them really did a bad job at trying to prove them.

So, what will be the next conjecture you're proving ?

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Re: Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Fri Aug 13, 2010 9:15 pm UTC

Doraki

it will be Goldbach's.

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Re: Revised Simple Proof of Beal's Conjecture

Postby jaap » Fri Aug 13, 2010 9:59 pm UTC

jaap wrote:
MrAwojobi wrote: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.
And 29 = 132 + 73 is derived from the right hand side of which Pythagorean triple?

MrAwojobi wrote:Yes jaap

because 12^2 + 5^2 = 13^2


I see. What about 65 - 25 = 882 ?

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Re: Revised Simple Proof of Beal's Conjecture

Postby imatrendytotebag » Fri Aug 13, 2010 10:04 pm UTC

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.


I don't see this. Certainly A + B can be a positive integer raised to a positive integer power without A and B sharing a common prime factor, and probably so can A^2 - AB + B^2. I'm not saying that what you claimed is false, but I just don't see your reasoning.
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Re: Revised Simple Proof of Beal's Conjecture

Postby Nitrodon » Fri Aug 13, 2010 10:27 pm UTC

MrAwojobi, given that 2252 + 2722 = 3532, can you derive an expression of the form 2252 + B^y = C^z with y,z > 2 in the same manner as the triples you mentioned earlier? If you can't, please explain why your method works for the triples you mentioned earlier but not this one.

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Re: Revised Simple Proof of Beal's Conjecture

Postby MrAwojobi » Fri Aug 13, 2010 10:37 pm UTC

jaap

one of the previous examples you gave was flawed and this one is also flawed because you are deviating from Beal's equation and Pythagoras equation i.e. subtraction instead of addition.

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Re: Revised Simple Proof of Beal's Conjecture

Postby jaap » Sat Aug 14, 2010 12:36 am UTC

MrAwojobi wrote:jaap

one of the previous examples you gave was flawed and this one is also flawed because you are deviating from Beal's equation and Pythagoras equation i.e. subtraction instead of addition.

Ok, write it like this then:
25 + 882 = 65
A=2, B=88, C=6
x=5, y=2, z=5
In your 'proof' of Beal you never explain why there need not be a common factor when y=2. You only say these exceptions are somehow related to Pythagorean triples, so what triple is this derived from?

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Re: Revised Simple Proof of Beal's Conjecture

Postby njperrone » Sat Aug 14, 2010 6:15 am UTC

MrAwojobi wrote:Doraki

it will be Goldbach's.


For fun, disprove it.

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Re: Revised Simple Proof of Beal's Conjecture

Postby gmalivuk » Sat Aug 14, 2010 2:53 pm UTC

MrAwojobi wrote:Doraki

it will be Goldbach's.
Didn't you read your other Beal's Conjecture trainwreck? That one's already been proven! Too bad you're too late.

Now don't do this again or you'll lose posting access.
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