mersenneforum.org Beal's conjecture ..........not
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2022-04-10, 16:22   #1
Awojobi

Feb 2019

2·47 Posts
Beal's conjecture ..........not

PROOF OF BEAL'S CONJECTURE
Attached Files
 PROOF OF BEAL.pdf (157.4 KB, 47 views)

 2022-04-10, 16:56 #2 Dobri   "Καλός" May 2018 5278 Posts It is stated in the article that "... every term in each of the 2 brackets must be integers and not irrational numbers." However, please note that the product of two irrational numbers can be an integer.
 2022-04-10, 18:01 #3 Awojobi   Feb 2019 5E16 Posts The fact that the product of 2 irrational numbers can produce a rational number is not too relevant in my argument because I am talking about numerical values here. Take for instance the numerical value of square root of 2. This numerical value can be written to any desired number of decimal places which will not be the exact value. Hence, the numerical value of square root of 2 is only an approximation.
 2022-04-10, 22:37 #4 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 112·13 Posts Recommending 6 (six) more problems what you could easily attack: https://en.wikipedia.org/wiki/Millennium_Prize_Problems all of them worth 1 million bucks (USD).
2022-04-11, 00:33   #6
charybdis

Apr 2020

80410 Posts

Very impressive - your proof still works if you set y and z equal to 1 or 2. That means you've managed to prove the Generalized Beal Conjecture:

Quote:
 Generalized Beal Conjecture If A^x + B = C, where A, B, C and x are positive integers with x > 2, then A, B and C have a common factor > 1.

2022-04-11, 02:44   #7
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

22×5×269 Posts

Quote:
 Originally Posted by charybdis Very impressive - your proof still works if you set y and z equal to 1 or 2. That means you've managed to prove the Generalized Beal Conjecture:
Cool! So, as a numerical example, A = 2, B = 5 C = 13 , x = 3 . 2 ^ 3 + 5 = 13, so {2,5,13} share a common factor bigger than 1.
Neat!

 2022-04-11, 02:56 #8 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 72×73 Posts If we allow one of the three exponents be 2 (but of course cannot be 1), the other two exponents must be >= 3, do there exist infinitely many solutions other than 2^3+1^n=3^2?
2022-04-11, 03:10   #9
chalsall
If I May

"Chris Halsall"
Sep 2002

3·3,529 Posts

Quote:
 Originally Posted by sweety439 If we allow one of the three exponents be 2 (but of course cannot be 1), the other two exponents must be >= 3, do there exist infinitely many solutions other than 2^3+1^n=3^2?
Dude... Why do you pollute other's threads? Is it not enough to simply talk to yourself?

2022-04-11, 03:14   #10
chalsall
If I May

"Chris Halsall"
Sep 2002

1058710 Posts

Quote:
 Originally Posted by VBCurtis Cool! ... Neat!
Is there actually something there?

It would be an excellent example of an outside idea being interjected if there was. I'm (clearly) unqualified to do anything but stupidly ask.

But... I would welcome an answer from those who /actually/ know how this stuff works.

Rather than the usual noise...

2022-04-11, 05:26   #11
axn

Jun 2003

5,387 Posts

Quote:
 Originally Posted by chalsall Is there actually something there?
No, that was sarcasm. The thread is where it belongs - in Misc. Math.

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