Beal's conjecture ..........not

[Awojobi]

PROOF OF BEAL'S CONJECTURE

[Dobri]

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.

[Awojobi]

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.

[R. Gerbicz]

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).

[Awojobi]

I know about them already. Restrict your comments to my proof.

[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:

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.

[VBCurtis]

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!

[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?

[chalsall]

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?

[chalsall]

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...

[axn]

Originally Posted by chalsall

Is there actually something there?

No, that was sarcasm. The thread is where it belongs - in Misc. Math.