Prove The Beal Conjecture and Get $1M

Just saw this on slashdot and thought someone here would be interested in the challenge.

According to the article, a banker in TX is offering $1M for a proof of the Beal Conjecture. From the slashdot article:
"The Beal Conjecture states that the only solutions to the equation A^x + B^y = C^z, when A, B and C are positive integers, and x, y and z are positive integers greater than two, are those in which A, B and C have a common factor."

Note that the common factor must be a prime number, therefore 1 does not count as the common factor. (Which is unfortunate, because for a few seconds I thought I had proven it true and was going to be the recipient of $1,000,000.)

The Wikipedia article states that if a counterexample exists, at least one of the terms (A, B, C, x, y, z) must be greater than 1000.

Quote: JB

if a counterexample exists, at least one of the terms (A, B, C, x, y, z) must be greater than 1000.

Seems someone tried by brute force already :)

I have what is probably a basic mathematics question:

Given the above equation (A^x +B^y = C^z) does this mean that A and B and C are unique integers?

(Likewise with x and y and z.)

Or does this mean that they can be unique, but they don't have to be?


It's easier for a layman like me to understand Beal's conjecture when it is stated this way:

There are no positive integers A, B, C, x, y, z satisfying the equation

A^x +B^y = C^z

where x, y, z > 2 and A, B, C are co-prime (that is, gcd(A, B) = gcd(A, C) = gcd(B, C) = 1).


I too had read about this offer, (which is currently making the news) and so while at work today I wrote a program in an attempt to disprove the conjecture. The program generated integers, being careful to make sure that A & B, A & C, and B & C were all co-prime. (First I had to look up what co-prime was! I didn't know, or had long forgotten if I did ever know.)

I then generated exponents and plugged everything into the formula, hoping that my generated numbers would satisfy the equation. But I was unsure if A and B and C (and x and y and z) all HAD to be unique.

JB's post and the link he gives mentions this statement:

"By computerized searching, greatly accelerated by aid of modular arithmetic, this conjecture has been verified for all values of all six variables up to 1000. So in any counterexample, at least one of the variables must be greater than 1000."

Six variables... but I'm still not sure if one (or more) of the variables can be identical. Does anyone know for sure?

The Wiki page gives an example where A, C, x, and y are the same (3³ + 6³ = 3⁵). But obviously, any example which disproves the conjecture must have A, B, and C values which are not identical.

So it does. Good work. That should answer that.

Just looking at it, my amateur mind wonders whether there might be a way to construct a family of impossible objects analogous to the Frey curve...

Quote: EdCollins

But I was unsure if A and B and C (and x and y and z) all HAD to be unique.

What do you mean by "unique" ? A number is "unique" when satisfying exactly what property ?

The conjecture in the quoted form is simple:

"A^x + B^y = C^z with A,B.C positive intergers and x,y,z positive intergers greater two => A,B,C have a common factor"

A common factor q would be: A/q, B/q, C/q are integer.

By unique I meant different. I wanted to know if the integers could be the same value, or if they all had to all be different values.

Well, since A, B, and C are pairwise coprime, obviously no two of them can be identical.

My guess would be that two of x, y, and z can be the same (although of course all three can't be the same; I have a truly marvelous proof of this, but it wouldn't fit in this parenthetical).

I have discovered a truly remarkable proof which this margin is too small to contain.

It is much easier to get the million dollars from Beal by playing poker.

If you are not familiar with him, this is a good gambling story.

The-Professor-Banker-Suicide-King

5 -10 years ago I could easily see this thread leading me off of an hours (or days) long treasure hunt. Now, I think I'll stick to something easier... like trying to solve lunar poker.

Quote: 24Bingo

Well, since A, B, and C are pairwise coprime, obviously no two of them can be identical.

My guess would be that two of x, y, and z can be the same (although of course all three can't be the same; I have a truly marvelous proof of this, but it wouldn't fit in this parenthetical).

Yea, like an idiot I realized that once I thought about it a little more. Duh.

Thanks.