Beal’s Conjecture – a generalized Fermat’s Last Theorem.

Banker Andrew Beal offers $1m for maths solution.

**BEAL’S CONJECTURE**:

If

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.

http://metro.co.uk/2013/06/05/texas-banker-offering-1m-for-answer-to-30-year-old-maths-problem-3829105/

### Like this:

Like Loading...

This is an excellent post on Beal’s Conjecture.

I have a really stupid question.

Based on the criteria:

If

A^{x} + B^{y} = C^{z}

A) where A, B, C, x, y and z are positive integers and

B) x, y and z are all greater than 2,

then A, B and C must have a common prime factor.

Why can’t it just be disproved as such:

2^3 + 2^3 ≠ 2^3

(or any combination where A=B=C && x=y=z)

?

Clearly, it can’t be that simple, but what criteria am I missing? Is it just implied that well duh, you can’t have A=B=C AND x=y=z ?

In your examples, it’s okay to have A=B=C,

and for x=y, and y=z

but not x=y=z.

Wait …

that has to do with Fermat’s Theorem being proven that

A^n + B^n ≠ C^n for integers greater than 2.

Never mind.

It’s late. I haven’t studied math since high school.

It is a million dollar question…