Prove this conjecture and win $100,000

(or give a counterexample)

If A^x + B^y = C^z where A,B,C,x,y,z are positive integers and x, y, z are all greater than 2, then A, B, and C must have a common prime factor.
This is the Beal conjecture.

http://www.math.unt.edu/~mauldin/beal.html

Originally posted by smaia
(or give a counterexample)

If A^x + B^y = C^z where A,B,C,x,y,z are positive integers and x, y, z are all greater than 2, then A, B, and C must have a common prime factor.
This is the Beal conjecture.

http://www.math.unt.edu/~mauldin/beal.html

27^4+162^3=9^7

common prime is 1

Originally posted by uzless
27^4+162^3=9^7

common prime is 1

Sadly, the common prime is 3 🙁

Originally posted by uzless
27^4+162^3=9^7

common prime is 1

Is 1 really a prime?

Originally posted by FabianFnas
Is 1 really a prime?

Nope.

Originally posted by AThousandYoung
Nope.

So what does "common prime is 1" really mean?

Originally posted by FabianFnas
So what does "common prime is 1" really mean?

Nothing. The statement should be

"common prime is <prime number>"

or

"there is no common prime"

Originally posted by TheMaster37
Nothing. The statement should be

"common prime is <prime number>"

or

"there is no common prime"

Okay. Thanks.