Math Conjectures

Beal Conjecture

OPENMajorConjectureProposed 1993 · Full conjecture

Canonical statement

If positive integers A,B,C,x,y,zA,B,C,x,y,z satisfy
Ax+By=Cz,x,y,z>2, A^x+B^y=C^z,\qquad x,y,z>2,
then gcd⁡(A,B,C)>1\gcd(A,B,C)>1.
View source LaTeX
If positive integers \(A,B,C,x,y,z\) satisfy
\[
  A^x+B^y=C^z,\qquad x,y,z>2,
\] then \(\gcd(A,B,C)>1\).
NT-DIO-012:[BealPrizeOfficial][Mauldin1997Beal][DarmonGranville1995]
Number theoryAlgebraic geometry

Notes

The conjecture asserts that whenever positive integers satisfy Ax+By=CzA^x+B^y=C^z with all three exponents greater than 22, the bases must share a common prime factor, i.e. gcd⁡(A,B,C)>1\gcd(A,B,C)>1. It was proposed in 1993 by Andrew Beal, who attached a monetary prize to it; the problem and prize were described by Mauldin [Mauldin1997Beal], and the prize is now administered by the American Mathematical Society [BealPrizeOfficial]. Since the case x=y=zx=y=z with coprime bases is Fermat's Last Theorem, the conjecture is a natural strengthening of that statement to mixed exponents.

Progress is organised by exponent triple. Darmon and Granville proved that for each fixed triple (x,y,z)(x,y,z) with 1/x+1/y+1/z<11/x+1/y+1/z<1 the equation has only finitely many solutions in coprime integers [DarmonGranville1995], and many individual triples and infinite subfamilies have since been settled completely. The abcabc conjecture would imply that only finitely many primitive solutions exist in total, but even this would not by itself exclude every potential counterexample.

No argument is known that handles all exponent triples with collectively coprime bases at once, and the conjecture remains open.

References (3)

The boxed statement is the canonical open formulation — not a stronger variant or a related research program. The status reflects the catalog's last review; do your own literature search before investing serious effort.