Beal's conjecture is that for integers p,q,r all > 2, then for any solution in positive integers A,B,C of A^p + B^q = C^r, at least 2 of A,B,C have a common factor. But there seem to be plenty of examples of (*) A^p + B^q = C^r where exactly one of p,q,r is equal to 2, with the other two of them still > 2, and A,B,C being pairwise relatively prime positive integers. * * * One example is 10^2 + 3^5 = 7^3. Wikipedia mentions another such, 7^3 + 13^2 = 2^9. And of course 1^p + 2^3 = 3^2, which shows that the exponent 2 can appear on the right side of (*). I wonder several things: * Do there exist methods to find all solutions to an equation like A^2 + B^5 = C^7 ??? (Is this easier if at least one solution is known?) * What is known about the set of exponents {p,q,r} = {2,s,t} such that at least one nonzero solution exists to (*) in pairwise relatively prime A,B,C > 0 ? E.g., are there infinitely many triples {p,q,r} = {2,s,t} for which a solution to (*) exists? * Is it known that for a given set of exponents {p,q,r} = {2,s,t} there can be only a finite number of triples A,B,C satisfying (*) ? * Are there examples other than 1^p + 2^3 = 3^2 where the exponent 2 appears on the right side of (*) ? --Dan Sometimes the brain has a mind of its own.