3^8-18^3=3^6

3^14-162^3=3^12

104^3-91^3=13^5

1352^3-1183^3=13^8

??? Flt is solvable in Gaussian numbers.

(a+√b)^3-(-a+√b)3=c^3

(3+√93)^3-(-3+√93)^3=12^3

Summation of [(n+1)^3-n^3]=(c+1)^3

n=0 to c to upper limit.

Regarding the Beal conjecture itself, a search in which the exponents are all kept really low (<20) seems more interesting – the density of such powers is so much higher than that of high powers that you’d be doing most of your existing search, and you could take the bases up to very large numbers in compensation.

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