Beal's conjecture ..........not

[charybdis]

Originally Posted by sweety439

If we allow one of the three exponents be 2 (but of course cannot be 1), the other two exponents must be >= 3, do there exist infinitely many solutions other than 2^3+1^n=3^2?

No, by the Generalized Beal Conjecture as proved by Awojobi. Did you not read what I wrote??

The real answer is "yes". But you'll have to figure out for yourself why that is.

[Dr Sardonicus]

Originally Posted by sweety439

If we allow one of the three exponents be 2 (but of course cannot be 1), the other two exponents must be >= 3, do there exist infinitely many solutions other than 2^3+1^n=3^2?

There are infinitely many solutions to A^3 + B^3 = C^2 in positive integers A, B, C with gcd(A,B) == 1.

There are two types of solutions: those in which A + B and A^2 - A*B + B^2 are each 3 times a perfect square (e.g. 1^3 + 2^3 = 3^2) , and those with each a perfect square (e.g. 56^3 + 65^3 = 671^2).

I leave it as an exercise to work out formulas for each case.

[sweety439]

Does there exist a positive integer N, such that if the sum of the three exponents is >=N (and none of the three exponents is 1, and at most one of the three exponents is 2), then there exist only finitely many solutions other than 2^3+1^n=3^2? If so, find the smallest such positive integer N

(I also think that 2^3+1^n=3^2 will be the only solution if N is enough large)

[sweety439]

Originally Posted by Dr Sardonicus

There are infinitely many solutions to A^3 + B^3 = C^2 in positive integers A, B, C with gcd(A,B) == 1.

There are two types of solutions: those in which A + B and A^2 - A*B + B^2 are each 3 times a perfect square (e.g. 1^3 + 2^3 = 3^2) , and those with each a perfect square (e.g. 56^3 + 65^3 = 671^2).

I leave it as an exercise to work out formulas for each case.

I found that sequence is OEIS: A099426, also numbers n such that n^2 is in A202679