 mersenneforum.org Beal's conjecture ..........not
 Register FAQ Search Today's Posts Mark Forums Read  2022-04-11, 14:49   #12
charybdis

Apr 2020

23×3×31 Posts Quote:
 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.

Last fiddled with by charybdis on 2022-04-11 at 14:51 Reason: oops   2022-04-11, 15:32   #13
Dr Sardonicus

Feb 2017
Nowhere

26×7×13 Posts Quote:
 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.   2022-04-12, 04:40 #14 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·1,723 Posts 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) Last fiddled with by sweety439 on 2022-04-12 at 04:42   2022-04-12, 04:56   #15
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

1101011101102 Posts Quote:
 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   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post didgogns Miscellaneous Math 1 2020-08-05 06:51 Awojobi Miscellaneous Math 50 2019-11-02 16:50 Arxenar Miscellaneous Math 1 2013-09-07 09:59 Joshua2 Math 54 2009-10-19 02:21 Joshua2 Open Projects 0 2009-04-20 06:58

All times are UTC. The time now is 05:42.

Sun Jun 26 05:42:30 UTC 2022 up 73 days, 3:43, 1 user, load averages: 1.33, 1.10, 0.99