p^a+q^b=r^c
How many distinct solutions of the equation above are there, subject to the following constraints:
p, q, & r distinct primes
a, b, & c distinct positive integers,
each
more than one
None of the powers exceeds 1111.
We can start by comparing Ady's version with the Fermat-Catalan conjecture.
Fermat-Catalan:
(1) p,q,r need only be relatively prime, not prime numbers.
(2) a,b,c must satisfy 1/a+1/b+1/c<1; in effect the smallest power must be at least 2 and if another power is then 3, then the third cannot be less than 7, since 1/2+1/3+1/6=1.
In Ady's version, to meet the primality requirement, exactly one of p,q,r must be a power of 2 (by o+o=e, o+e=o), and exactly one of a,b,c must be 2, because any Ady version solution with a,b,c >=3 would be a counter-example to the related Beal Conjecture, none of which have been found, despite extensive research.
In fact, all of the known Fermat-Catalan solutions contain at least one second power, and all of the known prime Fermat-Catalan solutions contain at least one power of 2. These are:
1^{4}+2^{3}=3^{2} 1 is not prime
2^{5}+7^{2}=3^{4} OK
7^{3}+13^{2}=2^{9} OK
2^{7}+17^{3}=71^{2} OK
3^{5}+11^{4}=122^{2} 122 is not prime
33^{8}+1549034^{2}=15613^{3} no primality
1414^{3}+2213459^{2}=65^{7} no primality
9262^{3}+15312283^{2}=113^{7} only 113 is prime
17^{7}+76271^{3}=21063928^{2} only 17 is prime
43^{8}+96222^{3}=30042907^{2} only 43 is prime
So the solutions to the Ady version - up to the limits of current computation - are 2^5+7^2=3^4, 7^3+13^2=2^9, and 2^7+17^3=71^2.
KS is probably right about the power limit, but it doesn't really matter one way or the other.
Edited on September 25, 2023, 7:56 am
|
|
Posted by broll
on 2023-09-24 07:31:37 |
Know thy conjectures
