What is the 11th set of six numbers that satisfy a^m + b^n = c^k?

Board 8

Board 8 » What is the 11th set of six numbers that satisfy a^m + b^n = c^k?
foolm0r0n
01/08/2024
12:42 PM
#1
I'm trying to get my phd, really need help here
_foolmo_
he says listen to my story this maybe are last chance
Fiop
01/08/2024
12:46 PM
#2
a=0
b=0
c=0
m,n, and k = some positive integers

That's more than 11 sets of 6 numbers to satisfy that.
"so is my word...It will not return to me empty, but will accomplish what I desire and achieve the purpose for which I sent it." - Isaiah 55:11
foolm0r0n
01/08/2024
12:49 PM
#3
Fiop posted...
a=0
b=0
c=0
m,n, and k = some positive integers

That's more than 11 sets of 6 numbers to satisfy that.
Forgot to mention (1/m + 1/n + 1/k) needs to be less than 1. Also a, b, c need to be coprime.
_foolmo_
he says listen to my story this maybe are last chance
Sceptilesolar
01/08/2024
01:07 PM
#4
What does '11th set' mean? In what ordering?

For instance, (1, 2, 3, x, 3, 2) is a valid solution to this for any x, so is the 11th set the 11th possible x? I guess that would be 13, since m=1 and m=2 would violate the other condition. Am I wrong there?

Edit: Er no I guess m=7 would be the first valid m where 1/m + 1/2 + 1/3 < 1. So 17?
Just killing time until the world ends.
Xeybozn
01/08/2024
01:09 PM
#5
a=1
b=2
c=3
m=17
n=3
k=2

Pfft, easy.
Congrats to 2020 GotD Guru champ azuarc!
FoolFantastic
01/08/2024
01:10 PM
#6
I'm assuming none of the numbers can be the same, and that has two 2s and two 3s
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foolm0r0n
01/08/2024
01:17 PM
#7
Sceptilesolar posted...
What does '11th set' mean? In what ordering?
Ascending order of the value of c^k

FoolFantastic posted...
I'm assuming none of the numbers can be the same, and that has two 2s and two 3s
m,n,k can be the same as long as they satisfy the inequality

But a,b,c need to be coprime, and 1 is not coprime with 2 and 3
_foolmo_
he says listen to my story this maybe are last chance
Sceptilesolar
01/08/2024
01:21 PM
#8
Of course 1 is coprime with 2 and 3. There may be a missing stipulation though.
Just killing time until the world ends.
foolm0r0n
01/08/2024
02:57 PM
#9
Oh I thought 1 was an exception, but I guess it counts.

And I misread the current solutions, but apparently 1,2,3 has infinite solutions, so I should've excluded that too.
https://en.wikipedia.org/wiki/Fermat%E2%80%93Catalan_conjecture#Known_solutions

So I'm looking for the 10th set of numbers with all those rules and also excluding that first infinite equation.
_foolmo_
he says listen to my story this maybe are last chance
Sceptilesolar
01/08/2024
03:07 PM
#10
Even if you had another solution, that seems less like a Ph.D and more the result of sufficiently dedicated and powerful computing. I don't think anyone is getting 30042907^2 in their head.
Just killing time until the world ends.
foolm0r0n
01/08/2024
03:09 PM
#11
I would think so too, but apparently it's hard enough to compute that these are the only ones they have found. Probably needs a new computation trick.
_foolmo_
he says listen to my story this maybe are last chance
Board 8 » What is the 11th set of six numbers that satisfy a^m + b^n = c^k?