Anonymous at Sat, 7 Sep 2024 22:20:53 UTC No. 16366694

>>16366613
I have proof there are no solutions but the proof is too long to post on 4chan

Anonymous at Sat, 7 Sep 2024 22:46:18 UTC No. 16366744

>>16366694
Is this your last theorem?

Anonymous at Sun, 8 Sep 2024 00:58:31 UTC No. 16367054

>>16366613
p=14(mod15)
q=13(mod15)
That's as far as I got

Anonymous at Sun, 8 Sep 2024 22:39:49 UTC No. 16368838

>>16367054
Forgot my reasoning:
>Apply Fermat's little theorem (mod3)
>Apply Fermat's little theorem (mod5)
>Combine congruences

Anonymous at Sun, 8 Sep 2024 23:07:51 UTC No. 16368877

>>16368838
Multiply by p^3q, and apply Euler's theorem mod 7
>q=p^3(1-q) mod 7
(p,q)=(1,4),(2,4),(4,4) (mod 7)
So q = 52 mod 105,
p=14,28,56 mod 105

Anonymous at Mon, 9 Sep 2024 16:02:07 UTC No. 16369892

First, let's find the cubic residue of p^3 and q^5 in mod 9:
0,1,8 and 0,1,2,4,5,7,8 respectively
We know 76 is 4 mod 9, so let's find what x1 - x2 we can choose for the result to be 4: we get 0 - 4 and 8 - 5 (p^3 - q^5) in mod 9
Let's do the process again in mod 11:
0,1,10 and ALL 10 residues mod 11
possible combinations to get 10 mod 11 (76 mod 11): 0 - 1, 1 - 2 and 10 - 0

So we have 6 possibility for congurency mod 99:
choose one of the two mod 9, and one of the three mod 11. Let's use the chinese remainder theorem on every possibility: 0 mod 9 and 0 mod 11 = 0 mod 99, 4mod 9 and 1 mod 11 we get 67 mod 99. The new equation is 0 - 67 mod 99 congurent to 76 mod 99 [x]. Ill skip the details and show the 6 possibilities:
0 - 67 = 9 =/= 76
45 - 13 =/= 76
54 - 22 =/= 76
44 - 23 =/= 76
89 - 68 =/= 76
98 - 23 =/= 76

We can't ever get to a clean difference of 76 following the cubic and quintic residues. So no solutions.

Anonymous at Mon, 9 Sep 2024 16:03:15 UTC No. 16369894

>>16369892
Cubic and quintic residue*

Anonymous at Mon, 9 Sep 2024 16:09:57 UTC No. 16369902

>>16369892
Oops huge mistake the correct x1-x2 is 0 - 5 not 0 -4 this DOES give us a correct answer when used with 0 - 1 from the mod 11 case which is
0 - 23 = 76 BUT since p is the power of a prime it cant ever be a multiple of 99 (0 mod 99) so the result still stands