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Table with 4 consecutive primes prime(k), prime(k+1), prime(k+2) and prime(k+3) in a row such that prime(k)^x + prime(k+1)^x == prime(k+2) (mod prime(k+3)) has no solution x.
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%I #15 Dec 05 2022 04:43:43

%S 3,5,7,11,17,19,23,29,23,29,31,37,31,37,41,43,53,59,61,67,67,71,73,79,

%T 71,73,79,83,79,83,89,97,83,89,97,101,97,101,103,107,107,109,113,127,

%U 109,113,127,131,113,127,131,137,127,131,137,139,131,137,139,149,137

%N Table with 4 consecutive primes prime(k), prime(k+1), prime(k+2) and prime(k+3) in a row such that prime(k)^x + prime(k+1)^x == prime(k+2) (mod prime(k+3)) has no solution x.

%C Note overlapping primes between successive quadruples.

%C This is a rewriting of A082371 with prime(A082371(n)) building the first column in the table.

%e For prime 17, 17^x + 19^x == 23 (mod 29) has no solutions, which constitutes the 2nd row.

%e 3, 5, 7, 11;

%e 17, 19, 23, 29;

%e 23, 29, 31, 37;

%e 31, 37, 41, 43;

%e 53, 59, 61, 67;

%e 67, 71, 73, 79;

%e ...

%o (PARI) \\ No solutions to prime(i)^x+prime(i+1)^x ~= prime(i+2) mod prime(i+3)

%o noanpbn(m,n) = { for(p=1,m, f=0; for(x=0,n, if((prime(p)^x+prime(p+1)^x-prime(p+2))%prime(p+3)==0,f=1) ); if( f==0,print1(p" ")) ) }

%Y Cf. A082371.

%K easy,nonn,tabf

%O 1,1

%A _Cino Hilliard_, May 11 2003