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First of quadruple of consecutive primes p1,p2,p3,p4 such that the congruence p2^x - p1^x == p3 (mod p4) has no solution.
1

%I #15 Mar 18 2018 06:27:39

%S 3,13,53,59,61,71,73,97,109,127,137,149,151,179,197,239,241,277,283,

%T 293,311,313,389,401,419,431,433,439,457,463,467,491,499,503,541,547,

%U 557,563,569,577,601,619,641,643,653,673,743,769,773,797,853,881,887,907,911,919,929,971,991,1021,1031

%N First of quadruple of consecutive primes p1,p2,p3,p4 such that the congruence p2^x - p1^x == p3 (mod p4) has no solution.

%H Robert Israel, <a href="/A082376/b082376.txt">Table of n, a(n) for n = 1..10000</a>

%e For the prime quadruple 3,5,7,11, 5^x-3^x == 7 (mod 11) has no solutions.

%p Res:= NULL: count:= 0:

%p p1:= 2: p2:= 3: p3:= 5: p4:= 7:

%p while count < 100 do

%p found:= false;

%p for x from 1 to p4-2 do

%p if p2 &^ x - p1 &^ x - p3 mod p4 = 0 then found:= true; break fi

%p od:

%p if not found then Res:= Res, p1; count:= count+1 fi;

%p p1:= p2: p2:= p3: p3:= p4: p4:= nextprime(p4);

%p od:

%p Res; # _Robert Israel_, Mar 18 2018

%Y Cf. A082371.

%K easy,nonn

%O 1,1

%A _Cino Hilliard_, May 11 2003

%E Name clarified by _Robert Israel_, Mar 18 2018