Difference between prime and square of prime

Lallouet philip.lallouet at orange.fr
Wed Mar 19 14:43:31 CET 2008


11,13,31,17,19,37,23,41,43,29,31,73,0,37,79,41,43,61,47,89,67,53,71,73,59,61,79,0,67,109,71,73, 907,0,79,97,83,101,103,457,107,109,127,113,131,157,0,137,139,0,127,241,131,

149,151,137,139,157,0,257,163,149,151,193,0,157,199,0,163,181,167,281,211,173,191,193,179,151,199,0,227,229,191,193,211,197,199,241,0,317,223

 

In this sequence, a(n) is the least prime <10^6 such that a(n)-2*n is the square of any other prime. If no 

 

prime < 10^6 satisfying the condition has been found, a(n) has been given the value 0.

 

Example  :  a(1)=11 as 11-2*1=3^2

 

I wonder if this sequence is of some interest for publishing in  EOIS.



But I conjecture that for the terms of the index for which no solution has been found, no solution exists.

 

In other words I conjecture that among the infinity of prime numbers there is no pair (p,q>p^2) such

 

that  (q-p^2 ) = 26, or, 56, 68, 86,110,116 134 ,146,152,176 ,194 ,200,206,212,218, , , 

 

Note that all these numbers are of the type 6*k-4

 

If someone may prove that and define a formula, the sequence of the  concerned terms k should be 

 

certainly interesting ( 5,10,12,15,19,20,23 ,25,26,30,33,34,35,36,40, , , , ) 

 

Moreover  this sequence seems to be a duplicate of A059324 (6*p+5 = composite) , remark which 

 

may lead to the proof  that I am unable to find.

 

I wish your help and should be happy and grateful if this proof is found.

 

Best regards

Philippe LALLOUET

 

 
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