1,1

For twin primes and k=1, p(n)*p(n+1)+k is always a square. This follows from the fact that for any number x, x(x+2) + 1 = x^2+2x+1 = (x+1)^2. Since twin primes differ by 2, the product of twin primes + 1 is a square.

Note that the product of the special case of the first 2 consecutive primes 2 and 3 will produce infinitely many squares. 6+3 = 9, 6+10 = 16. 6+k = y^2 or k=y^2 - 6 for y > 4. This leaves us the cases for p(n) > 2 to prove the instances of k such that p(n)*p(n+1) + k != y^2.

Case k=2: Let x = p(n) and x+2m = p(n+1) since the next prime is a multiple of 2 away from the current prime. Now assume x^2+2mx + 2 = y^2.

Completing the square and rearranging terms, we have x^2 + 2mx + m^2 = y^2 -2 + m^2 or (x+m)^2 = y^2 - 2 + m^2 = z^2. Then y^2-z^2 = 2 - m^2. So m=1 is the only possibility.

This gives y^2-z^2 = 1 or y-z= and y+z=1, impossible.

This contradicts the assumption x^2+2mx+2 = y^2 so there are no consecutive primes such that p(n)*p(n+1)+k = y^2.

Case 5: Using the arguments for Case 2, c. so m = 1,2 are the only ppossibilities and y^2-z^2 = 4 or y^2-z^2 = 1 have no integer solutions.

Case 7: y^-z^2 = 7 - m^2. m = 1,2. y^2-z^2 = 6 has no integer solutions. For y^2-z^2 = 3 we have y-z = 1 y+z = 3 y = 2, z=1. Then x^2-2xm+7 = y^2 becomes x^2-2x+3 = 0 which has no integer solution.

Let us consider a working case for k = 14. y^-z^2 = 14 - m^2. m = 1,2,3. For m=1 y-z = 1 y+z = 13 y = 7 Then substituting m,y into x^2 + 2mx + 14 = y^2 we get x^2+2x + 14 = 49. Completing the square we get (x+1)^2 = 49-14+1 = 36 and x=5. So 5*7+14 = 49. I do not see a general proof for all cases that p(n)*p(n+1) + k != y^2.

(PARI) primesq2(n) = {local(x); for(x=1, n, if(primesq(10000, x)==0, print1(x", ") ) ) } primesq(n, m) = { local(c, k, x, p1, p2, j); c=0; for(k=m, m, for(x=1, n, p1=prime(x); p2=(prime(x+1)); y=p1*p2+k; if(issquare(y), c++; \ print1(k", "); break; ) ) ); c; }

Adjacent sequences: A129813 A129814 A129815 this_sequence A129817 A129818 A129819

Sequence in context: A028726 A047269 A039027 this_sequence A137708 A122806 A122546

easy,nonn,uned

Cino Hilliard (hillcino368(AT)hotmail.com), May 20 2007

There is probably no proof that this sequence is correct. - njas, May 24 2007