From: tpiezas@uap.edu.ph (Tito Piezas III)
Subject: The j-function and primes
Date: 8 Jan 2001 10:42:48 -0500
Newsgroups: sci.math
Summary: Prime values of quadratic polynomials and the j-function

Hello all,

Maybe someone can give an explanation for the question at the bottom
of this post.

First, the form:
       n^2 + n + p
is prime for n={0, (p-2)} IFF the field Q{sqrt(1-4p)} has class number
1.  Thus, p can ONLY be the index m of the last six Heegner numbers:
7,11,19,43,67,& 163, which have form 4m-1.

The form:
      2n^2 + p
is prime for n={0, (p-1)} IFF the field Q{sqrt(-2p)} has class number
2, permitting ONLY the four primes p = 3,5,11,& 29.

Second, the j-function j(q) is an algebraic or rational number, even
an INTEGER, at certain special values of q.  Let j(q)=
j(-e^-pi(sqrt(H))), where H are the six aforementioned Heegner
numbers.

          H        j(q)
          7        -15^3
         11        -32^3
         19        -96^3
         43       -960^3
         67      -5280^3
        163    -640320^3  

The prime factorization of -{j(q)}^1/3 is then

        15 = 3x5
        32 = 2^5
        96 = (2^5)x3
       960 = (2^6)x3x5
      5280 = (2^5)x3x5x11
    640320 = (2^6)x3x5x23x29

The largest ODD prime factors for each even j(q) are 3,5,11,&29,
respectively. (The set looks familiar?)

Question:  Why are the largest odd prime factors of the even 
j-function for numbers of class number 1, be the same four primes
(3,5,11,&29) of the field Q{sqrt(-2p)} with class number 2?  Is it
coincidence, or is there a reason?

Sincerely,

Tito Piezas III