I am impressed by your all results. I think yhat everyone has won.

I continue my explanation. (see my 2 previous mails)

As a prime 4k+1 can be considered connected to a "quasi-congruent"

number d=ab/2 it is natural to ask whether there is a elliptic curve

attached y^2=x^3-d^2x.

In fact as this not really a congruent number the elliptic curve

shall be modified. For some heuristic reasons there are two families

which hit : E1 y^2=8u(x^3-d^2x)+v and E2 y^2=8u(x^3+d^2x)-v, the

parameters u and v depending of the prime p.

Now, in the particular case of primes p(d)=4d^2+1, the parameters

v=1 is good for both E1 and E2 but u for E2 is u=1 and for E1 is u=

(2e)^2 where e is a natural number depending of the continued

fraction of sqrt(p). It is too long to explain the details...

Now put x=p/2 in E1 and E2 and you get:

E1 z1=32d^4+8d^2+1

E2 z2=4d(2d^2+1)

Here is our polynomials:

p(d)=4d^2+1

q(d)=z2+1=4d(2d^2+1)+1

r(d)=z1=32d^4+8d^2+1

I have immediately remarked that for small parameters d the numbers

p,q,r are often prime together.

I don't know exactly why but now you know that it is connected to:

- congruent numbers

- continued fractions

- elliptic curves with rational solutions

It requests further deep investigations to undersand this phenomena.

You can find an more detailed overview (in french) on my website

http://perso.wanadoo.fr/hh-mouvement.com/
Note number 4

Enjoy primes