- --- In primenumbers@yahoogroups.com,

"mikeoakes2" <mikeoakes2@...> wrote:

> Using a non-sieving script, I have gone up to a=1.8*10^10,

I tested all with p[3] < 10^11, finding

> for all 4 starting p, and have found no 10/12's, just your

> 9/12 and 2 new ones, viz.

> [237862725*x - 713588164, [3, 9]]

> [4939393834*x - 9878787657, [2, 9]]

> [7484387043*x - 52390709290, [7, 9]]

7/12: 2336

8/12: 60

9/12: 3 (as above)

Apologies for setting a puzzle that now seems far too hard :-(

David - --- In primenumbers@yahoogroups.com,

Kevin Acres <research@...> wrote:

> x=a*x+b either is a square or has a square as a major factor.

Suppose that we want to start with a square and get a square.

Then we must solve the Diophantine equation

y^2 = a*x^2 + b

For any pair (a,b), Dario will tell us all the solutions:

http://www.alpertron.com.ar/QUAD.HTM

David