Browse Groups

• ... Using a non-sieving script, I have gone up to a=1.8*10^10, for all 4 starting p, and have found no 10/12 s, just your 9/12 and 2 new ones, viz.
Message 1 of 143 , Dec 4 4:52 AM
View Source
>
> Kevin Acres <research@> wrote:
>
> > Seems very difficult to get a 8/12
>
> I found it rather easy to get a 8/12:
>
> p=7;for(k=1,8,print(factor(p));p=96328365*p-674298544);
> Mat([7, 1])
> Mat([11, 1])
> Mat([385313471, 1])
> Mat([37116615999606371, 1])
> Mat([3575382933574921687714871, 1])
> Mat([344410792240175811180613435317371, 1])
> Mat([33176528504850823203577211921154930229871, 1])
> Mat([3195840747248174368104654975623363340731873292371, 1])
>
> and not much harder to get a 9/12:
>
> p=3;for(k=1,9,print(factor(p));p=237862725*p-713588164);
> Mat([3, 1])
> Mat([11, 1])
> Mat([1902901811, 1])
> Mat([452629409458306811, 1])
> Mat([107663664748893631236931811, 1])
> Mat([25609172680658279861161720490056811, 1])
> Mat([6091467598816933241588548501453247755681811, 1])
> Mat([1448933082303802516987335475350335971266609083806811, 1])
> Mat([344647171299431744552466406656001243790997338184040724431811, 1])
>
> but then it did indeed start to get tough.

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

Mike
• ... 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
Message 143 of 143 , Jan 7, 2011
View Source
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: