Bernhard Helmes wrote:

> you did not used a solution of pell for a=3 with x^2-ay^2=1

There is an infinity of solutions to x^2-3*y^2 = 1,

with integers x and y.

Here are a few:

q=quadunit(12);r=1;for(k=1,20,r*=q;print([k,real(r),imag(r)]));

[1, 2, 1]

[2, 7, 4]

[3, 26, 15]

[4, 97, 56]

[5, 362, 209]

[6, 1351, 780]

[7, 5042, 2911]

[8, 18817, 10864]

[9, 70226, 40545]

[10, 262087, 151316]

[11, 978122, 564719]

[12, 3650401, 2107560]

[13, 13623482, 7865521]

[14, 50843527, 29354524]

[15, 189750626, 109552575]

[16, 708158977, 408855776]

[17, 2642885282, 1525870529]

[18, 9863382151, 5694626340]

[19, 36810643322, 21252634831]

[20, 137379191137, 79315912984]

How do you intend to prove that one pair of Pell's /infinite/

series does not correspond to my choice, modulo p?

That might take you some time :-?

David