Sushma

@OldJohn, thank you nice of you, Bye will in next chat

@OldJohn, have a good day(night), thanks

@OldJohn yes really

no, it is a joy, gothrough once

I also read a paper by Lenstra and stevenhagen on"Mersenne primes and artin reciprocity", which is nice,

I didn't finish the full book; the last chapter is still remaining, of complex multiplication, rest I finished, but initially I didn't understand, now slowly knowing that subject

yes I read that book, this is the side effect of that);

last year, I had run the command in nzmath to get a representation of a prime no. with125 digits, but this is not a huge number I do not know why this time it is taking such a long time

no, when I was without any answer I got your answer, so will wait

why pity?

yes I think so

ok, I can install Gap and check

or a command exists

in Gap

ok, so should I code it

yes sir, don't take trouble please

i was using nzmath, for python developed in tokyo metropolin university, for these type of problem,

ok, thanks

I meant for 21

as 442263555, 97162880914

yes, for that I got

yes

ok I will check

ok, quite possible, but that prime is cong. to 1 mod 35, so I thought, a soln might exist

Sir, I tried the same code in pari/gp 2.5.5 but for n=35, it didn't give

hi

Frob_{7} (2+\rt{2}) mod(-1+2\rt{2}) is identity, hence the Artin symbol is trivial is right?

hi, finally 1 question:

ok

ok @fretty nice of you to give some exposure, on Artin map, Frobenius elements, would you mind if I ask some doubts in future

not really all, but some from discussions

thank you very much,

then

multiply u^{6} to 2^3, subtract -1

hence the Artin symbol of ((2+\sqrt{2})^{7}-1)/1+sqrt{2}is trivial right

so, we need to evaluate not substitute for sqrt{2} is it so

so

can I substiute for $\sqrt{2}$ as 4 or 3, while evaluating mod 7

ok

yes

I had many doubts cleared, thanks

thank you very much for all the help you gave thanks for your time too

in fact was working out

sorry to disturb you, this was the point at which I was doubting, whether to substitute 4 or 3 for \sqrt{2}, to simplify the calculations, and reprsent the numbers in the form $x^2+7y^2$

tbh, in the Numerator (\sqrt{2}) it is an associate of $2+\sqrt{2}$ in the earlier example

say (2+\sqrt{2})^{7}-1)/(1+\sqrt{2}) mod 7 case, can I substitute sqrt{2}=4 and proceed?

or should we evaluate in some case

is it obvious from a=b mod r and s|r, so a=b mod r?

yes right

ok, good