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%3 N_20200604150102b_ColinWright    By ColinWright 2020/06/04 @ 15:01:02b -------------------------------- That means that 29 (say) factors in C.        (select only this node)     N_20200604150344a_ColinWright    By ColinWright 2020/06/04 @ 15:03:44a -------------------------------- ... by which I mean Z[i], the Gaussian Integers.        (select only this node)     N_20200604150102b_ColinWright->N_20200604150344a_ColinWright N_20200604150102e_ColinWright    By ColinWright 2020/06/04 @ 15:01:02e -------------------------------- These are all clearly related, and there are connections to be had, some obvious, others likely less so.        (select only this node)     N_20200604150650a_jsiehler    By jsiehler 2020/06/04 @ 15:06:50a -------------------------------- What I don't know, or can't bring to mind, is whether "-1 is a quadratic residue mod p" and "p is a sum of two squares" are related in a direct and important way or if it's really just a coincidence that "p is congruent to 1 mod 4" characterizes both.        (select only this node)     N_20200604150102e_ColinWright->N_20200604150650a_jsiehler N_20200604150830a_jsiehler    By jsiehler 2020/06/04 @ 15:08:30a -------------------------------- In the first case, the more relevant fact is that 29 factors in Z[i], the Gaussian Integers.        (select only this node)     N_20200604150102e_ColinWright->N_20200604150830a_jsiehler N_20200604151453a_ColinWright    By ColinWright 2020/06/04 @ 15:14:53a -------------------------------- Yes, you're right. I was too focused on using the Gaussian Integer factorisation with the sqrt(-1) from Z/(41Z) ... being a field makes factorising trivial.        (select only this node)     N_20200604150344a_ColinWright->N_20200604151453a_ColinWright N_20200604150830b_jsiehler    By jsiehler 2020/06/04 @ 15:08:30b -------------------------------- Viz., (5+2i)(5-2i).        (select only this node)     N_20200604150830c_jsiehler    By jsiehler 2020/06/04 @ 15:08:30c -------------------------------- This is a ring with interesting factorizations because not everything is a unit.        (select only this node)     N_20200604150830b_jsiehler->N_20200604150830c_jsiehler N_20200604150830d_jsiehler    By jsiehler 2020/06/04 @ 15:08:30d -------------------------------- I can't think of any ring that uses the "square root of -1" in Z/41Z in the same way.        (select only this node)     N_20200604150959c_jsiehler    By jsiehler 2020/06/04 @ 15:09:59c -------------------------------- It isn't really taking advantage of the fact that 9^2 = -1 (mod 41).        (select only this node)     N_20200604150959c_jsiehler->N_20200604151453a_ColinWright N_20200605121931a_ColinWright    By ColinWright 2020/06/05 @ 12:19:31a -------------------------------- So if p=a^2+b^2, then looking at the Gaussian Integers, Z[i], we can see that p=(a+bi)(a-bi).        (select only this node)     N_20200605121931a_ColinWright->N_20200604150102b_ColinWright N_20200604150959a_jsiehler    By jsiehler 2020/06/04 @ 15:09:59a -------------------------------- Both C and Z/41Z are fields, so saying that things "factor nontrivially" in either of those systems is sort of true, depending on what you mean by non-trivial, but not really the point.        (select only this node)     N_20200604150959b_jsiehler    By jsiehler 2020/06/04 @ 15:09:59b -------------------------------- Everything factors lots of ways in a field, because everything's a unit. For every non-zero x, there is a y such that 29=x*y (mod 41), and that's nothing special about 29.        (select only this node)     N_20200604150959a_jsiehler->N_20200604150959b_jsiehler N_20200604150959a_jsiehler->N_20200604151453a_ColinWright N_20200604152030a_ColinWright    By ColinWright 2020/06/04 @ 15:20:30a -------------------------------- There is a proof that p=a^2+b^2 that uses the existence of sqrt(-1) in a non-trivial way.        (select only this node)     N_20200604152030b_ColinWright    By ColinWright 2020/06/04 @ 15:20:30b -------------------------------- Proof:        (select only this node)     N_20200604152030a_ColinWright->N_20200604152030b_ColinWright N_20200604152030c_ColinWright    By ColinWright 2020/06/04 @ 15:20:30c -------------------------------- If p=4n+1 is a prime number, then there must be an integer m such that m^2+1 is divisible by p.        (select only this node)     N_20200604152030b_ColinWright->N_20200604152030c_ColinWright N_20200604152030d_ColinWright    By ColinWright 2020/06/04 @ 15:20:30d -------------------------------- Z[i] is a UFD.        (select only this node)     N_20200604152030e_ColinWright    By ColinWright 2020/06/04 @ 15:20:30e -------------------------------- p does not divide i, hence divides neither m+i nor m-1.        (select only this node)     N_20200604152030d_ColinWright->N_20200604152030e_ColinWright N_20200604152030f_ColinWright    By ColinWright 2020/06/04 @ 15:20:30f -------------------------------- But p divides their product, which is m^2+1.        (select only this node)     N_20200604152030e_ColinWright->N_20200604152030f_ColinWright N_20200604152030g_ColinWright    By ColinWright 2020/06/04 @ 15:20:30g -------------------------------- So p cannot be a prime element in Z[i], and must have a nontrivial factorization.        (select only this node)     N_20200604152030f_ColinWright->N_20200604152030g_ColinWright N_20200604152030h_ColinWright    By ColinWright 2020/06/04 @ 15:20:30h -------------------------------- Considering the norm (which is multiplicative), the factorisation can only have two factors.        (select only this node)     N_20200604152030g_ColinWright->N_20200604152030h_ColinWright N_20200604152030i_ColinWright    By ColinWright 2020/06/04 @ 15:20:30i -------------------------------- So the factorisation must be of the form p=(x+yi)(x-yi), hence p=x^2+y^2.        (select only this node)     N_20200604152030h_ColinWright->N_20200604152030i_ColinWright N_20200604145656a_ColinWright    By ColinWright 2020/06/04 @ 14:56:56a -------------------------------- Thinking about the primes congruent to 1 mod 4 ...        (select only this node)     N_20200604150102a_ColinWright    By ColinWright 2020/06/04 @ 15:01:02a -------------------------------- An odd prime p is the sum of two squares iff p=1 (mod 4).        (select only this node)     N_20200604145656a_ColinWright->N_20200604150102a_ColinWright N_20200604150102c_ColinWright    By ColinWright 2020/06/04 @ 15:01:02c -------------------------------- There exists a square root of -1 mod p iff p=1 (mod 4) (Again, limited to odd primes).        (select only this node)     N_20200604145656a_ColinWright->N_20200604150102c_ColinWright N_20200604150102a_ColinWright->N_20200604150102e_ColinWright N_20200604150102a_ColinWright->N_20200605121931a_ColinWright N_20200604150102d_ColinWright    By ColinWright 2020/06/04 @ 15:01:02d -------------------------------- That means that 29 (say) factors non-trivially mod 41 (say).        (select only this node)     N_20200604150102c_ColinWright->N_20200604150102d_ColinWright N_20200604150102c_ColinWright->N_20200604150102e_ColinWright N_20200604150102d_ColinWright->N_20200604150959a_jsiehler N_20200604150650b_jsiehler    By jsiehler 2020/06/04 @ 15:06:50b -------------------------------- The usual proofs I know for those two things don't seem to have much to do with each other but maybe I'll learn something here.        (select only this node)     N_20200604150650a_jsiehler->N_20200604150650b_jsiehler N_20200604150650b_jsiehler->N_20200604152030a_ColinWright N_20200604150830a_jsiehler->N_20200604150830b_jsiehler N_20200604150830c_jsiehler->N_20200604150830d_jsiehler N_20200604150959b_jsiehler->N_20200604150959c_jsiehler N_20200604152030c_ColinWright->N_20200604152030d_ColinWright