mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2020-07-14, 12:54   #23
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

11100010000002 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
Hey clueless. You can't even do simple arithmetic. Nor do you bother to study any
math. The result makes you look stupid every time you open your mouth.

Consider any prime q that does NOT divide (3571-1). Take, e.g. q = 11

note that 1148^11 = 23 mod 3571.

23 is an 11'th order residue mod 3571. I think we all safely know (except perhaps you)
that 11 is finite. You should now be asking "what is special about 11?"

If you had bothered to take my earlier hint about cubic residues modulo a prime
that that is 1 mod 6 vs. primes that are -1 mod 6 you might have avoided this
latest erroneous assertion. I will give a further hint: Only 1/3 of the residues
less than p are cubic residues of p when p = 1 mod 6. But when q = -1 mod 6,
they ALL are. Learning WHY is directly tied into Lagrange's Theorem. It is also
tied into the Sylow theorems. [Ask yourself how many subgroups there are of size
(p-1)/3]

Go learn some mathematics. In particular learn Lagrange's Theorem. Learn
Euler's Theorem for quadratic reciprocity. Study its generalization. Learn what
a primitive root is. Read and study the Sylow theorems.

Consider the following:

Prove or disprove:

For prime p,q, x^q = a mod p always has a solution for every a when q does not divide p-1.

Then ask: What happens if q | (p-1)???


Go read and study Nick's excellent introduction [in this forum] to number theory.
I think we all know that you will ignore this advice.

Finally STFU until you can be bothered studying at least some of this subject.
If you want to post mindless numerology go to the misc.math sub-forum or
open your own sub-forum in the blogorrhea.

One might also want to ask:

When does x^q = a mod p have a single root, when does it have multiple roots,
and when does it split completely for given a, p, q???

Welcome to the wonderful world of Galois groups.

Note that this question also arises during study of the Special Number Field Sieve.

Last fiddled with by R.D. Silverman on 2020-07-14 at 12:56
R.D. Silverman is offline   Reply With Quote
Old 2020-07-14, 13:06   #24
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

2×1,787 Posts
Default

Reading the last posts, it occurred to me to wonder, given a prime p, how large can the smallest q be (in terms of p), that does not divide p-1.

One answer is, "of order ln(p) at most." I am sure that, given a lower bound for p (say 1000 or 1040 or something), a constant C near 1 could be given for which q is at most C*ln(p).

This is a consequence of PNT, though it might be possible to get by with less, e.g. some of Chebyshev's estimates which predate proofs of PNT.
Dr Sardonicus is offline   Reply With Quote
Old 2020-07-14, 13:33   #25
JeppeSN
 
JeppeSN's Avatar
 
"Jeppe"
Jan 2016
Denmark

25×5 Posts
Unhappy

Quote:
Originally Posted by R.D. Silverman View Post
Hey clueless. You can't even do simple arithmetic. Nor do you bother to study any
math. The result makes you look stupid every time you open your mouth.
[...]
Finally STFU until you can be bothered studying at least some of this subject.
If you want to post mindless numerology go to the misc.math sub-forum or
open your own sub-forum in the blogorrhea.
Silverman, in my opinion many of your posts are unnecessarily offensive. It is nice that you spend time to help people who do not have the mathematical insight you have, but you do it in a way that is rude and much too condescending. If you are unable to write in a polite and friendly manner, no matter how stupid other participants may seem to you, I think you should s*** t** f*** u*. /JeppeSN

Last fiddled with by Uncwilly on 2020-07-14 at 14:24 Reason: expurgated
JeppeSN is offline   Reply With Quote
Old 2020-07-14, 14:14   #26
Uncwilly
6809 > 6502
 
Uncwilly's Avatar
 
"""""""""""""""""""
Aug 2003
101×103 Posts

2×11×397 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
Finally STFU until you can be bothered studying at least some of this subject.
That sort of language does not belong in the Number Theory forum.

If you can't be civil, refrain from posting. There is no reason you have to post. If you do choose to get involved, I would suggest that you give an OP 2 rounds of comments. If they are hopeless after that, state so politely, then no longer engage.

Last fiddled with by Uncwilly on 2020-07-14 at 14:22
Uncwilly is offline   Reply With Quote
Old 2020-07-14, 14:26   #27
Uncwilly
6809 > 6502
 
Uncwilly's Avatar
 
"""""""""""""""""""
Aug 2003
101×103 Posts

221E16 Posts
Default

Quote:
Originally Posted by JeppeSN View Post
I think you should s*** t** f*** u*. /JeppeSN
That sort of language does not belong in the Number Theory forum. You will be joining R.D. Silverman in time away from posting.
Uncwilly is offline   Reply With Quote
Old 2020-07-16, 04:36   #28
devarajkandadai
 
devarajkandadai's Avatar
 
May 2004

22×79 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
One might also want to ask:

When does x^q = a mod p have a single root, when does it have multiple roots,
and when does it split completely for given a, p, q???

Welcome to the wonderful world of Galois groups.

Note that this question also arises during study of the Special Number Field Sieve.
Ok so I have been hasty.here is a summary of my contributions to number theory:
Euler's generalization of Fermat's theorem- a further generalization
(ISSN #1550 3747- Hawaii international conference on mathematics and statistics-2004)
The theorem: let f(x) = a^x + c
where a belongs N and is fixed, c belongs to Z and is fixed and x belongs to N. Then a^(x +k*f(x)) + c is congruent to 0 (mod f(x)).
Here k belongs to N.
Proof is based on Taylor's theorem.
Applications: 1) finding some factors of very large rational integers when expressed in an exponential form 2)finding impossible prime factors of exponential functions ( see A 123239 of OEIS)
Other contributions to number theory: a) Universal exponent generalization of Fermat's theorem(Hawaii international conference-2006) b)ultimate generalisation of Fermat's theorem(planetmath .org-2012)
c) modified Fermat's theorem in order to accommodate Gaussian integers as bases(mersenneforum .org-recent)
d)A theorem a la Ramanujan (AMS-BENELUX-1996)
Also search for "akdevaraj" on youtube.
e) a property of Carmichael numbers conjectured in '89 and proved by Carl Pomerance (generalised conjecture proved by Maxal- see A 104016 and A 104017 on OEIS)

Last fiddled with by devarajkandadai on 2020-07-16 at 04:40 Reason: A minor correction
devarajkandadai is offline   Reply With Quote
Old 2020-07-16, 07:32   #29
preda
 
preda's Avatar
 
"Mihai Preda"
Apr 2015

1,291 Posts
Default

Quote:
Originally Posted by Uncwilly View Post
That sort of language does not belong in the Number Theory forum.

If you can't be civil, refrain from posting. There is no reason you have to post. If you do choose to get involved, I would suggest that you give an OP 2 rounds of comments. If they are hopeless after that, state so politely, then no longer engage.
Let's cut him some slack. If he's right, knowledgeable and informative, I'm willing to let him choose the manner of expressing himself just to hear his ideas.
preda is online now   Reply With Quote
Old 2020-07-16, 09:47   #30
paulunderwood
 
paulunderwood's Avatar
 
Sep 2002
Database er0rr

2×1,723 Posts
Default

Quote:
Originally Posted by devarajkandadai View Post
Ok so I have been hasty.here is a summary

The theorem: let f(x) = a^x + c
where a belongs N and is fixed, c belongs to Z and is fixed and x belongs to N. Then a^(x +k*f(x)) + c is congruent to 0 (mod f(x)).
Here k belongs to N.
Let a=2, x=4, c=3 and k=1.

Then f(2) = 2^4 + 3 = 19

2^(4+1*19)+3 = 2^23 + 3 = 16 mod 19 ???

However if you are saying: Let f(x)=a^x+c. For all a in N and for all c in Z then there exists a k such that f(x+k*f(x))=0 mod f(x) for all x in N; that may be a different matter.

Last fiddled with by paulunderwood on 2020-07-16 at 10:19
paulunderwood is offline   Reply With Quote
Old 2020-07-16, 13:35   #31
kriesel
 
kriesel's Avatar
 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

110028 Posts
Default

I suggest that a moderator edit out the deliberate nastiness from the thread. Something along the lines of "(redacted abusive content)" would appear, twice in https://www.mersenneforum.org/showpo...7&postcount=22
and also in such quoted or original content in other posts as in 23 and 25.
That sort of deliberately abusive language does not belong anywhere in the mersenne forum.

To originate it, as RDS did, seems to me a greater issue, than to object to it as JeppeSN did, mimicking RDS to give RDS back a little taste of his own vitriol.

I suggest RDS spend his time off reading Dale Carnegie's "How to Win Friends and Influence People" and https://www.mersenneforum.org/showpo...00&postcount=1 and employ them upon return.

Last fiddled with by kriesel on 2020-07-16 at 13:44
kriesel is online now   Reply With Quote
Old 2020-07-16, 13:48   #32
Uncwilly
6809 > 6502
 
Uncwilly's Avatar
 
"""""""""""""""""""
Aug 2003
101×103 Posts

2·11·397 Posts
Default

Quote:
Originally Posted by kriesel View Post
To originate it, as RDS did, seems to me a greater issue, than to object to it as JeppeSN did, mimicking RDS to give RDS back a little taste of his own vitriol.
RDS posted it via an initialism. Jeppe did it in plain language that was expurgated by a moderator. Answering vitriol with vitriol just covers everyone with vitriolic acid.

Please PM RDS with your suggestions for him.
Uncwilly is offline   Reply With Quote
Old 2020-07-16, 14:36   #33
kriesel
 
kriesel's Avatar
 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

2·5·461 Posts
Default

Quote:
Originally Posted by Uncwilly View Post
RDS posted it via an initialism. Jeppe did it in plain language
The meaning is the same. Except for those rare readers of such a tender age as not to have encountered it fully spelled out before.

Quote:
that was expurgated by a moderator.
Missed that detail on first read.
Quote:
Answering vitriol with vitriol just covers everyone with vitriolic acid.
I'm all in for less vitriol. None would be a good level. I stand by the claim that to originate it is more serious than to reflect it, just as throwing the first punch defines who is at fault.
kriesel is online now   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
A tentative definition devarajkandadai Number Theory Discussion Group 24 2018-10-29 19:34
Tentative conjecture devarajkandadai Number Theory Discussion Group 10 2018-07-22 05:38

All times are UTC. The time now is 09:52.

Mon Oct 26 09:52:42 UTC 2020 up 46 days, 7:03, 0 users, load averages: 1.55, 1.76, 1.85

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.