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A tentative question

There seems to be no non-residues higher than quadratic order;is this related to Fermat's last theorem?

 Happy5214 2020-07-10 13:01

If I'm understanding you right, you're saying there are no, say, cubic non-residues. Well 4 is a cubic non-residue modulo 7 (there is no [i]x[/i] where [i]x[/i]^3=4 (mod 7)).

 R.D. Silverman 2020-07-10 14:02

[QUOTE=Happy5214;550172]If I'm understanding you right, you're saying there are no, say, cubic non-residues. Well 4 is a cubic non-residue modulo 7 (there is no [i]x[/i] where [i]x[/i]^3=4 (mod 7)).[/QUOTE]

Exercise for the O.P.

If p = 1 mod 6, how many cubic residues does it have? how many non-residues?
What about p = -1 mod 6? Did you bother at all to even try to answer your own question?

Do us all a favor. Don't come back until you take the time to actually do some reading

Your question could have been answered with 15 seconds of thought. But of course
you could not be bothered.

 retina 2020-07-10 14:07

[QUOTE=R.D. Silverman;550175][QUOTE=Happy5214;550172]If I'm understanding you right, you're saying there are no, say, cubic non-residues. Well 4 is a cubic non-residue modulo 7 (there is no [i]x[/i] where [i]x[/i]^3=4 (mod 7)).[/QUOTE]

Exercise for the O.P.

If p = 1 mod 6, how many cubic residues does it have? how many non-residues?
What about p = -1 mod 6? Did you bother at all to even try to answer your own question?

Do us all a favor. Don't come back until you take the time to actually do some reading

Your question could have been answered with 15 seconds of thought. But of course
you could not be bothered.[/QUOTE]I think you quoted the wrong person.

 R.D. Silverman 2020-07-10 15:13

[QUOTE=retina;550176]I think you quoted the wrong person.[/QUOTE]

Huh? I did not quote anyone. I simply made a follow-on post. My post was addressed
to the O.P.

 xilman 2020-07-10 16:34

[QUOTE=R.D. Silverman;550181]Huh? I did not quote anyone. I simply made a follow-on post. My post was addressed
to the O.P.[/QUOTE]You quoted Happy5214 in the same way I'm quoting you.

You clicked on the "Quote" button, not the "Post Reply" button.

Technical point in that your text clearly addressed the OP, not the person you quoted in the technical sense.

 R.D. Silverman 2020-07-10 22:38

[QUOTE=xilman;550190]You quoted Happy5214 in the same way I'm quoting you.

You clicked on the "Quote" button, not the "Post Reply" button.

Technical point in that your text clearly addressed the OP, not the person you quoted in the technical sense.[/QUOTE]

Point taken. This has become an interesting debate over the word "quote".

What [i]I [/i] wrote was a follow-on. It was suggested by Happy5214 that the O.P. check whether
4 mod 7 was a cubic residue. I simply extended this to other primes that were:
firstly 1 Mod 6 (as 7 is) and then to all other primes. [And added some admonishment.]

From now on I will try to use the "post reply" button instead of "quote" to avoid confusion when I do not
intend to quote anything.

 retina 2020-07-10 22:45

[QUOTE=R.D. Silverman;550228]This has become an interesting debate over the word "quote".[/QUOTE]I have copied (some of) your text and placed into a box above. That is a quote of what you said. You didn't write my post, I wrote it, so I quoted you.

 mathwiz 2020-07-10 23:58

[QUOTE=R.D. Silverman;550175]Do us all a favor. Don't come back until you take the time to actually do some reading

Your question could have been answered with 15 seconds of thought. But of course
you could not be bothered.[/QUOTE]

I am quoting R.D. Silverman, but addressing the moderators here:

Why is this sort of rude, insulting tone permitted on the forums? Certainly this vitriol and condescension discourages people from asking questions and thereby learning. Even if the answers may be trivially obvious to Mr. Silverman, they are not to others.

 R.D. Silverman 2020-07-11 00:51

[QUOTE=mathwiz;550233]I am quoting R.D. Silverman, but addressing the moderators here:

Why is this sort of rude, insulting tone permitted on the forums? Certainly this vitriol and condescension discourages people from asking questions and thereby learning. Even if the answers may be trivially obvious to Mr. Silverman, they are not to others.[/QUOTE]

The O.P. has a very long history of crank and nonsensical posts combined with
a proven unwillingness to spend time learning anything about the subject. He also
fails to respond to questions posed to him.

Anyone who wants to ask questions needs to show that they have made at least a
minimal effort to answer the question for themselves. To do so otherwise is rude
in and of itself because it places a time requirement on others. Even a cursory
web search by the O.P. would have revealed the answer. Teachers should not
have to waste their time with students who are unwilling to do basic homework.

 Batalov 2020-07-11 00:52

Eh... There is a button for that.

[QUOTE=mathwiz;550233]I am quoting R.D. Silverman, but addressing the moderators here:

Why is this sort of rude, insulting tone permitted on the forums?.[/QUOTE]

As a purely gedanken experiment, I might submit that the opposite is also true:

[QUOTE]The comments like that (a comment from someone who has nothing for or against the argument and only stopped by to criticise the tone of voice in which two members who know each other for 10+ years talk to each other) - that discourages qualified answers. people who know the answers stop answering because someone will "report them", so they just stop. And then forum fills up with very helpful and kind answers from folks who don't know the answer and don't even know how to learn the answer.[/QUOTE]

 R.D. Silverman 2020-07-11 01:07

[QUOTE=R.D. Silverman;550236]The O.P. has a very long history of crank and nonsensical posts combined with
a proven unwillingness to spend time learning anything about the subject. He also
fails to respond to questions posed to him.

Anyone who wants to ask questions needs to show that they have made at least a
minimal effort to answer the question for themselves. To do so otherwise is rude
in and of itself because it places a time requirement on others. Even a cursory
web search by the O.P. would have revealed the answer. Teachers should not
have to waste their time with students who are unwilling to do basic homework.[/QUOTE]

"What have you done so far that makes you believe what you say? Please
show your work. Also tell us [i]why[/i] you might think there is a relation between
FLT and the existence of higher order residues."

Do you believe that such a request on my part would also be rude??

Sometimes in order to teach we need to see what efforts have been made so far so
that we can see where the student has been led astray. When a poster has shown
(historically) an unwillingness to respond to such requests then an admonishment is,
and should be, in order.

I do not wish to see this sub-forum turn into sci.math. For homework help there is
another sub-forum available. There is also a misc.math sub-forum.

 Uncwilly 2020-07-11 02:14

[QUOTE=R.D. Silverman;550236]Anyone who wants to ask questions needs to show that they have made at least a minimal effort to answer the question for themselves.[/quote]
That is your opinion, you are entitled to it. We don't have to share it.
[QUOTE]To do so otherwise is rude in and of itself because it places a time requirement on others.[/QUOTE]Again your opinion, but this time it is wrong. There is no requirement for others to responded. You have no obligation to respond.
[quote]Teachers should not have to waste their time with students who are unwilling to do basic homework.[/QUOTE]There is no "have to" here. You can just stroll by and not involve yourself. If you feel that the OP is acting irresponsibly, you can have less stress by not dealing with their posts.

[QUOTE=Proverbs 26:17 JPS]He that passeth by, and meddleth with strife not his own, is like one that taketh a dog by the ears.[/QUOTE]

 R.D. Silverman 2020-07-11 02:32

[QUOTE=Uncwilly;550243] You have no obligation to respond. <snip>
[/QUOTE]

It requires time to read the question, even if no response is given.

[QUOTE]
There is no "have to" here. You can just stroll by and not involve yourself. If you feel that the OP is acting irresponsibly, you can have less stress by not dealing with their posts.[/QUOTE]

Why do you presume that dealing with their posts is stressful? I call it amusement.
And silence gives an implied consent that troll questions are OK.

If you want to turn this sub-forum into sci.math, I feel sorry for you. A failure to
perform even a little diligence will turn this forum into sci.math. Cranks, trolls,
and people unwilling to learn should go somewhere else. Questions posed
by people who are unwilling to put in an effort should be discouraged.

Yes, this is my opinion.

 R.D. Silverman 2020-07-11 02:41

[QUOTE=R.D. Silverman;550244]It requires time to read the question, even if no response is given.

Why do you presume that dealing with their posts is stressful? I call it amusement.
And silence gives an implied consent that troll questions are OK.

If you want to turn this sub-forum into sci.math, I feel sorry for you. A failure to
perform even a little diligence will turn this forum into sci.math. Cranks, trolls,
and people unwilling to learn should go somewhere else. Questions posed
by people who are unwilling to put in an effort should be discouraged.

Yes, this is my opinion.[/QUOTE]

And there is another valid reasons for not simply answering such questions:

"Give a man a fish and he eats for a day. Teach a man to fish and he eats for a

 Uncwilly 2020-07-11 03:21

According to various of your previous posts, you have assessed the OP as being one that doesn't like fish.

A tentative question

[CODE][/CODE][QUOTE=devarajkandadai;550153]There seems to be no non-residues higher than quadratic order;is this related to Fermat's last theorem?[/QUOTE]
Sorry;just proved that 23 is a non-residue of 7919 upto infinite order. This was done with aid of my paper "Euler's generalization of Fermat's theorem ( a further generalization)- Hawaii international conference ,2004.
Verification : pari code -
Is=Mod(17,7919)^7922 = =23

A tentative question

[QUOTE=devarajkandadai;550153]There seems to be no non-residues higher than quadratic order;is this related to Fermat's last theorem?[/QUOTE]
Sorry;just proved that 23 is a non-residue of 7919 upto infinite order. This was done with aid of my paper "Euler's generalization of Fermat's theorem ( a further generalization)- Hawaii international conference ,2004.
Verification : pari code -

Also is(n)=Mod(17,7919)^n==23
select(is,[1..7922]==23

 LaurV 2020-07-13 14:26

Try again, please. At least, if you don't test the pieces of code by yourself, you should carefully check that all the parentheses match... :lol:

Sorry;just proved that 23 is a non-residue of 7919 upto infinite order. This was done with aid of my paper "Euler's generalization of Fermat's theorem ( a further generalization)- Hawaii international conference ,2004.
Verification : pari code -
Is=Mod(17,7919)^7922 = =23[/QUOTE]
This example is not correct.correct example:
23 is non-residue of 3571 upto infinite order.I leave it to pari experts like Charles to verify.

 Dr Sardonicus 2020-07-14 12:08

[QUOTE=devarajkandadai;550527]This example is not correct.correct example:
23 is non-residue of 3571 upto infinite order.I leave it to pari experts like Charles to verify.[/QUOTE]

Er, ah, 23 is a cubic residue mod 3571.

The cube roots of Mod(23,3571) are Mod(34,3571), Mod(35,3571), and Mod(3502,3571).

 R.D. Silverman 2020-07-14 12:27

[QUOTE=devarajkandadai;550527]This example is not correct.correct example:
23 is non-residue of 3571 upto infinite order.I leave it to pari experts like Charles to verify.[/QUOTE]

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 [i]generalization[/i]. 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.

 R.D. Silverman 2020-07-14 12:54

[QUOTE=R.D. Silverman;550557]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 [i]generalization[/i]. 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.[/QUOTE]

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.

 Dr Sardonicus 2020-07-14 13:06

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 [i]not[/i] 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 10[sup]40[/sup] 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.

 JeppeSN 2020-07-14 13:33

[QUOTE=R.D. Silverman;550557]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.[/QUOTE]
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 [I]you[/I] should s*** t** f*** u*. /JeppeSN

 Uncwilly 2020-07-14 14:14

[QUOTE=R.D. Silverman;550557]Finally STFU until you can be bothered studying at least some of this subject.[/QUOTE]
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 [B]you[/B] [U][I]have[/I][/U] 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.

 Uncwilly 2020-07-14 14:26

[QUOTE=JeppeSN;550564]I think [I]you[/I] should s*** t** f*** u*. /JeppeSN[/QUOTE]That sort of language does not belong in the Number Theory forum. You will be joining R.D. Silverman in time away from posting.

[QUOTE=R.D. Silverman;550562]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.[/QUOTE]

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)

 preda 2020-07-16 07:32

[QUOTE=Uncwilly;550566]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 [B]you[/B] [U][I]have[/I][/U] 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.[/QUOTE]

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.

 paulunderwood 2020-07-16 09:47

[QUOTE=devarajkandadai;550745]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.
[/QUOTE]

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.

 kriesel 2020-07-16 13:35

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 [URL]https://www.mersenneforum.org/showpost.php?p=550557&postcount=22[/URL]
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 [URL]https://www.mersenneforum.org/showpost.php?p=548500&postcount=1[/URL] and employ them upon return.

 Uncwilly 2020-07-16 13:48

[QUOTE=kriesel;550759]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.
[/QUOTE]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.

 kriesel 2020-07-16 14:36

[QUOTE=Uncwilly;550763]RDS posted it via an initialism. Jeppe did it in plain language[/QUOTE]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.[/QUOTE]Missed that detail on first read.[QUOTE]Answering vitriol with vitriol just covers everyone with vitriolic acid.[/QUOTE]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.

 paulunderwood 2020-07-16 15:01

[QUOTE=devarajkandadai;550745]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.
[/QUOTE]

What does Taylor's Theorem have to do with this? Please elaborate.

[QUOTE=paulunderwood;550768]What does Taylor's Theorem have to do with this? Please elaborate.[/QUOTE]
Recall that Taylor's theorem can be stated as
f(x+h) =f(x) + hf'(x)..h^2/2!f''(x)......
Just replace h by f(x) and you get the required proof.

 paulunderwood 2020-07-18 09:23

[QUOTE=devarajkandadai;550914]Recall that Taylor's theorem can be stated as
f(x+h) =f(x) + hf'(x)..h^2/2!f''(x)......
Just replace h by f(x) and you get the required proof.[/QUOTE]

I still don't see it. Taylor's Theorem is defined for real numbers. You state x is in N. What is the derivative of f(x)? Can you give the first three terms of f(x+h)?

 ewmayer 2020-07-19 00:01

[QUOTE=kriesel;550759]I suggest that a moderator edit out the deliberate nastiness from the thread.[/QUOTE]

The crap-flinging should've been nipped in the bud - now it's way too much work to filter the small amount of useful content after post #3. But hey, any fellow mod looking for an Augean-stables-like challenge is welcome to it.

 R.D. Silverman 2020-07-21 13:46

[QUOTE=paulunderwood;550919]I still don't see it. Taylor's Theorem is defined for real numbers. You state x is in N. What is the derivative of f(x)? Can you give the first three terms of f(x+h)?[/QUOTE]

Of course the O.P. fails to respond to the question.
(about his function being nowhere differentiable).
He would have to acknowledge that his claimed theorem is false
and that the so-called proof is phony.

Noone here seems to think that the failure to respond is rude.

Based on the lack of responses it seems that it is OK to make a
false claim then run away and hide when asked a question about that
claim. Noone has commented on his lack of response.

Further, while I am reluctant to say the word 'plagiarism', I will
note that "paulunderwood" failed to give propper attribution
to the [i]source of his comment/question[/i] (made above).
I suggested in a PM to him that the claimed use of Taylor's theorem
was erroneous [i]prior[/i] to his public post. He failed to give attribution.

As for why the O.P. failed to respond I can only guess:

Derivatives do not exist so the claim of a proof by Taylor's theorem is phony.

Does anyone here have any academic integrity?

Trolls and cranks are OK. Failure to respond to polite questions is OK.
Plagiarism is OK. But if I criticize a troll you all shout "rude".
Some moderator with too much power will ban someone over the use
of 'STFU', but ignore repeated crank posts and trolling as well
as lack of any kind of technical integrity.
[Note: See below!!!!]

[b]Finally[/b], the O.P. asked in a misc math "Group Theory" thread why there was no
group theory discussion. This is a simple, innocent discussion.
Ernst, in a typical moderator [i]hypocritical[/i] post then wrote:

"Why - you want to give yourself a chance to look like an idiot in an exciting new thread?
Seriously, dude, just shut up already. Here, I'll help you by giving you a nice 7-day timeout."

In the original thread, Ernst also said:
"crap-flinging should've been nipped in the bud".

But I suppose that since he is a moderator it is OK for him to call the OP an idiot.
It is OK for [b]him[/b] to tell the OP to shut up. It is OK for him to call the original thread
"crap flinging".

And in this instance all the OP did was to ask an innocent question! And for that he got banned!???
He did [b]not[/b] get banned for his repeated (over many years) trolling false claims in the math forum, but for simply
asking a question in misc. math.

Can you say "abuse of moderator power"?? Can you say "hypocrite"???
Can you say "screwed up priorities"??? Can you say "plagiarism"???

 Uncwilly 2020-07-21 14:15

[QUOTE=R.D. Silverman;551184]Noone here seems to think that the failure to respond is rude.

Based on the lack of responses it seems that it is OK to make a
false claim then run away and hide when asked a question about that
claim. Noone has commented on his lack of response.[/QUOTE]
If the OP does not respond to answers, then everyone else should sit on their hands.
This is an internet forum, not a face to face conversation. In a face to face conversation not responding [U]might[/U] be rude. On a forum, there can be many reasons not to respond.

So, do you want trolls dealt with or not? You say that the mods have ignored it, then that they acted to harshly (this is the same person that you identified as a troll).

You can criticize trolls, but you must do so civilly. The language used, whether in words or initials should be appropriate for a high school classroom.

No one is forcing you to be here. You it is your choice. If you want to remain here and be active, you need to watch your language. You have been warned and banned before. Therefore you get less slack as a repeat offender. You are not ignorant of the need for civility.

 R.D. Silverman 2020-07-21 15:01

[QUOTE=Uncwilly;551188]If the OP does not respond to answers, then everyone else should sit on their hands.
This is an internet forum, not a face to face conversation. In a face to face conversation not responding [U]might[/U] be rude. On a forum, there can be many reasons not to respond.
[/QUOTE]

When someone makes a bold claim and another person then asks for clarification
the OP should be obligated to respond. I call that common courtesy. You claim
that people here should be courteous.

You are hiding behind what is really going on. The OP posted a bold faced lie
and then deliberately ignored a question because he had no valid response. Yet you
excuse the behavior as possibly "having many reasons not to answer".

[QUOTE]
So, do you want trolls dealt with or not? You say that the mods have ignored it, then that they acted to harshly (this is the same person that you identified as a troll).
[/QUOTE]

Ernst acted overly harshly [b]not[/b] in response to a troll post, but rather to a reasonable question. And he used similar language for which I am criticized. And noone said
a thing about it. It is hypocricy of the worst kind.

[QUOTE]
You can criticize trolls, but you must do so civilly. The language used, whether in words or initials should be appropriate for a high school classroom.
[/QUOTE]

You obviously have limited experience in high school classrooms. You also seem
more concerned about superficial language than the content of troll posts and about

[QUOTE]

No one is forcing you to be here. You it is your choice. If you want to remain here and be active, you need to watch your language. You have been warned and banned before. Therefore you get less slack as a repeat offender. You are not ignorant of the need for civility.[/QUOTE]

I agree about the civility. The issue is the question: [i]what is civility[/i]? I claim
trolling and crank posts and failure to respond is highly uncivil. You disagree.
You claim "STFU" is uncivil. I claim that it is mere hyperbole. I claim that
calling someone "clueless" is an appropriate response to repeated false posts
in combination with a deliberate [b]unwillingness[/b] to learn or study.

And I see that you totally ignored my comments about plagiarism. You seem to
think that it is OK as implied by your silence.

And you ignored my comments about Ernst's abuse of power and the lack of

I call this hypocricy on your part. You see what you want to see and ignore the rest.

As for "forcing you to be here" we've all heard that argument before. i.e. "our country.
love it or leave it. You are not forced to be here". It is a crock.

Silence in response to trolling only gives consent. And it encourages
further trolling. And failure to respond allows less knowledgeable people to believe
that nonsense claims might be true.

You seem to want this forum to be a "feel good social club" and do not seem to care
about any kind of academic integrity. I think it stinks. "Feel good" polite responses
to trolls and cranks DO NOT WORK. They keep offending. But you seem to think
it is OK.

Get the trolls and cranks out of this sub-forum and demand that they only post
in the misc math or blogorrhea forums.

 Xyzzy 2020-07-21 15:21

Bob:

Grow up and stop acting like a petulant child.

The world is not perfect or fair. This forum isn't either.

You lack discipline and class.

Post one more time in a manner that is not civil and you will be banned forever.

And this time there will be no second/third/fourth/etc. chance.

F O R E V E R

Over the years you have most likely alienated yourself from colleagues, friends and family with your aberrant behavior.

Do you want to lose what you have here as well?

You can change how you act. You can change the way you are. It takes time but it can be done. It requires discipline.

Fake it until you make it.

If you want help everyone here will bend over backwards to assist you.

[URL]https://fablesofaesop.com/the-tree-and-the-reed.html[/URL]

 S485122 2020-07-21 15:29

[QUOTE=R.D. Silverman;551194]...
Get the trolls and cranks out of this sub-forum and demand that they only post in the misc math or blogorrhea forums.[/QUOTE]Which, as I understood it, was a deal between the moderators and you. As I understand it, this thread should have been moved to the Misc. sub-forum after post #1 (or 3 if no moderator wanted to look into the content of post #1.)

Jacob

 Dr Sardonicus 2020-07-21 19:06

I was under the impression that not responding to posts by trolls was entirely appropriate. I have, in fact, been reminded of this by other Forumites when unable to resist the temptation to respond to our resident trolls from time to time, and I have been able to return the favor occasionally.

One of the smilies was part of said reminders,
[center]:dnftt:[/center]

 ryanp 2020-07-21 20:08

[QUOTE=R.D. Silverman;551184]failed to give propper attribution[/QUOTE]

Uh oh. What do I have to do with all this?

 JeppeSN 2020-07-21 23:02

[QUOTE=R.D. Silverman;551184]
[b]Finally[/b], the O.P. asked in a misc math "Group Theory" thread why there was no
group theory discussion. This is a simple, innocent discussion.
Ernst, in a typical moderator [i]hypocritical[/i] post then wrote:

"Why - you want to give yourself a chance to look like an idiot in an exciting new thread?
Seriously, dude, just shut up already. Here, I'll help you by giving you a nice 7-day timeout."

In the original thread, Ernst also said:
"crap-flinging should've been nipped in the bud".

But I suppose that since he is a moderator it is OK for him to call the OP an idiot.
It is OK for [b]him[/b] to tell the OP to shut up. It is OK for him to call the original thread
"crap flinging".

And in this instance all the OP did was to ask an innocent question! And for that he got banned!??? [/QUOTE]

I also find it problematic that he got that response ("look like an idiot", "just shut up"), and got a 7-day quarantine, because he asked for a group theory subforum. A bad and silly question, but not in itself something you would be banned for? But maybe there was an accumulation of low quality posts from him leading up to this (so the group theory request was just the straw that broke the camel's back)? /JeppeSN

 jnml 2020-07-22 08:11

[QUOTE=Xyzzy;551199]Bob:

Post one more time in a manner that is not civil and you will be banned forever.

And this time there will be no second/third/fourth/etc. chance.

F O R E V E R
[URL]https://fablesofaesop.com/the-tree-and-the-reed.html[/URL][/QUOTE]

I see his posts as direct and honest. But more importantly, informed. Even when they're coming with no sugar coating

I object to threatening him with a life-long ban for that.

 ewmayer 2020-07-22 22:42

[QUOTE=ryanp;551226]Uh oh. What do I have to do with all this?[/QUOTE]

Ha, ha - thanks, Ryan, best laugh I've had all day. (Even it came a day late).

[QUOTE=JeppeSN;551239]I also find it problematic that he got that response ("look like an idiot", "just shut up"), and got a 7-day quarantine, because he asked for a group theory subforum. A bad and silly question, but not in itself something you would be banned for? But maybe there was an accumulation of low quality posts from him leading up to this (so the group theory request was just the straw that broke the camel's back)? /JeppeSN[/QUOTE]

Yes, precisely the latter case - you have to factor in the poster's history. As I noted in the Mod subforum yesterday:
[quote]The above occurred after I'd spent over an hour I really couldn't well spare on a busy weekend reading through the OP's previous thread that turned into a right royal crapfest. So Devaraj caught me in an especially grumpy mood. I am, after all, a mere human like the rest of you. To employ the kind of initialism [R.D. Silverman] dearly loves, BFD.[/quote]
IMO, Devaraj has a long history of posting patently nonsensical and "your snippet of code here is clearly wrong"-afflicted noisy claims. IOW, wasting the time of his betters. If one of the actual mathematics experts hereabouts wants to start a group theory thread, great. But I fear with the above poster's sketchy history of math posts and the fact that an inordinate number of threads started by him rapidly degenerate into the aforementioned "right royal crapfests", IMO he would do us, and the forum, a favor by listening more and posting less, with occasional posts that show some genuine effort to actually master the material in question. One can hope he will use the time-out I gave him to do so.

 ewmayer 2020-07-23 04:46

[QUOTE=jnml;551264]I see his posts as direct and honest. But more importantly, informed. Even when they're coming with no sugar coating

I object to threatening him with a life-long ban for that.[/QUOTE]

The "informed" part is not the issue. The "repeatedly and gratuitously rude" part is. And see my note above about a poster's history. No one on this forum has a longer history of abusive posts, driving-away newcomers, getting repeatedly warned, short-term-banned and even "permanently" (later and ill-advisedly reversed) banned by the mods than RDS. His earlier perma-ban was lifted *only* on the condition that he show drastic behavior betterment. He did ... for a while.

And to put the "informed" part in context - I know for a fact that there are absolutely world-class number theorists who have declined an invitation to come post here and help improve the discussion, specifically because RDS is allowed to run amok and give his unworthy-of-a-5-year-old petulance free rein.

 Uncwilly 2020-07-23 12:49

[QUOTE=ewmayer;551339]His earlier perma-ban was lifted *only* on the condition that he show drastic behavior betterment. He did ... for a while.[/QUOTE]And that is why the seeming Draconian ban happened for the current incident. If you know you are on a short leash and challenge it, expect a reaction.

 LaurV 2020-07-24 06:39

When I get in an argument with Uncwilly about things which were posted in the past in the forum and by who, like [URL="https://www.mersenneforum.org/showthread.php?p=549758"]recently happened[/URL], can I tell him STFU? (i.e. "search the forum, Uncwilly!", by the same logic like RTFM or STFF :razz:, hehe, we are learning "one acronym a day"... )

 ewmayer 2020-07-24 23:31

[QUOTE=LaurV;551406]hehe, we are learning "one acronym a day"...[/QUOTE]

Sorry to disappoint, but those are all initialisms :). Acronyms are words (neologisms), as in pronounceable-as-words, formed from the initials of other words, e.g. radar, laser, fubar. It is debatable which type ROFL is, since it could be plausibly pronounce "roffle" - based on that I've added it to my personal mental LOA (List of acronyms, pronounced just like the second part of the Hawaiian mountain Mauna Loa), as I have your forum handle, Mr. "Larv".

Ha, it just occurred to me that Paul L's forum handle is a proper acronym when the 'x' is pronounced the Chinese way: "shillman", sounds like someone trying to hard-sell you a dodgy used car.

 a1call 2020-07-25 02:06

[QUOTE=ewmayer;551499]

Ha, it just occurred to me that Paul L's forum handle is a proper acronym when the 'x' is pronounced the Chinese way: "shillman", sounds like someone trying to hard-sell you a dodgy used car.[/QUOTE]
Turn out there is much more to it than pronouncing it as "sh":

[url]https://resources.allsetlearning.com/chinese/pronunciation/The_%22j%22_%22q%22_and_%22x%22_sounds[/url]

Can you smile when you pronounce "sh"?

ETA Russian pronunciation of x should also work.
[url]http://masterrussian.net/f17/pronunciation-x-7686/[/url]

 xilman 2020-07-25 08:17

[QUOTE=ewmayer;551499]Sorry to disappoint, but those are all initialisms :). Acronyms are words (neologisms), as in pronounceable-as-words, formed from the initials of other words, e.g. radar, laser, fubar. It is debatable which type ROFL is, since it could be plausibly pronounce "roffle" - based on that I've added it to my personal mental LOA (List of acronyms, pronounced just like the second part of the Hawaiian mountain Mauna Loa), as I have your forum handle, Mr. "Larv".

Ha, it just occurred to me that Paul L's forum handle is a proper acronym when the 'x' is pronounced the Chinese way: "shillman", sounds like someone trying to hard-sell you a dodgy used car.[/QUOTE]Most of you know my love of irony, paradoxes and self-referential humour.

TLA is accurately self-referential. ETLA is not.

ETLA is an acronym. TLA is not.

 ewmayer 2020-07-25 22:14

[QUOTE=xilman;551539]ETLA is an acronym. TLA is not.[/QUOTE]

Only because you're not pronouncing TLA properly. Think an l-instead-of-r-speech-impaired version of "tra". Putting words in Michael-Palin-as-a-lisping-version-of-Pontiuth-Pilate-in-Life-of-Bwian's mouth to illustwate: "I have a vewy gweat fwiend in Wome, hith name ith Bigguth Dickuth and he liketh to pwanth about singing 'tla-la-la'."

[Yes, I know I'm mixing my lisp types in the above quote ... I pwead poetic wicense!]

 kriesel 2020-07-29 15:27

[QUOTE=R.D. Silverman;551184]Your priorities are screwed up.

[B]Finally[/B], the O.P. asked in a misc math "Group Theory" thread why there was no
group theory discussion. This is a simple, innocent discussion.
Ernst, in a typical moderator [I]hypocritical[/I] post then wrote:

"Why - you want to give yourself a chance to look like an idiot in an exciting new thread?
Seriously, dude, just shut up already. Here, I'll help you by giving you a nice 7-day timeout."

In the original thread, Ernst also said:
"crap-flinging should've been nipped in the bud".

But I suppose that since he is a moderator it is OK for him to call the OP an idiot.
It is OK for [B]him[/B] to tell the OP to shut up. It is OK for him to call the original thread
"crap flinging".

And in this instance all the OP did was to ask an innocent question! And for that he got banned!???
He did [B]not[/B] get banned for his repeated (over many years) trolling false claims in the math forum, but for simply
asking a question in misc. math.

Can you say "abuse of moderator power"?? Can you say "hypocrite"???
Can you say "screwed up priorities"??? Can you say "plagiarism"???[/QUOTE]
RDS is a very educated intelligent person. I believe he can do better than the recent vitriol, and he's done so in the past.
And RDS raises some valid points about standards for some and not others.

Watching from the outside, I suspect being a moderator is a thankless job. (Do moderators take a vacation from it?) Differences of opinion are inevitable.

Something that has concerned me for a while now is that there does not appear to be any recourse for an ordinary user mistreated by one or more moderators, or any visible effective restraint upon the excesses of a moderator.

I consider what RDS refers to by Ewmayer as both uncharacteristic of Ernst, and mild compared to what I've seen from at least one other moderator.

But it's the tall nail that gets pounded down, no good deed goes unpunished, bad things happen to the bearer of bad tidings, and whistleblowers get retaliation not thanks for identifying an issue.
What I've observed is ridicule, insults, even threats of violence by moderators go unaddressed, and even supported by other moderators piling on.

That is something that ought be addressed effectively.
Instead we have moderators piling on and supporting each other in their excesses.
There is a better way.
The problem won't get addressed while scapegoating is used instead of recognizing the problem.

[B]On the other hand:[/B]
Where would we be if the following people had been driven away?
Error (identified the Jacobi symbol check for LL testing)
Preda (creator of gpuowl)
R. Gerbicz (discovered the GEC for PRP, and a TF NF verification method)
Patnashev ([URL="https://mersenneforum.org/showthread.php?t=25323&highlight=patnashev"]PRP proof built upon the GEC[/URL], which will almost eliminate double-checking effort)
Prime95 (author of prime95/mprime and collaborator in other software; creator of GIMPS)
Scott Kurowski (creator of PrimeNet)
James Heinrich (creator/administrator of mersenne.ca, helps with PrimeNet server scripts)
creators of the various other gpu GIMPS programs

Who have we already driven away, whose potential contributions and consequent loss to the project, we'll never know?

We can do better Let's do that.

 xilman 2020-07-29 16:45

[QUOTE=kriesel;551882]Watching from the outside, I suspect being a moderator is a thankless job. (Do moderators take a vacation from it?) Differences of opinion are inevitable.[/quote]Very very occasionally we receive thanks.

AFAICT, being a mod is almost certainly a life sentence. I recall only one ever being relieved of the task, and that was at his explicit request.

[QUOTE=kriesel;551882]Something that has concerned me for a while now is that there does not appear to be any recourse for an ordinary user mistreated by one or more moderators, or any visible effective restraint upon the excesses of a moderator.[/QUOTE]There is a recourse --- make a private appeal to mods (in plural, note, and a PM can have up tpo 10 recipients) who appear not to be too deeply involved from an emotional viewpoint.

Perhaps it was before you became a subscriber, or perhaps it occurred when you were not paying close enough attention, but I know for a fact that a supermod was given a 1-week ban, and that sentence was visible to all. At a lower level of severity, actions of mods have been criticized in public. I know this for a fact, having been on both sides of such criticism.

 LaurV 2020-08-29 08:55

[QUOTE=kriesel;551882]
[B]On the other hand:[/B]
Where would we be if the following people had been driven away?
[/QUOTE]
That list misses many other contributors from before you became a member here (msft, flash, bdot, judger, kracker, etc, (sorry for not mentioning all))

[B]On the other hand:[/B]
Where would we be if the following people had not been driven away?
(insert long interminable list of cranks and idiots here)

 Dr Sardonicus 2020-08-29 10:54

[QUOTE=kriesel;551882]<snip>
Something that has concerned me for a while now is that there does not appear to be any recourse for an ordinary user mistreated by one or more moderators, or any visible effective restraint upon the excesses of a moderator.
<snip>
What I've observed is ridicule, insults, even threats of violence by moderators go unaddressed, and even supported by other moderators piling on.
<snip>
Instead we have moderators piling on and supporting each other in their excesses.
[/QUOTE]You made a complete jackass of yourself with frivolous complaints to this effect back in April.

I advised you to refrain from that sort of thing.

Now, here you are, making an even bigger jackass of yourself by [i]repeating[/i] the [i]same[/i] frivolous complaints.

If you think that repeating the same nonsense over and over will result in others accepting it as true, or even acquiescing to it, think again.

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