[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 (35711). 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 (p1)/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 p1. Then ask: What happens if q  (p1)??? 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 subforum or open your own subforum 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. 
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 p1.
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. 
[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 subforum or open your own subforum 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 
[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. 
[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 statistics2004) 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 conference2006) b)ultimate generalisation of Fermat's theorem(planetmath .org2012) c) modified Fermat's theorem in order to accommodate Gaussian integers as bases(mersenneforum .orgrecent) d)A theorem a la Ramanujan (AMSBENELUX1996) 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) 
[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. 
[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. 
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. 
[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. Please PM RDS with your suggestions for him. 
[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. 
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