20200711, 01:07  #12  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}×13×23 Posts 
Quote:
Suppose that I had asked (in response to the first post) "What have you done so far that makes you believe what you say? Please show your work. Also tell us why 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 subforum turn into sci.math. For homework help there is another subforum available. There is also a misc.math subforum. Last fiddled with by R.D. Silverman on 20200711 at 01:08 

20200711, 02:14  #13  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
29DD_{16} Posts 
Quote:
Quote:
Quote:
Quote:


20200711, 02:32  #14  
"Bob Silverman"
Nov 2003
North of Boston
1110100110011_{2} Posts 
It requires time to read the question, even if no response is given.
Quote:
And silence gives an implied consent that troll questions are OK. If you want to turn this subforum 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. 

20200711, 02:41  #15  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}×13×23 Posts 
Quote:
"Give a man a fish and he eats for a day. Teach a man to fish and he eats for a lifetime". 

20200711, 03:21  #16 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
7·1,531 Posts 
According to various of your previous posts, you have assessed the OP as being one that doesn't like fish.

20200713, 13:23  #17  
May 2004
2^{2}·79 Posts 
A tentative question
Quote:
Verification : pari code  Is=Mod(17,7919)^7922 = =23 

20200713, 13:55  #18  
May 2004
2^{2}·79 Posts 
A tentative question
Quote:
Verification : pari code  Also is(n)=Mod(17,7919)^n==23 select(is,[1..7922]==23 Last fiddled with by devarajkandadai on 20200713 at 14:00 Reason: Corrected pari code 

20200713, 14:26  #19 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2768_{16} Posts 
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...
Last fiddled with by LaurV on 20200713 at 14:28 
20200714, 04:33  #20  
May 2004
100111100_{2} Posts 
Quote:
23 is nonresidue of 3571 upto infinite order.I leave it to pari experts like Charles to verify. 

20200714, 12:08  #21  
Feb 2017
Nowhere
7×853 Posts 
Quote:
The cube roots of Mod(23,3571) are Mod(34,3571), Mod(35,3571), and Mod(3502,3571). 

20200714, 12:27  #22  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}×13×23 Posts 
Quote:
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 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 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. 

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