[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] Some additional thoughts: 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 [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 subforum turn into sci.math. For homework help there is another subforum available. There is also a misc.math subforum. 
[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] 
[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 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. 
[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 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.[/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 lifetime". 
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 nonresidues higher than quadratic order;is this related to Fermat's last theorem?[/QUOTE]
Sorry;just proved that 23 is a nonresidue 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 nonresidues higher than quadratic order;is this related to Fermat's last theorem?[/QUOTE]
Sorry;just proved that 23 is a nonresidue 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 
[QUOTE=devarajkandadai;550437]Corrected pari code[/QUOTE]
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: 
[QUOTE=devarajkandadai;550435][CODE][/CODE]
Sorry;just proved that 23 is a nonresidue 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 nonresidue of 3571 upto infinite order.I leave it to pari experts like Charles to verify. 
[QUOTE=devarajkandadai;550527]This example is not correct.correct example:
23 is nonresidue 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). 
[QUOTE=devarajkandadai;550527]This example is not correct.correct example:
23 is nonresidue 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 (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. 
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