Talk:Fermat's little theorem

message: I think it should be great to include a proof of the theorem in the article. If it is then sorry.  — Preceding unsigned comment added by Santropedro1 (talk • contribs) 04:05, 14 June 2013 (UTC)
thread_title:  Proof? 

message: I want to introduce the usage of p^k|(x^p-x)^k.Should it belongs to Fermat's little theorem?--Tttfffkkk (talk) 00:13, 22 September 2015 (UTC)
replies: text: What you are describing here are direct computational applications of Fermat's little theorem (the first is no more than an application of the binomial theorem). The other applications mentioned in the article were notable mathematical results and I fail to see how these formulae would be considered notable. References written in other languages are not prohibited, but should be limited to material that is not available in English or of significant notability in its own right (neither of which would be true in this case). Also, this reference is a newly published (2015) primary source and the Wikipedia preference is for reliable secondary sources. I do not see how this addition could be improved to make it a reasonable contribution to this article. Bill Cherowitzo  (talk) 19:00, 23 September 2015 (UTC)
replies: text: Thanks for replying.Everyone would prefer using Euler's theorem than copying this long summation. The only advantage of this method is that the degree of divisor pk is lower than \varphi(p^k).So that at least you can use this method after using Euler's theorem(If not,this long summation would be totally useless).You can make the degree much more lower by x(x+1)...(x+n-1)(reach to Kempner function),but the formula of remainder is too big.--Tttfffkkk (talk) 01:16, 24 September 2015 (UTC)
thread_title: Division form section

message: The discussion in the preceding section has pointed to me that the sentence before Fermat's citation was wrong. For fixing it, I had to quote original Fermat's statement of the theorem and to provide a translation. This shows undoubtedly that Fermat used "progression" for geometric progression, and thus translating "progression" by "series" is a blatant mistake, probably caused by the ignorance of mathematics by the original translator. Therefore I have replaced "series" by "progression" in Ore translation. Nevertheless I strongly support replacing the whole translation, as suggested in the preceding thread. D.Lazard (talk) 16:57, 20 February 2016 (UTC)
replies: text: I have finally found a reliable secondary source which translates the final phrase as, "...; I would send you a demonstration of it, if I did not fear going on for too long." I will make the change in the article. There is however, one problem. This translation, and every other translation I have seen, uses " ... generally true for all series and for all prime ...". I clearly see that Fermat uses "progression" here, so I agree with you, but am at a loss as to why this appears so uniformly (other uses of "progression" in the letter are translated as "progression", so this seems to be intentional). The author of this translation (Michael Mahoney) was criticized by Weil for making some translation errors in a review of the book, but this does not explain the consistent use of "series" by other translators.Bill Cherowitzo  (talk) 19:31, 20 February 2016 (UTC)
replies: text: I cannot answer to your question, but I may guess something: most sentences, where "progression" appear, contain mathematical statements that give hints for the meaning of the word. But here, taking this sentence out of the context, a reader, who is unable to understand that Fermat's little theorem is related with geometric progressions, cannot understand the word, and thus uses the word of the common language that seems (wrongly) to make more sense. I guess that many readers have copied this mistake only because the preceding sentence, that states the theorem, is rarely cited (I agree that it is rather difficult to understand, for a modern mathematician). This leads naturally to the wrong presentation, in the preceding version of the article, of what Fermat did wrote D.Lazard (talk) 19:57, 20 February 2016 (UTC)
thread_title:  Around Fermat's citation 

message: The article claims that since each test would reveal a non-prime with probability 3/4, conducting enough random tests brings the probability of revealing a non-prime arbitrarily close to 1. That logic would be fine if there were infinitely many tests to choose from. But each test starts with choosing a number a between 1 and p, so there are fewer than p tests to choose from. Therefore, the best probability of revealing a non-prime is about 1 - (1/4)^p, not 1 as the article says. Admittedly in practice, we only need to use the primality test on large numbers, so the probability of revealing a non-prime with repeated tests is quite close to 1. 
We should edit the incorrect statement. I haven't yet done so because I have yet to find an authoritative source. CautiousTiger (talk) 20:21, 1 January 2018 (UTC)
replies: text: Miller–Rabin primality test#Accuracy says: "It can be shown that for any odd composite n, at least 3/4 of the bases a are witnesses for the compositeness of n". That's the reason for this in Fermat's little theorem#Miller–Rabin primality test: "If  is not prime, the probability that this is proved by the test is higher than 1/4." It means that if you test 1/4 of all a without finding a witness then n is certain to be prime. Making this many tests would be extremely slow compared to other primality proving methods so it's not done in practice. PrimeHunter (talk) 23:25, 1 January 2018 (UTC)
replies: text:  Ah, thanks for clearing up my confusion. Should we delete this section of the talk page now? CautiousTiger (talk) 01:48, 2 January 2018 (UTC)
replies: text:  no, there is no reason not to leave it as-is. --JBL (talk) 02:16, 2 January 2018 (UTC)
thread_title:  Miller-Rabin: k non-conclusive tests 

message: I removed the conjecture, dated this month, as we can find a contradiction in Carmichael number.  — Arthur Rubin  (talk) 04:32, 15 May 2019 (UTC)
replies: text: Thank you!  I was in the process of removing it when you just barely beat me to it.  All the edits to insert it had no edit summary and the conjecture appears to be recent original research. — Anita5192 (talk) 04:47, 15 May 2019 (UTC)
thread_title:  Caceres conjecture 

message: A,B,p,q,r are whole numbers.
p is a prime number.
Is A^{p-1}-1 divisible by p.   \frac{A^{p-1}-1}{p}=?
\frac{A^{p-1}-1^{p-1}}{p}=\frac{A^{p-1}-B^{p-1}}{p}
\frac{A^{p-1}-B^{p-1}}{p}=\frac{(p+q)^{p-1}-(p+r)^{p-1}}{p}
\frac{A^{p-1}-1^{p-1}}{p}=\frac{(p+q)^{p-1}-\{p+(-p+1)\}^{p-1}}{p}
\frac{A^{p-1}-1}{p}=\frac{\sum_{k=0}^{p-1}\binom{p-1}{k}p^{p-1-k}q^k-\sum_{k=0}^{p-1}\binom{p-1}{k}p^{p-1-k}(-p+1)^k}{p}
\frac{A^{p-1}-1}{p}=\sum_{k=0}^{p-1}\binom{p-1}{k}p^{p-2-k}q^k-\sum_{k=0}^{p-1}\binom{p-1}{k}p^{p-2-k}(-p+1)^k
\frac{A^{p-1}-1}{p}=\sum_{k=0}^{p-1}\binom{p-1}{k}p^{p-2-k}\{q^k-(-p+1)^k\}
The following is worth mentioning..
\binom{p-1}{p-1}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}=
\frac{\binom{p-1}{p-1}\{q^{p-1}-(-p+1)^{p-1}\}}{p}
\{q^{p-1}-(-p+1)^{p-1}\}
 is divisible by p
.
Wim Coenen (talk) 18:03, 7 January 2023 (UTC)
thread_title:  Fermat's little theorem unravelled. 

message: A,B,p,q are whole numbers.
p is a prime number.
A\neq qp
You choose a random number B.
Then there is always a number  A=B+p
A=B+p
\frac{A^{p-1}-B^{p-1}}{p}=\frac{(B+p)^{p-1}-B^{p-1}}{p}
\frac{A^{p-1}-B^{p-1}}{p}=
\frac{\textstyle\sum_{k=0}^{p-1}
\displaystyle\tbinom{p-1}{k}{B^{p-1-k}p^k-B^{p-1}}}{p}
\frac{A^{p-1}-B^{p-1}}{p}=
\frac{\textstyle\sum_{k=1}^{p-1}
\displaystyle\tbinom{p-1}{k}{B^{p-1-k}p^k+B^{p1}-B^{p-1}}}{p}
\frac{A^{p-1}-B^{p-1}}{p}=
\frac{\textstyle\sum_{k=1}^{p-1}
\displaystyle\tbinom{p-1}{k}{B^{p-1-k}p^k}}{p}
\frac{A^{p-1}-B^{p-1}}{p}=
\textstyle\sum_{k=1}^{p-1}
\displaystyle\tbinom{p-1}{k}{B^{p-1-k}p^{k-1}}
A^{P-1}-B^{p-1} is divisible by p.
Wim Coenen (talk) 19:36, 7 January 2023 (UTC)
thread_title:  Adding to Fermat's little theorem unravelled.