See this proof:
http://cr.yp.to/talks/2003.03.23/slides.pdf
Anyone else has a better solution?
I tried Wilson's Theorem, but it's still too lengthy and full of arithmetic calculation.
I tried Miller-Rabin Test, but to prove \(41\) is another prime is another daunting task.
I tried Pocklington primality test, but there's still plenty of calculations to work on.
Help!
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Top NewestSieve of Eratosthenes is the fastest way, but not the elegant way I think. – Marc Vince Casimiro · 2 years, 5 months ago
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I may seem a bit dumb here, but isn't a prime number a number which isn't divisible by any numbers but \(1\) and itself? – Satvik Golechha · 2 years, 5 months ago
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83 is a prime number because it is not divisible by 2, 3, 5, 7 and is less than 10^2. That takes just 5 operations to show.
Why are you looking for something else? – Agnishom Chattopadhyay · 2 years, 5 months ago
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Yes, I'm looking for something else, \(5\) operations is getting a little tedious, don't you think? – Pi Han Goh · 2 years, 5 months ago
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I think that maybe he wanted a proof based on mathematical logic for the primality of \(83\) and not a computational one (albeit computational methods are the most efficient for small values). Maybe he's considering a general solution to test the primality of a number \(n\) in the most efficient and logical way possible that does not rely on heavy computations. Maybe he considered \(n=83\) to get the ball rolling.
I think I used too many "maybe"s in a single comment. :P – Prasun Biswas · 2 years, 5 months ago
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Why mot try AKS? – Agnishom Chattopadhyay · 2 years, 5 months ago
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If I remember correctly, that is a deterministic algorithm that requires you to check for as many values of \(a\) as possible [because there are infinitely many numbers coprime to a given prime] to ensure that the number taken is prime and we also have to keep \(x\) as a free variable, so the computation requires extensive use of binomial theorem when larger numbers are to be tested.
How do you expect someone to expand a binomial to the power \(83\) quickly? Ain't nobody got time for that! :3
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A formal mathematical proof is completely different from a computational (computer-assisted) one. Your approach works if someone were to check the primality using a computer program but that doesn't work in pure mathematics. Atleast, that's what I think! – Prasun Biswas · 2 years, 5 months ago
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Interesting observation, I totally forgot about AKS. I was considering Lucas Primality Test too, but then I got stumped when I have yet to prove that \(41\) is another prime, but even if I had proven that \(41\) is prime, finding a value of \(1<a<83\) such that \(a^{82} \equiv 1 \pmod {83} \) and \( a^{ 82/q} \not\equiv 1 \pmod {83} \) will be another painful task. But still not as tough as Sieve of Eratosthenes. Thank you! – Pi Han Goh · 2 years, 5 months ago
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I think the simplest thing to do here is trial division because of the small input. – Agnishom Chattopadhyay · 2 years, 5 months ago
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There is no reason to not accept a computer assisted proof as formal/pure. After all, the computer algorithms are proven to be mathematically working.
Why'd you expect to run the algorithm by hand? – Agnishom Chattopadhyay · 2 years, 5 months ago
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But it doesn't guarantee the primality. You can be extra sure by testing for several coprime values but you can never formally state that the number is prime without a rigorous proof. And no matter how much you say, a computer has its limits too.
Take the \(3n+1\) conjecture as an example. Many people have proved the validity of the conjecture for extremely high values of \(n\) but it is still an open unsolved problem in mathematics awaiting for a rigorous proof. – Prasun Biswas · 2 years, 5 months ago
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Excuse me!! The AKS test is deterministic, it definitely guarantees primality and has been rigorously proven. You have misunderstood the algorithm.
(I do not claim that I know the algorithm fully but there are definitely \( \phi (n) \) such integers less than n and just using the integers less than n suffices to work because you are working in Z/nZ)
I did not claim that the Collatz Conjecture has been proven at all. The purpose of running through high numbers was to check for a counterexample, not prove it.
Computers can be and has been used to create rigorous proofs. Unfortunately, at least 80% of all people outside proffessional mathematics think otherwise. – Agnishom Chattopadhyay · 2 years, 5 months ago
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\(\mathbb{Z}/n\mathbb{Z}=\{[0],[1],[2,],\ldots,[n-1]\}\)
where, \([k]\) is an equivalence class in the subgroup \(\mathbb{Z}/n\mathbb{Z}\), i.e., the elements of the class give a residue of \(k\) modulo \(n\).
So, the elements that are considered in the subgroup are not limited to the totatives of \(n\). Now, I haven't yet read the full proof of AKS nor do I understand it properly yet but I'd like to ask if your claim is that the test works for the whole equivalence class just by testing for the smallest non-negative element of that class?
On second thoughts, maybe it does work since we are working with a congruence relation after all.
P.S - My arguments may sound stupid sometimes, don't get offended by that. We aren't fighting over here. We're just expressing our opinions and having a friendly discussion. :) – Prasun Biswas · 2 years, 5 months ago
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It's alright and I apologise for getting excited too. – Agnishom Chattopadhyay · 2 years, 5 months ago
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No need to get offended. I was merely expressing my opinion. – Prasun Biswas · 2 years, 5 months ago
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Sorry, but what is "AKS" ? – Azhaghu Roopesh M · 2 years, 5 months ago
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The best primality test on this planet and two others – Agnishom Chattopadhyay · 2 years, 5 months ago
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Thanks dude, I didn't know about it :) – Azhaghu Roopesh M · 2 years, 5 months ago
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Maybe.... – Pi Han Goh · 2 years, 5 months ago
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You look like a dude though. :P – Prasun Biswas · 2 years, 5 months ago
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\(\ddot\smile\) – Azhaghu Roopesh M · 2 years, 5 months ago
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I think it should be a "she" instead of "he" . – Azhaghu Roopesh M · 2 years, 5 months ago
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What the *$#@?? @Pi Han Goh is female? :o – Prasun Biswas · 2 years, 5 months ago
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MAIL
– Pi Han Goh · 2 years, 5 months agoLog in to reply
But your comment sounds \(\large LAIM\). :3 – Prasun Biswas · 2 years, 5 months ago
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I think you like to irritate people, don't you ? No offence, really :) – Azhaghu Roopesh M · 2 years, 5 months ago
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Well, doesn't my profile pic already answer that question? – Prasun Biswas · 2 years, 5 months ago
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Yeah, should have guessed so :) – Azhaghu Roopesh M · 2 years, 5 months ago
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:P – Prasun Biswas · 2 years, 5 months ago
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"Mail" ???? – Azhaghu Roopesh M · 2 years, 5 months ago
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It might be unrelated , but do you know of Bertrand Russell and Alfred North Whitehead who wrote a 360 page proof for proving something as obvious as \(1+1=2\) ? – Azhaghu Roopesh M · 2 years, 5 months ago
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I've heard of that one. Do you have a link to it?
Anyhow, which part of my proof is not valid? – Agnishom Chattopadhyay · 2 years, 5 months ago
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I don't have the link to the proof , but here's the link to the article which I read just over a week ago .
There's nothing wrong in your proof , since we just have to check that there's no number less than 83 that's a factor of 83.
But I think @Pi Han Goh wants to prove that 83 is prime using some other proof , I guess so . – Azhaghu Roopesh M · 2 years, 5 months ago
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I saw a video of terance tao on prime numbers . He provided a contradictory proof that there are infinety many primes. The same proof can be applied here too.
Taking 82 decomposing as 41 . 2 and then adding one will provide us with a number which can't be divisible by any other number other than 1 and itself – Megh Choksi · 2 years, 5 months ago
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I think you have misunderstood Euclid's argument
Multiplying two prime numbers and adding one does not guarantee a prime. Consider \( 2 \times 7 + 1 \) – Agnishom Chattopadhyay · 2 years, 5 months ago
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Yes now too I am not understanding it. – Megh Choksi · 2 years, 5 months ago
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Which part are you not able to comprehend > – Azhaghu Roopesh M · 2 years, 5 months ago
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Thanks – Megh Choksi · 2 years, 5 months ago
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Do you know the correct version lf the argument? – Agnishom Chattopadhyay · 2 years, 5 months ago
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Nope , I made it incorrect . – Megh Choksi · 2 years, 5 months ago
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I don't suppose you want a Computer Code , do you sis ? – Azhaghu Roopesh M · 2 years, 5 months ago
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What about Fermat's little theorem? – Joel Yip · 2 years, 1 month ago
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