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Is there a better way to prove that \(83\) is a prime number?

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!

Note by Pi Han Goh
2 years, 1 month ago

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Sieve of Eratosthenes is the fastest way, but not the elegant way I think. Marc Vince Casimiro · 2 years, 1 month 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, 1 month 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, 1 month ago

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@Agnishom Chattopadhyay Yes, I'm looking for something else, \(5\) operations is getting a little tedious, don't you think? Pi Han Goh · 2 years, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Prasun Biswas Why mot try AKS? Agnishom Chattopadhyay · 2 years, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Prasun Biswas 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, 1 month ago

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@Pi Han Goh I think the simplest thing to do here is trial division because of the small input. Agnishom Chattopadhyay · 2 years, 1 month ago

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@Prasun Biswas 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, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Prasun Biswas 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, 1 month ago

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@Agnishom Chattopadhyay \(\mathbb{Z}/n\mathbb{Z}=\{[0],[1],[2,],\ldots,[n-1]\}\)

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, 1 month ago

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@Prasun Biswas

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? Nope

It's alright and I apologise for getting excited too. Agnishom Chattopadhyay · 2 years, 1 month ago

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@Agnishom Chattopadhyay No need to get offended. I was merely expressing my opinion. Prasun Biswas · 2 years, 1 month ago

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@Agnishom Chattopadhyay Sorry, but what is "AKS" ? Azhaghu Roopesh M · 2 years, 1 month ago

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@Azhaghu Roopesh M The best primality test on this planet and two others Agnishom Chattopadhyay · 2 years, 1 month ago

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@Agnishom Chattopadhyay Thanks dude, I didn't know about it :) Azhaghu Roopesh M · 2 years, 1 month ago

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@Prasun Biswas Maybe.... Pi Han Goh · 2 years, 1 month ago

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@Pi Han Goh You look like a dude though. :P Prasun Biswas · 2 years, 1 month ago

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@Pi Han Goh \(\ddot\smile\) Azhaghu Roopesh M · 2 years, 1 month ago

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@Prasun Biswas I think it should be a "she" instead of "he" . Azhaghu Roopesh M · 2 years, 1 month ago

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@Azhaghu Roopesh M What the *$#@?? @Pi Han Goh is female? :o Prasun Biswas · 2 years, 1 month ago

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@Prasun Biswas

MAIL

Pi Han Goh · 2 years, 1 month ago

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@Pi Han Goh But your comment sounds \(\large LAIM\). :3 Prasun Biswas · 2 years, 1 month ago

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@Prasun Biswas I think you like to irritate people, don't you ? No offence, really :) Azhaghu Roopesh M · 2 years, 1 month ago

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@Azhaghu Roopesh M Well, doesn't my profile pic already answer that question? Prasun Biswas · 2 years, 1 month ago

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@Prasun Biswas Yeah, should have guessed so :) Azhaghu Roopesh M · 2 years, 1 month ago

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@Azhaghu Roopesh M :P Prasun Biswas · 2 years, 1 month ago

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@Pi Han Goh "Mail" ???? Azhaghu Roopesh M · 2 years, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Azhaghu Roopesh M 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, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Agnishom Chattopadhyay 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, 1 month ago

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@Megh Choksi 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, 1 month ago

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@Agnishom Chattopadhyay Yes now too I am not understanding it. Megh Choksi · 2 years, 1 month ago

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@Megh Choksi Which part are you not able to comprehend > Azhaghu Roopesh M · 2 years, 1 month ago

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@Agnishom Chattopadhyay Thanks Megh Choksi · 2 years, 1 month ago

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@Megh Choksi Do you know the correct version lf the argument? Agnishom Chattopadhyay · 2 years, 1 month ago

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@Agnishom Chattopadhyay Nope , I made it incorrect . Megh Choksi · 2 years, 1 month ago

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I don't suppose you want a Computer Code , do you sis ? Azhaghu Roopesh M · 2 years, 1 month ago

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What about Fermat's little theorem? Joel Yip · 1 year, 9 months ago

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