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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Pi Han Goh
3 years, 8 months ago
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Top NewestSieve of Eratosthenes is the fastest way, but not the elegant way I think.
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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?
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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?
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Yes, I'm looking for something else, \(5\) operations is getting a little tedious, don't you think?
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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
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Why mot try AKS?
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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!
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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!
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I think the simplest thing to do here is trial division because of the small input.
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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?
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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.
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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.
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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. :)
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It's alright and I apologise for getting excited too.
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No need to get offended. I was merely expressing my opinion.
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Sorry, but what is "AKS" ?
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The best primality test on this planet and two others
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Thanks dude, I didn't know about it :)
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Maybe....
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You look like a dude though. :P
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\(\ddot\smile\)
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I think it should be a "she" instead of "he" .
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What the *$#@?? @Pi Han Goh is female? :o
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MAIL
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But your comment sounds \(\large LAIM\). :3
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I think you like to irritate people, don't you ? No offence, really :)
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Well, doesn't my profile pic already answer that question?
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Yeah, should have guessed so :)
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:P
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"Mail" ????
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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 \(1+1=2\) ?
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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?
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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 .
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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
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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 \( 2 \times 7 + 1 \)
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Yes now too I am not understanding it.
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Which part are you not able to comprehend >
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Thanks
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Do you know the correct version lf the argument?
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Nope , I made it incorrect .
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I don't suppose you want a Computer Code , do you sis ?
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What about Fermat's little theorem?
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