See this proof:

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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## Comments

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TopNewestI 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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Sieve of Eratosthenes is the fastest way, but not the elegant way I think.

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I don't suppose you want a Computer Code , do you sis ?

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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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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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article which I read just over a week ago .

I don't have the link to the proof , but here's the link to theThere'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 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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I think it should be a "she" instead of "he" .

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@Pi Han Goh is female? :o

What the *$#@??Log in to reply

## MAIL

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Maybe....

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Why mot try AKS?

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primality test on this planet and two others

The bestLog in to reply

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How do you expect someone to expand a binomial to the power \(83\) quickly? Ain't nobody got time for that! :3

Image

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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Why'd you expect to run the algorithm by hand?

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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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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

proveit.Computers

can beandhas beenused to create rigorous proofs. Unfortunately, at least 80% of all people outside proffessional mathematics think otherwise.Log in to reply

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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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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!

Interesting observation, I totally forgot about AKS. I was consideringLog in to reply

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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 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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What about Fermat's little theorem?

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