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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$\large LAIM$. :3

But your comment soundsLog in to reply

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Maybe....

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$\ddot\smile$

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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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$a$ 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.

If I remember correctly, that is a deterministic algorithm that requires you to check for as many values ofHow 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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$\mathbb{Z}/n\mathbb{Z}=\{[0],[1],[2,],\ldots,[n-1]\}$

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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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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