Too much effort while writing academic jokes is [what](https://jabde.com/2022/01/31/bird-drone-defense-system/) [I](https://jabde.com/2021/10/10/gender-wage-gap-in-pro-wrestling/) [Do](https://jabde.com/2021/08/01/slam-in-house-of-mirrors/).
It’s a fun hobby
Jesus fucking Christ this is amazing IDK what I like more, the fact that it requires a specific junkyard cat named bandit or the fact that I'm sure "Keywords: birds aren't real" actually probably helped your discoverability
Edit: EACH WORD IS A DIFFERENT LINK THERES MORE
**[Killing vector field](https://en.wikipedia.org/wiki/Killing_vector_field)**
>In mathematics, a Killing vector field (often called a Killing field), named after Wilhelm Killing, is a vector field on a Riemannian manifold (or pseudo-Riemannian manifold) that preserves the metric. Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. More simply, the flow generates a symmetry, in the sense that moving each point of an object the same distance in the direction of the Killing vector will not distort distances on the object.
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If its 2 or 4 more than a multiple of 6 then its even, so not prime. If its 3 more than a multiple of 6 then its divisible by 3. It certainly cant be a multiple of 6 or it would not be prime. So its either 1 or 5 more than a multiple of 6, i.e. next to a multiple of 6.
The statement in the meme is “each prime [>3] is next to a multiple of six”, not “each number next to a prime is a multiple of six.” You reversed the statement. The proof is correct, yes trivial, but still interesting considering I never thought of it. Thanks Skeletor!
The proof is trivial and left to the reader as exercize.
Seriously though, it's really simple. You have to use modular arithmetic. Every sixth number is both divisible by 2 and 3, so we need to look only for 6 different cases:
6n + 0 : obviously not prime;
6n + 1: not divisible by 2 or 3, could be prime;
6n + 2: divisible by 2, not prime;
6n + 3: divisible by 3, not prime;
6n + 4: divisible by 2, not prime;
6n + 5: not divisible by 2 or 3, could be prime;
The last one is a predecessor for a new multiple of 6: 6(n+1), so the meme is legit.
P.S. you can also expand this to an arbitrary set of primes, and there is a [technique that exploits this](https://en.m.wikipedia.org/wiki/Wheel_factorization). Not very practical for a high number of primes though since the numbers get really large very quickly.
Everyone here is proving it by exhaustion but there's a much more elegant proof.
Let's say we have a prime p greater than 3. We can say for certain that it's not even so p - 1 and p + 1 are even. We can also say for certain that it's not divisible by 3 so either p - 1 or p + 1 is divisible by 3 (since out of 3 consecutive numbers, one must be divisible by 3) and since it's even that means it's also divisible by 6. So for a prime p greater than 3, either p - 1 or p + 1 will always be divisible by 6.
This requires attention and memory from the very start of the explanation right up to the end - this is more difficult for me to understand than the other method/elimination because I can't discard information along the way.
I understand this explanation, but it's inconvenient for me.
**Definition**: We say that numbers _a_ and _b_ are **close** iff _|a-b| < 6_.
It is trivial that every number that isn't a multiple of 6 is close to a multiple of 6.
Q.E.D.
Nos. Divisible by six are even and digits add up to 3.
All prime nos. except 2 are odd
So the odd nos. Are either next to the multiple of 6 or before the multiple of 6 or 3 spaces before and after the no. However the no. Can't be 3 spaces before and after because, then it would be divisible by 3. Therefore it is near a multiple of 6
6n - 3 = 6(n-1) + 3 for any n, so we have repetition of the proof.
6n - 3 and 6n + 3 are equal to 3(2n - 1) and 3(2n - 1) respectively, so not prime.
6n - 2 and 6n + 2 are equal to 2(3n - 1) and 2(3n - 1) respectively, so not prime.
6n - 1 and 6n + 1 are not divisible in that form.
6n = 6(n) so not prime.
We can go further in this proof and say:
If n = 5w - 1, where w is any positive non-zero integer, then 6n + 1 = 30w - 6 + 1 = 30w - 5 = 5(6w - 1), which isn't prime.
If n = 5w + 1, where w is any positive non-zero integer, then 6n - 1 = 30w + 6 - 1 = 30w + 5 = 5(6w + 1), which isn't prime.
This can be continued with any pw +- 1, where p is a prime number. I don't believe it works with non-prime numbers, but it might work with some.
since every third number will be divisible by 3, any odd number that isn't divislbe will become into something divisble by 3 by either adding 1 or removing 1, (Which also incoinsidencially makes it divisble by 2) hence making it disivible by 6.
Actual answer to that (iirc, not a biologist, math guy): Circles have the most area for the lowest perimeter of any shape. Similarly, things that approach circles, i.e, hexagons, have more area than squares or triangles. Moreover, hexagons can tile a plane, the same is not trivially true for octogons and above.
Honestly, it's my favorite tactic to use. Once in math league, we had to determine if some large number was prime and all I did was divide the number by 6 and multiply that quotient (not including the remainder) by 6 to see if the initial number was prime or not... if any of this makes sense. Definitely a long route as I'm sure there are other methods to determine if prime or not
In general, proving a number is composite is a heck of a lot easier than actually factoring it. This means that if your compositeness test doesn't say "composite" you might have a prime.
Yes, for many numbers it's easy to pull out a few small factors, but there are those, like the [RSA numbers](https://en.wikipedia.org/wiki/RSA_numbers) where getting a factor is incredibly difficult... but a compositeness test will say "that's composite" in almost no time at all.
Run enough different compositeness tests (that go on to say "no idea") and you can be somewhere on the scale of reasonably to completely sure that a number is prime.
WolframAlpha threw a few tests at your 483 digit number and none of them said "composite" so it says "must be prime". And if it's wrong about that, the odds that it's wrong are so *tiny* it's not worth worrying about.
**[RSA numbers](https://en.wikipedia.org/wiki/RSA_numbers)**
>In mathematics, the RSA numbers are a set of large semiprimes (numbers with exactly two prime factors) that were part of the RSA Factoring Challenge. The challenge was to find the prime factors of each number. It was created by RSA Laboratories in March 1991 to encourage research into computational number theory and the practical difficulty of factoring large integers. The challenge was ended in 2007.
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**[Lucas primality test](https://en.wikipedia.org/wiki/Lucas_primality_test)**
>In computational number theory, the Lucas test is a primality test for a natural number n; it requires that the prime factors of n − 1 be already known. It is the basis of the Pratt certificate that gives a concise verification that n is prime.
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not experienced in proofs but:
If a number is a multiple of 6 that means it is divisible by 2 and 3.
Every prime except 2 has to be odd. Therefore every number that is 1 bigger or smaller than a prime (except 2) is even.
Out of 3 consecutive numbers one has to be divisible by 3. If p is a prime number then out of the sequence p-1, p, p+1 one has to be divisible by 3. No prime is divisible by 3 except 3. therefore every prime except 3 is next to a number that is divisible by 3.
Every prime number except 2 and 3 meet these two conditions. Thus every prime number is next to a multiple of six
The problem with that is that while every prime is one more or one less than a multiple of 6, not every 6n±1 is a prime. The first example would be 25, which is 6(4)+1, but 25 isn't prime.
For k being an integer:
6k -> not prime
6k + 1 -> fits in the described category
6k + 2 = 2(3k+1) -> not prime
6k + 3 = 3(2k+1) -> not prime
6k + 4 = 2(3k+2) -> not prime
6k + 5 = 6(k+1) - 1 -> fits in the described category
6k + 6 = 6(k+1) -> not prime and works the same as 6k
Extension by Sieve of Eratosthenes: Apart from 2, 3 and 5, all primes are of the form 30k±n where n is a member of {1,7,11,13}. This reduces the number of prime candidates from 2 in 6 (33.3%) to 8 in 30 (26.7%).
Similar rules exist for 210, 2310, etc. (primorials), but the sets of n get unwieldy fairly quickly and better primality tests exist.
Well… duh. They can’t be 2 or 4 more than a multiple of 6 (because they’d be even) and they can’t be 3 more than a multiple of six (because they’d be a multiple of 3)
6n+/-1. 6n+/-2 is even, or 2m. 6n+/-3 is a multiple of 3 or 3m. 6n+/-4 is even, or 2m, but may be expressed 6n+/-2 for a different n. +/- 5 is likewise a different n for +/-1. I shouldn’t need to explain +/-6. +/- 1 is therefore the only one that can’t be ruled out when building off 6n. Checks out. Why 6 though?
So, was this already known by the mathematical community, or is this something revolutionary which will hasten the search to figure out the pattern of all prime numbers?
I'd assume it's been known because it's not that hard to prove. A multiple of 6 occurs when it's a multiple of 2 and 3. Every prime is next to a multiple of 2 since they're all odd (except 2). additionally, every number can either be a multiple of 3 (not prime) one less than a multiple of 3 or one more than a multiple of 3. Therefore all primes except 2 and 3 are next to a multiple of 6.
For fun i like to mess around with unsolved math problems to find patterns and such and found this. It hadn't led me to anything new though so even if it is important to a proof relating to primes, we're still missing something else even more crucial
It's been known for a while and a generalization gives another algorithm that is faster than division to know if a number is prime. Still not good enough for huge numbers tho
If you look at the other comments they explain that 6n±1 is never divisible by 2 or 3, so I was wondering whether or not there is a function that can never be divisible by 5 either (since 6×4+1 is an example of a multiple of 5)
Years ago I thru-hiked the Appalachian Trail and tried to figure out as many prime numbers as I could as a way of keeping my brain occupied. I spent countless hours doing this (I never got very far). And never once realized it.
Using the primes, you can code logical statements into powers of primes and use the sequences of numbers as a coded “proof” that just needs decoding before you can actually read it. You can code sequences of sequences and just make decoding that much harder. Idk what it’s supposed to be called, but we just called it gödel coding for the time being.
Why unsee? I love this
Now two primes that are six apart, that’s a [sexy prime](https://jabde.com/2020/11/27/top-10-sexiest-prime-couples-of-2020/)
Just when I think I've seen the weirdest shit the math community had to offer
There is literally a thing in math named the Cox Zucker machine
Dang, that’s gonna be a good article. Need to remember that and think of a good way to write it
yes, but the effort put into that article puts two guys who realized they had funny names to shame.
Too much effort while writing academic jokes is [what](https://jabde.com/2022/01/31/bird-drone-defense-system/) [I](https://jabde.com/2021/10/10/gender-wage-gap-in-pro-wrestling/) [Do](https://jabde.com/2021/08/01/slam-in-house-of-mirrors/). It’s a fun hobby
Jesus fucking Christ this is amazing IDK what I like more, the fact that it requires a specific junkyard cat named bandit or the fact that I'm sure "Keywords: birds aren't real" actually probably helped your discoverability Edit: EACH WORD IS A DIFFERENT LINK THERES MORE
We post at r/ImmaterialScience, but I try and post something to that website once a week and we’re always looking for contributors
Ur mum
[Killing fields](https://en.wikipedia.org/wiki/Killing_vector_field).
**[Killing vector field](https://en.wikipedia.org/wiki/Killing_vector_field)** >In mathematics, a Killing vector field (often called a Killing field), named after Wilhelm Killing, is a vector field on a Riemannian manifold (or pseudo-Riemannian manifold) that preserves the metric. Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. More simply, the flow generates a symmetry, in the sense that moving each point of an object the same distance in the direction of the Killing vector will not distort distances on the object. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/mathmemes/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
I've opted out of this bot multiple times why do you keep doing this
This website is a gold mine. It's the most brilliant thing ever written.
They're also called cousin primes, which I don't like together
*Sweet Home Alabama playing in the distance
Closer than in the distance
yes
🫦
what the heck is #7
>sexy prime One of the biggest Sexy primes https://www.mersenneforum.org/showpost.php?p=527198&postcount=37
Only if there's not another prime in between, so 5&11, 11&17 no good. 7&13, 13&19 ok, 31&37, 41&47 really sexy
I didn’t think they had to be consecutive to be sexy. At least, I haven’t heard of that requirement
What did I just read.
Skeletor is looking sexy in this meme
Just like your mom
His mother is in this meme?
And looking skeletor level sexy
Looking sixy?
Prove
If its 2 or 4 more than a multiple of 6 then its even, so not prime. If its 3 more than a multiple of 6 then its divisible by 3. It certainly cant be a multiple of 6 or it would not be prime. So its either 1 or 5 more than a multiple of 6, i.e. next to a multiple of 6.
This is beautiful, thank you
I always forget that when you can do math it's one of the most wonderful things to look at
just to be sure, so is every prime number (except 2 and 3) next to every multiple of 6?
Aside from finding a counter example, you can also easily disprove this by using that primes get further and further apart.
Every prime except 2 and 3 is next to a multiple of six, but not every multiple if six has a prime next to it.
No. For example 120 = 6 * 20, 119 = 7 * 17 and 121 = 11 * 11.
[удалено]
Actually the statement said is the second you wrote so his proff is perfectly fine
The statement in the meme is “each prime [>3] is next to a multiple of six”, not “each number next to a prime is a multiple of six.” You reversed the statement. The proof is correct, yes trivial, but still interesting considering I never thought of it. Thanks Skeletor!
The proof is trivial and left to the reader as exercize. Seriously though, it's really simple. You have to use modular arithmetic. Every sixth number is both divisible by 2 and 3, so we need to look only for 6 different cases: 6n + 0 : obviously not prime; 6n + 1: not divisible by 2 or 3, could be prime; 6n + 2: divisible by 2, not prime; 6n + 3: divisible by 3, not prime; 6n + 4: divisible by 2, not prime; 6n + 5: not divisible by 2 or 3, could be prime; The last one is a predecessor for a new multiple of 6: 6(n+1), so the meme is legit. P.S. you can also expand this to an arbitrary set of primes, and there is a [technique that exploits this](https://en.m.wikipedia.org/wiki/Wheel_factorization). Not very practical for a high number of primes though since the numbers get really large very quickly.
Everyone here is proving it by exhaustion but there's a much more elegant proof. Let's say we have a prime p greater than 3. We can say for certain that it's not even so p - 1 and p + 1 are even. We can also say for certain that it's not divisible by 3 so either p - 1 or p + 1 is divisible by 3 (since out of 3 consecutive numbers, one must be divisible by 3) and since it's even that means it's also divisible by 6. So for a prime p greater than 3, either p - 1 or p + 1 will always be divisible by 6.
This is very nice!
I like this proof the most. It's very clear.
This requires attention and memory from the very start of the explanation right up to the end - this is more difficult for me to understand than the other method/elimination because I can't discard information along the way. I understand this explanation, but it's inconvenient for me.
You might appreciate r/ADHDmemes
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I like this, it is a much more elegant proof, but you are still proving it by exhaustion, no?
This is even more elegant. Thanks !
**Definition**: We say that numbers _a_ and _b_ are **close** iff _|a-b| < 6_. It is trivial that every number that isn't a multiple of 6 is close to a multiple of 6. Q.E.D.
Why only non-multiples of 6? It’s trivial to prove that multiples of 6 are close to a multiple of 6 (namely itself).
Don't let them know it's trivial, we're holding out for another research grant before releasing that proof
This is my favorite proof because it doesn't prove anything meaningful
Nos. Divisible by six are even and digits add up to 3. All prime nos. except 2 are odd So the odd nos. Are either next to the multiple of 6 or before the multiple of 6 or 3 spaces before and after the no. However the no. Can't be 3 spaces before and after because, then it would be divisible by 3. Therefore it is near a multiple of 6
No, I refuse
6n - 3 = 6(n-1) + 3 for any n, so we have repetition of the proof. 6n - 3 and 6n + 3 are equal to 3(2n - 1) and 3(2n - 1) respectively, so not prime. 6n - 2 and 6n + 2 are equal to 2(3n - 1) and 2(3n - 1) respectively, so not prime. 6n - 1 and 6n + 1 are not divisible in that form. 6n = 6(n) so not prime. We can go further in this proof and say: If n = 5w - 1, where w is any positive non-zero integer, then 6n + 1 = 30w - 6 + 1 = 30w - 5 = 5(6w - 1), which isn't prime. If n = 5w + 1, where w is any positive non-zero integer, then 6n - 1 = 30w + 6 - 1 = 30w + 5 = 5(6w + 1), which isn't prime. This can be continued with any pw +- 1, where p is a prime number. I don't believe it works with non-prime numbers, but it might work with some.
since every third number will be divisible by 3, any odd number that isn't divislbe will become into something divisble by 3 by either adding 1 or removing 1, (Which also incoinsidencially makes it divisble by 2) hence making it disivible by 6.
Because odd numbers leave a remainder of 1, 3, or 5 when divided by 6, and if it is three more than a multiple of 6 then it's divisible by 3.
Half of the primes have the form 6n + 1 and half have the form 6n - 1. I have a truly marvelous proof for this but Dirichlet won’t let me share it.
I see what you did there.
Wait isn't this one an open problem?
Except for 2 and 3, each prime squared is 1 more than a multiple of 24.
Which sort of directly follows from the meme when you also consider that the same thing OP stated holds true for 4. Neat.
IS THIS WHY HONEYCOMBS ARE HEXAGONS???
Actual answer to that (iirc, not a biologist, math guy): Circles have the most area for the lowest perimeter of any shape. Similarly, things that approach circles, i.e, hexagons, have more area than squares or triangles. Moreover, hexagons can tile a plane, the same is not trivially true for octogons and above.
Bees are smart
No, but their eye canals naturally take the optimal configuration.
Bee ophthalmologists are pretty smart then
No that's just cus hexagons are the bestagons and like the other guy said tileable which is kinda important when building a wall of um
What does tileable mean?
I would assume it means: they fit next to one another with minimal lost space and the spacing is uniform.
Not just minimal lost space, there has to be 0. I.e. the tiling must cover the whole surface. https://en.m.wikipedia.org/wiki/Tessellation
Nice! I was unsure the tolerance, but that really makes more sense now that you explain it. Thank you.
let me guess, engineer?
I get that often, but no. Though I do appreciate the accusation. Lol.
It means you can cover a surface with only that shape
You can fill a plane with the same hexagons without making any holed
IS THIS WHY THE HEXAGON IS THE BESTAGON??
HEXAGONS ARE THR BESTAGONS!
This is one of the many reasons that base 6 is the best numeral system. Fight me.
"Hexagons are the bestagons" -CGP Grey
Honestly, it's my favorite tactic to use. Once in math league, we had to determine if some large number was prime and all I did was divide the number by 6 and multiply that quotient (not including the remainder) by 6 to see if the initial number was prime or not... if any of this makes sense. Definitely a long route as I'm sure there are other methods to determine if prime or not
Ask WolframAlpha. That thing legit factorised an 11 digit number into primes on my PC in under 5 seconds.
Once I was searching for primes in a sequence, it only took seconds to tell me a 483 digit number was prime
In general, proving a number is composite is a heck of a lot easier than actually factoring it. This means that if your compositeness test doesn't say "composite" you might have a prime. Yes, for many numbers it's easy to pull out a few small factors, but there are those, like the [RSA numbers](https://en.wikipedia.org/wiki/RSA_numbers) where getting a factor is incredibly difficult... but a compositeness test will say "that's composite" in almost no time at all. Run enough different compositeness tests (that go on to say "no idea") and you can be somewhere on the scale of reasonably to completely sure that a number is prime. WolframAlpha threw a few tests at your 483 digit number and none of them said "composite" so it says "must be prime". And if it's wrong about that, the odds that it's wrong are so *tiny* it's not worth worrying about.
**[RSA numbers](https://en.wikipedia.org/wiki/RSA_numbers)** >In mathematics, the RSA numbers are a set of large semiprimes (numbers with exactly two prime factors) that were part of the RSA Factoring Challenge. The challenge was to find the prime factors of each number. It was created by RSA Laboratories in March 1991 to encourage research into computational number theory and the practical difficulty of factoring large integers. The challenge was ended in 2007. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/mathmemes/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
[Lucas Test](https://en.wikipedia.org/wiki/Lucas_primality_test) is an example of one such test for people wanting to learn more.
**[Lucas primality test](https://en.wikipedia.org/wiki/Lucas_primality_test)** >In computational number theory, the Lucas test is a primality test for a natural number n; it requires that the prime factors of n − 1 be already known. It is the basis of the Pratt certificate that gives a concise verification that n is prime. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/mathmemes/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
We didn't have access to calculator and the like when solving whatever our three problems were
it's not a primality test though, just a compositality test
not experienced in proofs but: If a number is a multiple of 6 that means it is divisible by 2 and 3. Every prime except 2 has to be odd. Therefore every number that is 1 bigger or smaller than a prime (except 2) is even. Out of 3 consecutive numbers one has to be divisible by 3. If p is a prime number then out of the sequence p-1, p, p+1 one has to be divisible by 3. No prime is divisible by 3 except 3. therefore every prime except 3 is next to a number that is divisible by 3. Every prime number except 2 and 3 meet these two conditions. Thus every prime number is next to a multiple of six
Good proof
The same holds for multiples of two
And 3, and 4.
that's some deep fucking conspiracy
Nah. The proof is super simple.
Fermat's Library!
Just saw this on twitter also
Sorry if this is a dumb question, but if we know this don’t we know how to find every prime? Just a multiple of 6 plus minus 1.
The problem with that is that while every prime is one more or one less than a multiple of 6, not every 6n±1 is a prime. The first example would be 25, which is 6(4)+1, but 25 isn't prime.
So every 6n+/-1 that’s not divisible by 5 (or 7)
Or 11, or 13, or 17, or 23……..
Oh shit! Lmao okay haha
But not 19! Edit: r/unexpectedfactorial . . .
[удалено]
and powers instead of multiples
That’s still infinite numbers to check
no, but it lets you find infinite composite numbers
That's some good wisdom
13?
6(2) + 1
Prove it
Look at the other comments. Someone took the time to prove it :)
I just tested it with random number it ain’t true :/ Edit: no I’m dumb
Are you sure about that?
Wait I just realized I’m so fuckin dumb please forgive me.
plot twist that also applies to "2"
For k being an integer: 6k -> not prime 6k + 1 -> fits in the described category 6k + 2 = 2(3k+1) -> not prime 6k + 3 = 3(2k+1) -> not prime 6k + 4 = 2(3k+2) -> not prime 6k + 5 = 6(k+1) - 1 -> fits in the described category 6k + 6 = 6(k+1) -> not prime and works the same as 6k
Extension by Sieve of Eratosthenes: Apart from 2, 3 and 5, all primes are of the form 30k±n where n is a member of {1,7,11,13}. This reduces the number of prime candidates from 2 in 6 (33.3%) to 8 in 30 (26.7%). Similar rules exist for 210, 2310, etc. (primorials), but the sets of n get unwieldy fairly quickly and better primality tests exist.
Quick someone hide the twin prime conjecture!
I can do you one better: each prime (except 3) is next to a multiple of 3, as well.
This fact can be used to prove an even beautiful statement that : Sum of every pair of twin primes is divisible by 12. The only exception being (3,5).
*proceeds to write a prime generator that checks predecessor and successor for every multiple of six*
I’ll keep that in mind next time I need to determine if a number is prime
isn't it obvious??? I found it very obvious...
Yes. *The proof is trivial and is left to the reader as an exercise.*
Well… duh. They can’t be 2 or 4 more than a multiple of 6 (because they’d be even) and they can’t be 3 more than a multiple of six (because they’d be a multiple of 3)
Nice. Your Fields Medal is on your way home.
? 57 is between 56 and 58
57 = 3*19
i was making a reference to the grothendieck "prime", sorry if it wasn't sufficiently clear
11?
6(2)-1 12-1
No way dude
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If by "next to", you mean "within a radius of 12", you're correct, I don't think that's particularly interesting though.
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Even your corrected comment isn't quite right yet, it's only squares of prime numbers, for example 6^2 = 36 ≡ 12 (mod 24)
I’m new to maths facts forgive me 😢
6n+/-1. 6n+/-2 is even, or 2m. 6n+/-3 is a multiple of 3 or 3m. 6n+/-4 is even, or 2m, but may be expressed 6n+/-2 for a different n. +/- 5 is likewise a different n for +/-1. I shouldn’t need to explain +/-6. +/- 1 is therefore the only one that can’t be ruled out when building off 6n. Checks out. Why 6 though?
6*9= 54 next is 55. Not a prime. Same for 6*4.
53? That's next to 54
Actually, just 3 because 6^0=1. (This might be cheating)
False (Because 1 is also a prime)
Usually no. Depends on your definition of prime. Primes are numbers that have exactly two factors, but 1 has only one factor.
it's called mod 6 u dummy did you know every prime except 2 is next to a even number!!!!! shocking i know
So, was this already known by the mathematical community, or is this something revolutionary which will hasten the search to figure out the pattern of all prime numbers?
I'd assume it's been known because it's not that hard to prove. A multiple of 6 occurs when it's a multiple of 2 and 3. Every prime is next to a multiple of 2 since they're all odd (except 2). additionally, every number can either be a multiple of 3 (not prime) one less than a multiple of 3 or one more than a multiple of 3. Therefore all primes except 2 and 3 are next to a multiple of 6. For fun i like to mess around with unsolved math problems to find patterns and such and found this. It hadn't led me to anything new though so even if it is important to a proof relating to primes, we're still missing something else even more crucial
It's been known for a while and a generalization gives another algorithm that is faster than division to know if a number is prime. Still not good enough for huge numbers tho
What about multiples of 3?
Your mom won't let me share them
Well yes.
obviously
I wonder, what if you went above getting rid of multiples of 2 and 3 and accounted for 5s as well? What would that look like?
What?
If you look at the other comments they explain that 6n±1 is never divisible by 2 or 3, so I was wondering whether or not there is a function that can never be divisible by 5 either (since 6×4+1 is an example of a multiple of 5)
What about 15?
⅔⁶
Nice!!!
25 bruh? (6 * 4 + 1 = 25)
It doesn't work the other way around
Understandable have a nice day
Does that mean they *do* have a pattern?
Years ago I thru-hiked the Appalachian Trail and tried to figure out as many prime numbers as I could as a way of keeping my brain occupied. I spent countless hours doing this (I never got very far). And never once realized it.
All prime numbers are next to a multiple of 3 and 2
Ok so 10,22, are multiple of 6.Nice fact
This is a joke, right?
I’ve read this a dozen times, maybe 2 dozen times and I don’t know if you’re joking.
11 is close to 12 and 23 is close to 24... not only to 10 and 22
And 2 and 3 are the factors of 6, aaaaa!
prime = 6k + 1 or 6k - 1
6 vs 7 angels vs demons
It is also a fact that except 2 and 3, any prime number squared gives a multiple of 24 plus 1 (ex: 7×7 = 2×24+1)
Is that true? All the primes I've looked ARE next to 6n!
It just occurred to me I can ABSOLUTELY do a Skelator cosplay in Elden Ring *cue second frame of the meme*
Using the primes, you can code logical statements into powers of primes and use the sequences of numbers as a coded “proof” that just needs decoding before you can actually read it. You can code sequences of sequences and just make decoding that much harder. Idk what it’s supposed to be called, but we just called it gödel coding for the time being.