Twin Proofs for Twin Primes - Numberphile
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- เผยแพร่เมื่อ 26 มี.ค. 2022
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Twin primes with James Maynard: • Twin Prime Conjecture ...
Ben Sparks on the Numberphile Podcast: • The Happy Twin (with B...
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Ben Sparks must be a great math teacher. Energy, big smile, almost childlike wonder.
Are you angry 😠
Hold on, He’s British. It’s maths teacher 🤣
He does Spark some energy in his explanations...
I'm sorry for the pun. I'm a math teacher, it is stronger than me.
He reminds me of my algebra teacher, Mr. Burke, but in a negative way. Mr. Burke was constantly putting down. Make a mistake, have an eyeroll and a question about how you ever made it this far. he did not inspire. My next teacher, Mr. Hughes, he was much more of a teacher, and I landed my first A in math class, not to mention awakening an interest in learning and math specifically.
Part of twins and in his prime.
The mod9 being equivalent to digital root is insane to me, despite being such a simple proof
I think the most simple proofs can be the most mind blowing!
It's actually just the more general version of the well-known divisibility trick.
Ah, but that only works in Base 10. There are equivalents in other bases (ex. 8, 16, etc.). The same approach works in using mod [Base - 1] at which point you'll realize that our representation of written numbers is largely arbitrary and you can represent them in significantly different ways. What's important is how a number is assembled (sums, multiples, etc.), not how we write it. A prime is a prime regardless of representation.
I agree, this proof is so elegant
@@Syrange13 exactly! If you know all multiples of 9=9 it’s not surprising. I learned that on Square One TV as a child. But it’s also nice to learn the “elegant” way to prove it 🙂
I love that Brady immediately tried to make sure it wasn't just a property true of all primes! Very easy thing to miss, he's clearly used to mathematical thinking now.
HE'S LEARNING!
~James Grime
At first I thought "isn't this a property of any two numbers that differ by 2?". Turns out it is not.
Brady's conjecture may have been false, but it was an important moment.
@@radadadadee same
Should've known the TAS researcher extraordinaire was also into recreational math
What you could do is show 3 by example. Say "Pick two twin primes, for example, 3 and 5. Multiply them together to get 15, and we calculate the digital root, and we get 6. Now I'm going to make a prediction. Choose another set of two twin primes..." You've subtly eliminated the bad case, AND you have shown that you can get different results. ;)
1000iq
Why don't the pair 3 and 5 work? Because 3 is the only prime number that is divisible by 3.
At 11:54 you say, in regards to the 2 twin primes "if they are prime they are not divisible by 3". That is false; 3 is prime, and 3 is divisible by 3. I'm not being critical, just analytical! I enjoyed the video!
@@GrowlinWillie lol in the same way, almost all the primes don't end in 5. Almost all. 🙂
@@chessandmathguy Given that the number of primes is infinite, I can confidently say 0% of primes end in five!
@@smergthedargon8974 of course there are 0 primes ending in 5! since any number ending in 120 is even
"Proof by thinking" is now my favorite kind of proof.
This is a nice way to check if two known primes are twin. No, wait-
I mean, you're not wrong.
@@delofon are u daft
😂
@@Wtahc Are you?
@@oz_jones nop
If you "want the 8 to appear" naturally, define your primes as 3k+2 and 3k+4 and multiply out to 9k^2 + 18k + 8. Of course, the demystifying bit is that generated pairs that AREN'T twin primes also yield 8 as a digital root anyway. For k=7, 23 x 25 = 575, 5+7+5 = 17, 1+7=8.
That is often the case tho, with the nice patterns in primes. There is a Matt Parker on numberphile video about prime squares mod 24, that also has that same problem.
Part of the proof was that of three integers in a row, one of them can be divided by 3. In the case of n-1, n and n+1 where the first and last are primes, it has to be n that can be divided by 3.
Could a similar conclusion be said about n, n+2 and n+4? Or would you 'move the n back' after the first step?
(Sorry, English is not my first language)
@@hierismail For n+2 and n+4, the number in the middle would be n+3. If n is divisible by 3, then so is n+3, so that doesn't really change anything.
For any set of three numbers in a row, one of them MUST be divisible by 3. So if n+2 and n+4 are both prime, and therefore not divisible by 3, then the one in the middle must be, just like in the n+1 and n+3 example.
@@kainotachi oh of course! The answer was right there, thank you
@@hierismail, if k is not divisible by 3, exactly one of n, n+k, n+2k is divisible by 3.
I love every episode of Ben Sparks on Numberphile.
An informed numberphile viewer might remember another video, "Squaring Primes" from Matt Parker here. And indeed,in both cases, the first proof was an exhaustive one, while the second proof was the "prettier".
There's also casting out nines with James Grime!
The simple identity a^2 - 1 = (a-1)(a+1) is indeed a powerful idea in quite some proofs in number theory.
@@epsi Well, at least it wasn't a "Parker square" of a solution ;-)
@@monkeybusiness673😂😁
Not only is the number between twin primes divisible by 3, it also has to be even, so its always divisible by 6.
Every number is divisible by 6! All you have to do is put a slash between the number you want to divide and the 6 and voila! e.g. - 1/6
(jk, I know what you meant)
@@ConsciousExpression But that's 6, not 6! = 720
@@mmmmmmmmmmmmm
Drat! Foiled in my pedantry!
@@mmmmmmmmmmmmm but 1! and 0! are equal to 1, and that seems like witch craft
Which also means all of the numbers between twin primes are abundant since 6 is perfect, the only exceptions are 4 and 6
The algebraic proof also shows why 3, 5 doesn't fit the pattern. Since (n-1) = 3 in this pair, the conjecture that n = 3k is wrong. Neat!
Yeah! It's like 3 is the new 2.
Matt Parker agrees (in his "4th dimension" book), in that he calls 5 the first honest to-goodness prime, since it has an actual non-prime below it. 2 and 3, he says, are primes merely "by default".
@@sternmg quite justified statement given that 4 is the first number that has even a chance to have other divisors
2 has a non-prime below it. 5 has a composite number below it.
Doesn't explain the reason 3 and 5 exist as twin primes at all, though. That's simply chalked up to the happenstance of the early primes being too small to need to fit the pattern.
I like Ben Sparks the best among all the brilliant mathematicians who appear on this channel. I still cannot forget the chaos theory video and the one with the mandelbrot set from Ben. Thank you.
Man, I've loved modular arithmetic ever since I first learned about it. It opens up so many unexpected doors in maths.
Math olympiad contests feel like 90% applied modular arithmetic.
Revolving doors? (:
The second proof tells you that the product of *any* pair of numbers that are either side of a multiple of three is congruent to 8 (mod 9). E.g. 14×16=224=8 (mod 9).
You get the twin primes result as a bonus since they are always either side of a multiple of three.
You can use the same proof to prove that all cousin primes (4 apart) will have a digital root of 5, and sexy primes (6 apart) greater than 7 will have a digital root of 4
We need a Numberphile extra on these just for the fun of it
All in favor of making "sexy primes" an official classification?
I'm definitely on board.
@@idontwantahandlethoughIf twin primes are 2 numbers apart, and primes 4 numbers apart are cousins I think primes 6 numbers apart should live in the same valley following the same logic. That would make them Saarland primes in Germany, and maybe Idaho primes in the US?
*running away
@@idontwantahandlethough I'm pretty sure it's already official. At the very least, it's appeared on Numberphile more than once.
You can actually go further! All primes >3 are in the form of either 6k-1 or 6k+1. This means that:
with twin primes: (6k-1)(6k+1)=36k^2-1; they give 35 mod 36
with cousin primes: (6k+1)(6(k+1)-1)=36k(k+1)-6k+6(k+1)-1=36k(k+1)+12k+5; these give 5 mod 12
with sexy primes: (6k+1)(6(k+1)+1)=36k(k+1)+6k+6(k+1)+1=12k+7 or (6k-1)(6(k+1)-1)=36k(k+1)-6k-6(k+1)+1=36(k+1)-12k-5; they give 7 mod 12
etc.
This means that in senary (my favorite number base):
twin primes give something that ends in: 55
cousin primes give something that ends in: 5, 25, 45
sexy primes give something that ends in: 11, 31, 51
etc.
A nice reason to like senary (there are many more, but they're irrelevant to this discussion)
That was lovely. I love primes. More primes. The "3" between twin primes was awesome
Actually it's 6, because the number has to be even, too, but it doesn't affect the proof.
@@TheImpressionist235 Thanks for clearing that up
@@SaturnCanuck and 6 also means every number between twin primes would be abundant except for 4 and 6
got a bit overexcited, I thought we had gotten a twin primes proof
this is still weird and cool though!
So did I unfortunately
same here
Expect the pair 3 and 5 ofcourse!
always love the ben sparks episodes, i knew some of the ideas behind it but the way he puts the information is always so interesting
The second proof showed that although "Brady's Conjecture" was false, it has a cousin - that the digital root of the product of two integers either side of a 3 is 8 - was true. I think that although the second proof is neater, it does also highlight that it's not *really* to do with them being twin primes but rather them hugging multiples of 3.
Doesn't work with 8, 9, 10
@@arashbeheshtiro7799 8x10 = 80 which has a digital root of 8
I like the little fact that the even number between twin primes is always divisible by three. Such a small thing but it means you can cut out two thirds of your search space when looking for twin primes.
Well, 0% of integers are perfect squares yet there is still an infinite number of them.
4 isn't
This is true for all primes larger than 3, not just twin primes.
Every number between twin primes is even in fact, so if you have a twin prime you know that the number in the middle is divisible by 2, 3 and 6
You can make this even more impressive. The product of two twin primes is also actually one less than a multiple of 36. You can proof this by doing the second proof and realizing that twin primes are always 1 mod 6 and 5 mod 6, which I believe was a numberphile video as well.
I was thinking this too. The same way you can argue that one of any three consecutive numbers is a multiple of 3, it's also true that at least one of those three numbers is a multiple of 2. If the first and third number are prime, then the one in the middle must be a multiple of both 3 and 2, and therefore is a multiple of 6.
@@RunstarHomer Unfortunately it's true for any triple of numbers where the middle is a multiple of 6. Take 23,24,25 (not a twin prime). 23x25=575, digital root 17=>8. So while its interesting, it doesn't have any way of telling you whether they are a twin prime or not.
@@qchronod You must be fun at parties.
@@qchronod "if a number is between a twin prime, then the number is divisible by 6" is only a necessary condition, at least it narrowed down the search by a little bit
@@edgarleft why are you saying that
This was so much fun. In uncertain times, it’s a relief to find that numbers remain reliable.
Until you add all the positive integers together.... ;-P
The 2nd proof shows that for any number n that is divisible by 3, the digital root of (n-1)*(n-+) is 8. None of of these numbers need to be prime. For example, n = 15 or n = 21.
But that's the point. If that is true for ANY number, then is true for Twin Primes too.
Particularizing the above for a pair of twin primes bigger than 3 (which always satisfy that they're separated by a multiple of 3), then you're done. It's is enough to prove a more general statement for any particular statement contained in it to hold true 🤷🏾♂️
Does n-2 and n+2 do anything for you?
@@salvadorjacome2694 THANK YOU! That's exactly what I was missing in the second proof. I felt like I was left hanging, and wasn't looping back around to the idea that ben was proving.
As my old (white-suited) mechanical mathematics teacher always repeated after setting up the mathematical model of a physical system; "and now we just let the algebra do the donkey work."
That part of me prefers the second proof too. State the problem accurately, and just watch the solution fall out. It's magical; like it's just meant to be there. (Which, of course it is!)
But now, as a software developer, I like the brute-force "try everything" approach a bit more too.
That's more like; "This will get an answer. Your equations may have missed something, but we can be sure this covers all cases. Because we actually look at all of them!"
For the proof in this video, of course, the distinction is mostly irrelevant.
But I'm pretty sure that NASA would use the first approach to send people into space, whereas the Met Office use the second to give you a weekly forecast. (Even though that sounds totally the wrong way around!)
Sort of. Point is; a very interesting topic, and the main thing communicated (hopefully) is that the proof is as much about communication as it is about fact.
my physics teacher in HS would say something similar. "Now we have done all of the physics and we have a math problem."
@@Broan13 Did he sneer?
Like the 'all science is either physics or stamp collecting' remark?
I love the jibes between the scientific disciplines. But only one can keep the scores accurately.
@@Varksterable haha no. He was a kind, fun guy with a PhD in astrophysics teaching us HS students physics and astronomy. He was a very even tempered guy that didn't pick sides much on the sciences and other related subjects
With programming, the best way to prove that something works is to just try it and see if it works.
@@vigilantcosmicpenguin8721 Look up 'SIL'.
So this actually proves that the digital root of the product of two integers, prime or not, separated by two is always eight if neither of those two integers is divisible by three.
Yes, prime-ness is not necessary, only not-divisible-by-3-ness. Also, your statement can be "iff": digitalRoot(n * (n+2)) = 8 if *and only if* both n and n+2 are not divisible by 3.
@@theadamabrams Nice!
*Simplest proof*
Twin primes greater than 3 are around a multiple of 2 (even number in the middle) and 3 (one of the (n-1), n and (n+1) has to be 0 mod 3).
Therefore, the *general formula of a twin prime pair* is (6k-1) and (6k+1) starting up with k >= 1 (5 and 7 are first such twin primes > 3).
Their product = (6k-1) x (6k+1) = 36k^2 - 1 = 9 x (2k)^2 - 1 = 9 x [ (2k)^2-1 ] + 8 = 9 m + 8 starting up with m = (2k)^2 - 1 = 2*2 - 1 = 3. [ starting at 7x5 = 35 = 9 x 3 + 8 ]
Therefore, *Single digit root of twin prime = (6k-1)(6k+1) mod 9 = 8*
Twin prime: the composite number between them is divisible by 6.
add them together and they are divisible by 12.
(cousin primes' composite is divisible by 3).
12:06 "We proved by thinking..." Is such a great line!
This bring back so much memory to me. Modulo arithmetic was fascinating to me when I was in middle school.
Very well presented. Thank you.
In Portuguese, the digital root operation is informally called "noves fora" which means "nines out". 😀
Some refer to 'casting out nines' in English. :)
12:05 proof by thinking is the best type of proof
I miss my twin brother, he passed several years ago 😢. I'll see him soon.
So you can pick any integer divisible by 3 and take the 2 numbers on either side.
123,456,788 and 123,456,790
Twin proofs for twin primes? More like "Every time I watch Numberphile I always have a great time!" Thanks again so much for making all of these very high-quality videos on so many different topics within math.
If 'p' is a prime number bigger than 3, then (p^2) -1 is always divisible by 24 with no remainder. I think numberphile has touched on this before, but this video has inspired me to investigate if another Modular arithmetic fact is available along that line if thinking 🤔
p = 6k+1 or 6k-1 for all primes p except 2 and 3.
Thus, p² -1 = 36k² + 12k = 12 * k * (3k+1)
If k is even, then 24 | 12k. If k is odd then 3k+1 is even and thus 24 | 12 * (3k+1)
It is prime time for twin primes
Nice way of teaching. Thanks for making numbers easy for us.
Great video and guest.
"... just kidding, I wrote a different number under each corner of the paper. Gotcha!"
This was very inspiring. I'm a teacher as well, and especially version one of the proof is accessible for students who don't know a lot of algebra yet. I like it a lot.
Loving the Klein Bottle on the shelf behind him. I have one too 😁🤙🏻
(Here is an algebraic proof to the Collatz conjecture)
The method to be applied has only two rules: -
1. For odd numbers multiply it by 3 and add 1
2. For even numbers, we keep on dividing till we get an odd number (if it's not 1 we apply rule 1 again)
Now, for all powers of 2, we keep applying rule 2 we will eventually get to 1, which is 2^0.
(So it has to be true for all powers of 2).
All numbers which are not powers of 2, are either odd or some even number which is the product of some power of 2 and an odd number (other than 1). Now, if it's an even number we keep applying rule 2, to get to that odd number (other than 1). So, in both cases we will hit an odd number other than 1.
Now, all odd numbers can be written in the form 2x + 1, where x is any whole number. So, when we apply rule 1 to it, we are actually converting it to an even number of the form 6x + 4. (3 multiplied to an odd must give an odd number, since 3 is odd. 1 added to an odd number is even. Since the number we started with was having the form 2x + 1, the even number formed by applying rule 1 will have the form 6x + 4.)
But, when we apply rule 2 to this even number, the algebra tells we will never get the earlier odd number back again. This is because 3x + 2 > 2x + 1(unless x is 0, in which case 2x + 1, would have been 1). Even if, 3x + 2 is an even number when we divide it by two all the numbers that we get will be less than 2x + 1. Which means we will get different odd numbers every time. All these odd numbers, will have the form 2x + 1, but the x must have a different value for each one of them.
This in turn means, the even numbers that we will get by applying rule 1 will be a different number each time. This process will continue until we hit an even number which is a power of 2, which will take us back to 1.
Therefore, all numbers on which the method is applied following the rules will eventually take us back to 1.
Hence proved 🙂🙂🙂
So to reduce a number mod m, all you have to do is convert it to base m+1, find the digital root, and convert back!
The moral of the story, to me, is that this is a property of any two numbers on either side of a multiple of 3, and twin primes (other than 3 and 5) are a subset of those pairs of numbers.
You can even generalize a little bit further: given that the primes are odd, the number encircled by them is also a multiple of 2, which leads to the statement that you can only find twin primes around multiples of 6.
Better generalization is all primes greater than 3 are of the form 6k ±1
Another banger video, Brad !!!
Amazing and amazing how easy it's actually is. Started instantly some research by my own and thougth about other number systems with different bases than 10. The digital root for numbers of any integral base is modulo(base-1). For some bases you got simply a different value for the digital root and some bases like 8 have several digital roots for this problem. For base 8 for example it is 1, 3 and 6. Base 7 it's just 5. Base 7 is interesting in an other way, too. The twins are always 1 and 5 modulo to 6. The gap between them can only be 0 modulo to 6. So, the number between prime twins is not only always divisible by 3 it's divisible by 6 either, but this is pretty clear, because the number between the twins has to be even, too and so it's 6 aswell. (6k-1)(6k+1)=36k^2-1=mod9=8
Thanks to the second proof, we know that the product of twin primes will always be one less than the square of the number in between. that's kinda cool
That’s true for any pair of numbers that are 2 apart.
(n - 1)(n + 1) = n² - 1
Thinking about how digital roots work (I used to love digital roots at the age of about 8), in base x, the digital root of any number y would by y mod (x-1). This works in all cases, except for binary.
So if you were to work out the digital root in hexadecimal of 173AB (just a random value), you would have to work out 173AB mod F, which when converted to decimal becomes 95147 mod 15, which gives us 2, so the digital root of 173AB in hexadecimal would be 2.
Just another to do it so we get an answer that's not 1-9, lets do 10D. 10D mod F in base 16 is the same as the decimal expression 269 mod 15. 269 mod 15 = 14. 14 in decimal is E in hexadecimal, so the hexadecimal digital root of 10D is E (although that one could have been solved by simply adding 1 and D together).
That was excellent. Thank you.
I've a question about the mod 9 digital root proof: The digital root of 9 is 9, but 9 mod 9 = 0. Is it treated as a special case or am I grasping it wrong?
It's a special case; see my comment (coming soon)
I'm 64. The algorithm suggested this for me after I had been researching my dyscalculia. 🤣
I really enjoyed the video though. This guy must be a great teacher.
I noticed that the digital root of the product primes with a difference of four seems to come out to 5 each time.
Twin primes are of the form 6n-1, 6n+1. Multiply then gives 36n^2-1. Digit sum of 36 is 0. 0 times n^2 digit sum is 0, so digit sum of 36n^2 -1 is -1, which is 8
If you do not believe all primes are of the form 6n-1 or 6n+1, think about 6n+2, 6n+3, 6n+4 and 6n+5 which is back to being of the form 6n-1.
The product of twin primes > 3, is congruent to 8 mod 9.
Is this true of any twin primes (other than 3 & 5, of course)? Yes, because it's true of any pair of integers that bracket a multiple of 6; and all twin prime pairs do that.
(6k+1)(6k-1) = 36k² - 1 = 9(4k²) - 1, which is congruent to -1, and therefore, to 8, mod 9.
The reason twin primes always flank a multiple of 6, is that any pair of integers that differ by 2, must include a multiple of either 2 or 3, unless they are ±1 mod 6.
Which is also why (3, 5) are exceptions to the twin-prime-product rule. (A multiple of 3 can't be prime unless it is 3 itself.)
Fred
you can go one further, since the middle number must also be even, the product of twin primes is one less than a multiple of 36
(6k+1)(6k-1)=36k^2-1
Absolutely mind blown
You know...I think I need to learn how to write proofs and videos like this are going to help get me there. Thanks.
Excellent video.
There's another infinite sequence of twining:
Step 1: Take a prime pair.
Step 2: Multiply those.
Step 3: Subtract that until you reach a twin prime.
for (3, 5), I did:
_(3, 5)_
(11, _13_ )
(137, _139_ )
and
( _18_ , _919_ , 19,079)
Every time I watch one of these, I feel obligated to figure out the proof before we arrive at it. I ended up with the algebra proof and was satisfied, but humorously, hadn’t even considered just checking the cases mod 9!
I did the same when they first introduced twin primes years back. I proved that there are infinitely many as an exercise. Sadly the proof doesn't fit in the youtube comment section :(
“Proved by thinking” -Ben Sparks
Such a great teacher
Thanks for giving mr a solution to my question
The second proof makes it clear that this is a property of any two numbers which are one more and one less than a multiple of 3, prime or not. For example 20 and 22 have this property.
I did 59 and 61, the product of which has a digit sum of 26. Then I found 4 digit twin primes and their 4 million + product... also had a digit sum of 26. !
PS I've just in the last couple of weeks been experimenting with the Goldbach Twin primes conjecture (every even number larger than 4208 is the sum of two twin primes - that's primes that happen to be twins, not twin prime pairs themselves), and I found out a lot of things about twin primes, but I never suspected that every digit root of their product would be 8. I'm now going to watch the remaining 13 minutes of this video and find out that I'm being an idiot.
Edited: 9:15 Yay! This was the first proof I ever did on my own (and in base k digital root of any number n is n mod k-1)
"find out that I'm being an idiot."
For the rescue of your honour: there are some mistakes that you can only perform with a certain level of expertise.
@0:28 I like his honesty
Brady, long time viewer first time commenter here, love the channel👍( also from Adelaide) in a unrelated topic i think it would be really awesome if you could do a video about how to visulise very large numbers. i think it is(for me anyway) hard to understand how vast some of the numbers are, even a number 100 digits long is so massive. it would be awesome to have some visulisation on this. Thanks for the awesome content👍
I see our resident "Maximus Mathmaticus" is back...and yes we are entertained!
He's the best math teacher I've ever seen
Very beautiful proof
Yes, if 2 is the difference between two primes (unless one of them is 2 or 3), ie if they're twin primes, then the product of their digiroots is always 8. (Bearing in mind this applies to other number pairs of the form 3k+2 and 3k+4). If the difference is 4, such as in the example given in the vid, 7 and 11, then the pd is always 5.
If 6, then there 's no single pd since 6 is a multiple of 3. The pd's are 1, 4, or 7. If 8 then 2, if 10 then 2. 12 is another multiple of 3, so 1 4 7 again. 14, it's 5, 16 it's 8 again. So a cycle of 8 5 2 2 5 8.
If the difference is 0, such as between 13 and 13, then since 0 is a multiple of 3, then it's 1 4 7, which are the digiroots of all perfect squares except for 3 and multiples.
I play with primes alot myself and have found something amazingly simple and answers many questions about the "randomness" of primes. If you'd like to see it, message me.
Great video. Ben is slightly mistaken at 13:05. As he showed, the product is 9k^2-1. To get +8 you have to rewrite it as 9(k^2- 1)+8
Let p, p+2 be twin primes
It is easy to see that P can only be {2,5,8} mod9 meaning that p+2 can be {4,7,1} mod9
Thus p(p+2) can be {8,8,8} mod9
So p(p+2) is 8 mod9
And thus the digital root is always 8
This is the same kinda mental wavelength to me as when Matt showed the prime/24 connection
Awesome stuff gentlemen. :-)
It would've been nice if Ben had a twin and they did the twin proofs for -seven brothers- twin primes
He does have a twin, but yes, it would have been nice if they'd done one proof each!
I'll ask my bro next time. :)
soooo so awesome!!!
The product is not only always (except 3*5) 8 mod9, but also 35 mod36
Digital root for base n is equivalent to mod (n-1). Should be simple to generalize it.
Wonderful!
Long time no see, Ben!
Thanks!
Would love to see Brady’s reaction when he’s involved in the questions. Would be awesome
I prefer that second proof, with (3*k+1)*(3*k-1). Very direct and clean and inarguable.
Is there anything interesting about the amount of steps involved in the summing to arrive at the digital root?
Interesting in that the number 8 has always, like ALWAYS, been my favorite number.
The number between twin primes is a multiple of 6.
Thus, the mod9 of twin primes product comes to 8, because the product is ((6x)-1)×((6x)+1)=(6x)^2 - 1; (6x)^2 is a multiple of 9; subtract 1 and get 8.
Twin primes (except for 3,5) must be separated by a multiple of 6 ; more generally, by a multiple of 3. So (3n - 1)(3n +1) = (3n)^2 - 1 = 9n^2 - 1~ 8 (mod 9).
THIS WORKS FOR COUSIN PRIMES, Digital root is 5 for all cousin prime products. Cousin primes are always (n-2)(n+2) but n is divisible by 3. Therefore, (3k-2)(3k+2) =- 9K^2 -4 =5mod9
Twin primes? Exciting
`mod9 = digital root` is the only mathematical concept I've personally discovered
The product of twin primes (excluding 3 and 5) is also equivalent to 35 (mod 36).
Proof: All twin primes other than 3 and 5 are in the form (6n - 1, 6n + 1) where n is a positive integer.
(6n - 1)(6n + 1) = 36n² - 1, which is equivalent to 35 (mod 36).
I love Ben Sparks
i find a clock to be a great intro to modular arithmetic. Everyone knows 11 + 2 = 1. Granted, there's no zero, but I find it an intuitive place to start.
James maynard is truly a legend
Wow, I would really like to say "thank you" for this extraordinary prime number video.
Why?
Because I like to see proofs for very non-obvious things, and here we get TWO proofs for a very non-obvious thing, which is actually really exciting.
While I don't question the results I do like to think about the potential for stagecraft trickiness - the "well I already wrote my prediction so it doesn't matter if I see see the calculations" thing *could* potentially be used to trick a mark in to making any prediction work, at least for a single game, since once you're watching the calculation could just give arbitrary operations on the fly until an opportunity arose to give one that reaches whatever "prediction" you wrote down...