Question

Playing with the Goldbach Conjecture and the Arithmetic Derivative.

Answer 1

*

Answer 2

So I noticed that since the goldbach conjecture says the sum of two primes is equal to every even number, so we could rewrite it as: \[\Large 2n=p+q=(pq)'\] So now we can evaluate that derivative. A quick remark is that pq has 4 divisors, 1, p, q, and pq so we know the only kinds of numbers that satisfy this are \[\Large \tau(pq)=4\]

Now let's make it a little more interesting, since we can write the first line by dividing through by 2 we have: \[\Large n = \frac{p+q}{2}=\frac{(pq)'}{2}\] So equivalently we can say every integer greater than 1 is the average of two primes. But in order to be an average we have to take two numbers that are equal distant from n to keep the same average: \[\Large n = \frac{(n+r)+(n-r)}{2}=\frac{p+q}{2}\] So here I've written "r" to mean radius away from n. So a quick example: \[\Large 8=\frac{(8-3)+(8+3)}{2}=\frac{5+11}{2}\] However we can rewrite this by noting p=n+r and q=n-r to get: \[\Large n= \frac{(n^2-r^2)'}{2}\] Which is now a new way of restating the Goldbach Conjecture, that there exists some radius when these are equal.

2 more things coming and then I'm back to playing around.

Answer 3

First thing to note, we can see this statement obviously holds true for primes when r=0: \[\Large \frac{(p^2-0^2)'}{2}=\frac{2p^{2-1}}{2}=p\] So that's good to know, we can average the same prime to get the original prime, no surprise there really.

The second thing to note is we can rewrite this another way: \[\Large n=\frac{(n^2-r^2)'}{2} \] as \[\Large n \frac{(n+r)'+(n-r)'}{2}+r \frac{(n+r)'-(n-r)'}{2}\] Which has a commn looking and slightly suggestive form of even/odd. At this point I'm just playing around, and thought it would be nice to say all this to help me get my thoughts straight. But if you're curious feel free to ask anything or help brainstorm new ideas. Just playing around at this point.

Answer 4

I suppose technically when I said \[\Large \tau (a)=4\] we can say \[\Large \tau(a)=3\] when p=q, however this implies the number is already prime so it's not interesting to me, and that entire case is covered as being the arithmetic derivative of a perfect square of a prime like I showed in the last post. So some fairly flimsy examples of this are \[\Large \frac{4'}{2}=2, \ \frac{9'}{2}=3 , \ \frac{25'}{2}=5 , \ etc...\]

Answer 5

Lots of people here for 30 minutes and no questions! Feel free to ask anything, I'd be more than happy to explain anything you might not have heard about before or explain anything I rushed over.

Answer 6

\[\Large n \frac{(n+r)'+(n-r)'}{2}+r \frac{(n+r)'-(n-r)'}{2}\]

should be

\[\Large n \frac{(n+r)'+(n-r)'}{2}+r \frac{(n-r)'-(n+r)'}{2}\]

right ?

Answer 7

and how did you get this relation
\[ n= \frac{(n+r)+(n-r)}{2} = \frac{(n^2-r^2)'}{2}\]
?

Answer 8

i was stuck at those two steps

Answer 9

Oh I don't know maybe I did write it slightly wrong, I will go through the derivation here then: \[\large \frac{(n^2-r^2)'}{2}= \frac{[(n+r)(n-r)]'}{2}= \\ \large \frac{(n+r)'(n-r)+(n+r)(n-r)'}{2} \\ \large \frac{n(n+r)'-r(n+r)'+n(n-r)'+r(n-r)'}{2} \\ \large n \frac{(n-r)'+(n+r)'}{2}+r \frac{(n-r)'-(n+r)'}{2}\]

Ahh so yes, I did write it down wrong, pesky negative sign lol.

Answer 10

im still wondering how u got that relation

Answer 11

\[ n= \frac{(n+r)+(n-r)}{2} = \frac{(n^2-r^2)'}{2}\]

This relation comes from the fact that (n+r) and (n-r) are both primes.

Answer 12

Got you so far!

Answer 13

For anyone confused, since I already typed this up: So if we replace n+r=p and n-r=q in here we have: \[\Large n = \frac{p+q}{2} = \frac{(pq)'}{2}\] and I guess an add on to earlier is that if \[\Large \tau (n^2-r^2)=4\] and \[\Large 2 \nmid (n^2-r^2)\] then it satisfies the goldbach equation \[\Large \frac{(n^2-r^2)'}{2}=n\]

Answer 14

assuming \(p \ne q\) we have \(\tau(pq) = \tau(p) \tau(q) = 2\cdot 2 = 4\)

Answer 15

Actually that might be even more confusing but really it's simple. It's only saying that (n^2-r^2) has only 2 different prime divisors and neither of them is 2. This ensures that it will be divisible by 2 since this is the general case as most primes are odd. Since an odd plus an odd is always even, it works out. The only time 2 is used is in showing that 2 is the average of two primes... both of which are 2 itself. XD

Answer 16

Hmm that is an interesting relation I didn't know that's how that function worked but I totally see it now! =D

As an extra thing, it seems like this is true? \[\Large 2 \nmid \tau(a) \implies \sqrt a \in \mathbb{Z}\]

Answer 17

that looks right as that just says all the exponents in prime factorization of \(a\) are even

Answer 18

\[\tau(3^4\cdot 5^2 ) = (4+1)(2+1) = 15 \]

but i feel we're digressing from goldbach
i dont see how this function is useful yet..

Answer 19

No you're right, we are definitely digressing. But I am not really sure how to move forward from here either lol. I feel like there is nothing really special about the even/odd derivative functions with the coefficients n and r. When the terms are both primes, then the r term is always 0.

Answer 20

I guess there isn't a whole lot of patterns we can work with in the arithmetic derivative. It just feels like a dead end.

Answer 21

This does sort of fit in with the twin prime conjecture when r=1 if that's of any interest.

Answer 22

this statement using derivatives looks cool... im not good with derivatives so idk how to use them cleverly.. but it feels like we can express the properties concisely using these

Answer 23

\[2n = (pq)'\]

proving existence of solution for above eqn is same as proving goldbach conjecture, i think ?

Answer 24

is it true that \(2n = a'\) always has a solution ?

Answer 25

Yeah, that's it. However I think expressing it in terms of a "radius" gives us the ability to make statements about symmetry. My intuition says there's no symmetry of the primes around any number so it should help.

If the goldbach conjecture is wrong, there is a number where when you look at the radius away it means that EVERY prime a radius away has a corresponding composite number on the other side a radius away. This seems kind of unlikely, but it gives us something to possibly look for.

Answer 26

I'm not sure about your statement, but here are some useful facts you can use\[\Large a = \prod_i p_i^{x_i} \\ \Large a' = a \sum_i \frac{x_i}{p_i}\] I'll try to figure it out.

Answer 27

Actually I know what you're saying is wrong because \[\Large 6' = 2+3=5\] However if we restrict ourselves a little this can change.

Answer 28

thats not what i was asking

Answer 29

\[2n = a'\]
Is every even number a derivative of some integer ?

Answer 30

thats like a weaker version of goldbach conjecture

Answer 31

Oh I see, so it allows for every even number to be the sum of nonprime numbers as well.I think we could probably prove that. Additionally you should kep in mind that we have things like this: \[\Large 16=39'=55'\]

Answer 32

im stuck at finding a solution for \(2 = a'\)

Answer 33

save 2, it looks like every other even integer equation is solvable

Answer 34

I don't think there are any solutions to that. The general way I sort of play around with is with primes, so try:
\[p' =1 \] doesn't work, how about 2 primes? \[(pq)' = pq'+p'q=p+q\] The smallest two primes you can have are 2 itself, so the minimum value p+q can have is 4. So in fact there are no solutions to these equations: \[a'=2 \\ a'=3\]

Answer 35

Oh makes sense still getting used to these creatures xD

Answer 36

This is in line with what the GC says, so maybe this is a good interesting strength of using arithmetic derivatives, who knows? Yeah it's weird haha.

Answer 37

As an aside, there are many interesting things going on with arithmetic derivatives that make them fun t play with, like this: \[\Large (p^p)' = pp^{p-1}=p^p\] In a way, it's like e^x haha.

Answer 38

Here, this might be something we can play with or think about:

If the twin prime conjecture is true then that means p and p+2 are primes so that implies that this has infinitely many solutions: \[\Large n''=1\]
we pick n=2p so that: \[\Large n =2p \\ \Large n' =2+p \\ \Large n''=1\] It's quite odd to see haha.

Answer 39

Oh if p and p+2 are primes and if there are infinitely many twin primes then thats equivalent to saying n''=1 has infinitely many solutions. pretty neat xD

Answer 40

Yeah actually the requirements for the goldbach conjecture are quite similar, maybe I can rephrase it to look like this.

Answer 41

interesting \(p^p\) is the only form of numbers that satisfy the eqn \(n' = n\)

Answer 42

Here's a problem I came up with on my own, it has kind of a dubious answer but as long as you know there's a trick I don't feel guilty about putting it: \[\Large n'=n^2\]

Answer 43

euler would plugin \(n = p^{ap}\) and see what happens hmm

Answer 44

We can rewrite the GC as \[\Large \frac{p+(p+2r)}{2}=n=p+r\]

So now we can rewrite the goldbach conjecture with q=p+2r as \[\Large (pq)'=2n \\ \Large [p(p+2r)]'=2n \\ \Large 2r+p+p(p+2r)'=2n\] And when r=1 we essentially have the twin prime conjecture wrapped up into this.\[\Large 2+p + p = 2n \\ \Large p+1=n \\ \Large (p+1)'=1= n'' \ (TPC)\] On the other hand if p+2r=q is prime, then it simplifies down to the goldbach conjecture\[\Large q+p=2n\]

So it definitely relates the two conjectures into a single formula which is nice. =D

Answer 45

how is (p+1)' = 1 ?

Answer 46

You have a good point not only is that wrong, I just noticed I wrote that part really weird too. I should have written that completely differently.

Since all twin prime numbers are p and p+2 then p+1 is the average of them, meaning that the number (p+p+2) is an even number satisfying the goldbach conjecture. So to rewrite it I should put:

Original equation
\[\Large \frac{[p(p+2r)]'}{2}=n\]

Twin prime condition r=1 \[\Large [p(p+2)]'=2n\]
\[\Large p'(p+2)+p(p+2)'=2n\] Since p and p+2 are primes, then \[\Large p+2+p=2n \\ \Large p+1=n\]

So maybe I need to work on this more so that it looks more like a differential equation that simplifies down to either one better.

Answer 47

I see... so `goldbach` and `n''=1 having infinitely many solutions` are not equivalent

goldbach implies that n''=1 has infinitely many solutions

not the other way. i misinterpreted previously..

Answer 48

im talking about goldbach and n''=1 equation

Answer 49

It's sort of sticky, because neither conjecture really proves the other.

If we prove the goldbach, that doesn't necessarily imply that there are infinitely many twin primes... Although all twin primes when added satisfy the goldbach conjecture, they are just special cases of radius=1.

If we prove the twin prime conjecture, all that means is that we will always be able to find an even number that satisfies the goldbach conjecture and that it will be the sum of twin primes, but not all goldbach numbers are the sum of twin primes.

So they are strongly related and in a sense it might be easier to prove both of these at the same time rather than separately as long as we can build up a perspective that makes sense.

The Goldbach conjecture is essentially saying that every even number has a "twin prime" relationship, except that they might not be n-1 and n+1 but further out. So twin primes truly seems to me to be the simplest part of this larger picture.

Answer 50

ohhhhh sorry so yes you are correct #_# I need to work on my reading comprehension skills.

Answer 51

no wait, you said twin primes ==> n''=1 has infinitely many solutions right ?

Answer 52

Yes, so I take that back. Goldbach true doesn't imply twin primes true, so it doesn't imply n''=1 is true either.

Answer 53

yeah so my previous two replies are just nonsense :P

Answer 54

p, p+2 are primes implies n''=1 has infinitely many solution

the converse is not necessarily true

Answer 55

Yes, when n=2p

Answer 56

Other conjectures like primes being separated by a prime number of numbers like 5 is satisfied by n=5p, so we have n'=5+p and then if 5+p is prime we have n''=1 again. But it fails if primes are composite distances apart like 4. So we'd have n=4p, n' = 4+4p unfortunately. Kind of weird, maybe we can establish _meaning_ if we can show that there are no primes that are composite number apart from each other... Well with some modifications.

I suppose I am really looking for some meaningful interpretations of arithmetic derivative since it seems up for grabs still.

Answer 57

*

Answer 58

An interesting pattern, start here: \[\Large a =2^4 \\ \Large a' = 2a\] \[\Large (pa)'=p'a+pa' = a+p2a = a(2p+1)\] So now I'm playing wth how long of chain can I go? \[(2a)'=5a \\ (5a)' = 11a \\ (11a)' = 23a \\ (23a)'=47a \\ (47a)' = 95a \\ (2a)''''=47a\] So I stopped since the fifth derivative gave me a composite number. We can start from another prime however: \[(3a)' = 7a \\ (7a)' = 15a\] Not so interesting lol. Anyways I think this concept of kind of using "eigenvalue" things might be useful.