Number Theory: How to find special rational numbers

You're naive.

What would you have done in my position? As I said, my paper violets arxiv's rule on relevance. And to get it into a journal, sure. I want to go and try to get it on my university's publication but to that I must first go through my professors and I personally think that they have really high standards. I don't want to give them a paper full of shit and then have them stop talking to me forever. My goal here is literally to showcase the theorems and see what people have to say.

It doesn't violate arxiv rules.
>I personally think that they have really high standards.
In mathematics the only standard you must pass is for your result to be correct and original (as in non-plagiarised, not necessarily significant or impressive).

engineer spotted

very nice user

there is a very high probability that the guy giving you advice has NEVER published a paper himself

the standards are:
>it is your own work
>it hasnt been done before
>the conclusions are correct from the given premises and of some significance
>the premises aren't too narrow or specific
>it has been written in the correct form and it has all the required secondary information such as department, sharing policy, peer review policy, funding

don't worry, if you are incapable of doing the translation to normie-language yourself, someone will just do it for you and take the fame from your discovery

>It doesn't violate arxiv rules.

It does, I know because I became obsessed one time ago and I read all of arxiv's rule. Here, I found the rule I mentioned earlier

>Material submitted to arXiv is expected to be of interest, relevance, and value to those disciplines. arXiv reserves the right to reject or reclassify any submission.

My paper is not relevant, unfortunately. I wish it were though. However, if this was an open problem that became famous then odds are someone would have solved it a long time ago. It is really simple.

Thank you. I worked really hard on this. Originally I had the mentality that perhaps I could solve this by finding solutions to the simultaneous diophantine equations that arise but then I saw a simpler way after I proved the second theorem.

In fact, the second theorem was the first I actually discovered. I first discovered the second theorem, then the third, then the first and finally the fourth. But I ordered it differently because really, they build on each other in the order I present in the paper. The reason I discovered them in a mixed order was because I was looking all over the place for solutions.

>there is a very high probability that the guy giving you advice has NEVER published a paper himself
I have never published a paper... on vixra.
Suit yourselves.

It is in the nature of research that most has little commercial value in the present but the few that do have enough value to cover the expenses for the rest. The problem being of course that one can never know in advantage where the gold is to be found.

Also some results turn out to be valuable much later.

You could rewrite it so it's more polished (read the guidelines) and submit it to this journal rose-hulman.edu/mathjournal/archives.php
It's specifically geared towards undergrads. As for the contents itself, I skimmed through your vixra file and it seems mostly ok. (You might want to upload it somewhere else though. Even your personal blog is better than vixra.)

When I was not even an undergrad, I wanted to put some shit in an academic journal to get academic street cred. I failed dramatically. It wasn't necessarily even not worthy of going to a journal with because it was definitely moderately interesting. I had typed it up in latex and attached my real name to it and everything, paying close attention to the guidelines for submission so I didn't look like a crank.

In hindsight, it was embarrassing that I wasted so much time on it without any real confirmation that this was something I should be trying to do.

Now I understand what you SHOULD do in a situation like that:

Just ask your damn professor. Tell them you would like to hear their thoughts on it.

How do you know it was correct and moderately interesting?

I knew it was undeniably interesting because I did enough research to be almost sure.

It was not a proof. This was not a mathematical journal, it was a computer graphics journal. I knew it worked because I had tested and built it. They were less than totally formal and encouraged the type of submission I had.

i feel like you're full of shit somehow.

How so? I gave you an example of the theorem in the OP. What flaws do you see in the argument?

Cool math, but your punctuation and writing needs some work.

give me a smaller example of a,b

"For this I present the dollowing definition:"

>>vixra.org/abs/1707.0392
>pdf download
>Vixra.org is a pre-print repository rather than a journal
It's mine now, brainlet. Enjoy your no fame.

Sure. But I will give you a different denominator because 123456789 is too big of a number, and actually the number I gave in the OP is the smallest solution for this denominator.

Instead let me use 12 as my denominator. Then consider the numbers
[math] a = \frac{1}{12} [/math] and [math] b = \frac{5185}{12} [/math]

Here they are in wolfram wolframalpha.com/input/?i=(1/12)^4 - (5185/12)^4 is an integer

Yeah, I found the typo. Fortunately I found it before sending it to my professor so I edited it. I have not submitted a modification for the vixra paper because as many people here say that vixra is bad, then maybe my professor will ask me to take it down so there is no reason to waste time submitting the edited version.

You can put it anywhere you want. The first version of these theorems are still under my name so I don't have to worry about that. You will just look silly by stealing from a vixra article kek.

My name is also L. Castillo. Checkmate, brainlet.

nice verry interesting! thanks for sharing

I know you are trolling but you must know that this paper is linked to me by more than just my name

(1-5185^6)/12^6 - int((1-5185^6)/12^6) = -1/96

Ah, yeah. Please do not misinterpret my results. The formula I used finds numbers such that the property holds up to the fourth power. It is not guaranteed that it will hold up to the 6th power.

However, what is interesting is that my method basically builds on top of itself. I obtained my formula for the fourth power by building on top of my formula for the 3rd power, and this can be repeated to find a formula for the fifth, sixth and so on powers.

In fact, I have some calculations that are in my notes but not in the paper. I have an example that works up to the sixth power.

Consider [math] a = \frac{7}{2}, b = \frac{39}{2} [/math]

The difference of powers will be an integer up to the sixth power. However, I know that it won't hold for the seventh power. But this is because of a reason outlined in my paper:

The set of solutions can be partitioned into the lines it belongs to. And each line has a limit for how far you can go before there are no more non-trivial solutions. Actually, in the last page of my paper I write down a sequence that describes this limit.

The point I just gave you belongs to the line [math] y = x + 16 [/math] and the limit it has is exactly 6. If you take any non-integer point in this line and take the difference of the 7th powers, the result will never be an integer.

I read it for fun, went a little over my head. pure math is insane. You really should do what other anons are saying though and ask a professor to look it over, and see if he can get you published in a reputable place. It definitely meets the requirements for being published.
t. biochem

make a formula to make a formula to the nth degree

quit making fun of our brainletism and tell us what you're up to.

I thought about this, and it is possible but I ultimately think it is meaningless.

I could build this ultra-generalized formula recursively but it would be ridiculously complicated and perhaps meaningless. I mean, just look at what you end up with my formula for n=4. It is already really insane and complicated.

I consider myself satisfied with what I achieved. As I said before, I could find formulas for n=5 and even general n, but that is actually meaningless. I already proved my method works, and it can keep working. If I went to just keep on going. I would just be repeating my method over and over again.

Heck, if you read my proof of the third theorem and the fourth theorem, they are basically the same argument, just with different meat. I think that unless I am able to find a new breakthrough that allows me to simplify my formulas then it is worthless to keep on going.

I already sent it to a professor of mine. He hasn't reponded yet so lets see how that goes. And I will keep the vixra just for the sake of the timestamp. I think it is important and as vixra does not take the rights of the paper and allows you to publish anywhere else I do not see it as a liability.

As for the math itself, it does look complicated if you do not have the right knowledge however you should know that if you ever get interested and learn about congruences and the chinese remainder theorem, you could perfectly come back and read my paper. In fact, you could potentially rediscover all my formulas once you know the right theorems. That said, I'm glad you felt like reading it.

if its only to the fourth power, then

pick a denominator d

then you can compute d^1 d^2 d^3 d^4

x^n - y^n will always have (x-y) as a factor

so if x and y are the integer parts of a and b, you just need to choose x,y such that d^1, d^2, d^3, d^4 are all factors of x-y

so there's your generalization to arbitrary n

Good observation, and in fact taking that (x-y) is a factor is a big part of my second theorem.

But you must see that your method would not produce all solutions. What I consider interesting about my method is that it presents you with a formula with some parameters. If you simply plug in any numbers in these parameters, you will find a solution.

No, it isn't. As I said, it does not generate all solutions.

To see this consider one of my previous examples, [math] a = \frac{1}{12} [/math] and [math] b = \frac{5185}{12} [/math]. This example works until n=4.

Now notice that [math] 12^4 = 20736 [/math] but [math] 1 - 5185 = -5184 [/math] so [math] 12^4 [/math] does not divide the difference a - b.

So the method presented in is valid, but it misses solutions. And technically it misses "almost all" solutions.

Pleas OP solve Goldbach.

wouldn't the expression (d^1d^2d^3d^4)(x-y) parameterize all solutions?

as in, you just multiply it out and then take the two terms as your actual x and y.

i guess that if you have something like d^1d^2 = d^3 it misses that one, but that can be easily checked for. it's just a factoring problem.

>And technically it misses "almost all" solutions

it definitely doesn't, since once the denominator is chosen, picking x-y such that d^4 appears as a factor can be done a countably infinite number of ways, and x,y are just integers.

how mean of you to tease us

Let me see if I understand what you are saying.

For simplicity lets say that [math] d=12 [/math] so [math] d^4 = 20736 [/math].

You are saying that you want to parametrize it as such: [math] 20736 \mid x - y [/math]. Therefore [math] x-y = 20736k, k \in \mathbb{Z} [/math].

Then you would set [math] x = 20736k + y [/math] right?

Well, I think it is not so simple because then given a certain [math] y [/math] you would still need to find a suitable [math] k [/math]. For example, if in the above expression you plug in [math] y= 1/99999999999 , k=1[/math] then the resulting numbers are not solutions:
wolframalpha.com/input/?i=(20736 + 1/99999999999)^4 - (1/99999999999)^4

So your formula cannot simply be used to plug in numbers. You need to find special numbers.

>Then you would set x=20736k+y right?

no. pick any integers x' and y', then take x = x'd^4, y=-y'd^4

so then you can always factor out d^4.

I am sorry, could you explain how this would yield non-trivial solutions?

If you take x' and y' integers and then set x = x'd^4 and y= -y'd^4 then your choice of x and y are integers, making them trivial solutions.

you said that a and b must be rational. as i've defined above a=x/d and b=y/d.

if a-b is an integer then the fractional parts of a and b must be the same, which is why we have a single d.

once you pick a denominator d, you're left to choose the two numerators x and y, which are integers.

I am sorry but I still don't understand

>a=x/d
But if x = x'd^4 then a = x'd^3, which is still an integer.

>if a-b is an integer then the fractional parts of a and b must be the same, which is why we have a single d.

my mistake, this one part doesn't follow, but i assume that you're given a denominator, so the point is moot anyway. once you have a denominator, you just have to choose the two numerators which are integers since a and b are rational

anyway, even if it doesn't generate all solutions, it will generate countably (infinite) many solutions.

True but it generates only integer solutions. These are the trivial ones, really.

But still if x = x'd^4 then a will be x' ^n d^(4n -1) which is still an integer except for when n=0. And when n=0 you would still need to look for a specifici x' to plug in.

nevermind, i see what you mean now.

but having d^4 as a factor of x-y does not exclude the possibility of x,y being non-integers. you could tack any rational r onto x=d^4x'r or y=d^4y'r to get non-integer x and y.

>but having d^4 as a factor of x-y does not exclude the possibility of x,y being non-integers

Yeah but I already explained how that is not enough to find a generating expression in > you could tack any rational r
You couldn't. Not every rational r will yield a solution. Again you need to find a very specific r to make it work.

ah, you're right.

i feel like there should be a pretty simply solution for this that doesn't involve more than basic algebra.

There may be, but if you classify solutions by the line they belong to then inevitably you need to consider congruences.

A different approach may yield different generators but I tried a couple of different things way back then and it was inevitable to get into congruences for [math] n>2 [/math].

For [math] n=2 [/math] a solution is possible using only algebra, because of the simplicity of the difference of squares formula. But after you get into differences of cubes and higher power, the algebraic approach is not so simple anymore.

well you've got me. if you're not trolling, then why upload to some shady arxiv knockoff?

...

>computer program
>research
lmao they were right to reject your submission.
you must prove the validity of your algorithm formally.
maybe you only tested it with cases that happened to work with no idea of its failure modes.

They didn't reject it, I never submitted it because they stopped publishing it when the editor retired years before. They never got around to updating the website. The editor was some professor at the University of Wisconsin Madison and he hasn't changed his website either. Also, I learned that they expect you to pay money and I would have had to apply for a fee waiver because I was a NEET. That is why OP needs to ask his professor what to do.

As for your speculation about how shit my paper was, you don't know what you are talking about.

It was about a lesser known family of fractals that could be computed extremely fast. These can be computed in parallel on a GPU, and incorporated into a ray tracer very easily. It also provided a technique/program for finding fractals that could be computed in this way, which were interesting by themselves.

It wouldn't have had to be Donald Knuth tier ""computer"" ""science"" masturbation

...

Because no connections and no relevance. Anyways, I am trying to fix that now. I already got my feedback, even got scrutinized a fucking lot lol so I sent it to my professor. Hopefully he sees it and decides to publish it somewhere.

After that then I may take it off vixra if and only if he asks me to. Otherwise I'll keep it for the timestamp.

Dont leave us hanging, what did your professor think?

bymp

Ah I didn't realize this thread was still up. He said he to see if we can publish it somewhere. I am still not sure what his plan is so I'm waiting for that still.

This seems trivial, I haven't read the paper so there may be some condition that you are satisfying that I was unaware of. Anyhow, If you want the denominator k, with a maximum power of n, then all you need is the numerators of a and b (a=A/k. b = B/k) to satisfy A-B=k^n. Simply realize that The denominator will always be a factor of k^n and that a^r-b^r=(a-b)* some number and you are golden.

I now wee that you are looking for the general solution after scrubbing through the thread.