To keep a long story short, I'm doing some bullshit maths

>>7779258
Do you mean [math]x^2+y^2[/math] ?

>>7779272
sorry, minus not plus.

>>7779272
>>7779274
no I mean x-y
Maybe even x^n-y^m if I can do some real sorcery

>>7779258
(x-y) is already factored, retard.

e^(ln(x-y))

e^(ln(x)/ln(y))

>>7779295
says you
I'm trying to solve the fucking prime distribution

>>7779310
Sure you are.

>>7779279
Well then you're retarded.

Also, factoring implies turning it into multiplication, not division. The closest thing to what you want to do is take
[math]x^2-y^2=(x+y)(x-y)[/math]
Which, if you look at it another way, is
[math](x-y) = (\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})[/math]

Have fun in school tomorrow!

>>7779308
Whoops

ln(e^x / e^y)

>>7779317
I didn't come here to do things someone else has already done

What I'm trying to do is take any number of the form 6m+1 or 6m+2 or 6m+3, and then do subtraction on them without losing track of their prime factors

the idea is to do this without breaking the laws of arithmetic (which might not be possible)

>>7779320
neat, but i want to do this without e

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>>7779333
actually wait;

Really I'm trying to generalize addition and subtraction for all numbers (6^n)m

Mostly because every prime number is of the form (6^n)m+1

>>7779333
you sound like an outsider who doesnt have any math education

>>7779342
>outsider
For sure
>no math education
i watched a a shitload of wildberger and this is what it did to me
take that as you will

>>7779345
yeah its pretty much impossible to deal with math outsiders since you dont even understand the words you use and outsiders also come up with weird analogies that are wildly incorrect but nearly impossible to dissuade them from using because they dont understand the fundamental concepts involved

>>7779353
I definitely understand the concepts; it'll just be harder for me to get you to know that I understand it.

So what are you not understanding?

>>7779363
I know you think you understand, like I said its impossible to dissuade outsiders

but from what you have written it is extremely obvious that you have no understanding of any of these topics whatsoever, and you should not expect to be taken seriously by even first year undergraduates

>>7779375
you sound like a really salty undergraduate

>>7779378
no one with an interest in math will take >>7779258
>>7779310
>>7779333
seriously. you don't know what you're doing

>>7779389
isn't the point of math that when you're presented with a problem, you figure out how to solve it?

I've told you what I'm trying to do. I've told you I don't want to use e. You shouldn't need any additional information.

But I guess you're just too busy with your undergraduate studies to do some real math

>>7779397
>real math
>the "point of math" is coming up will bullshit """"problems"""" and circlejerking on them

no

>>7779397
>isn't the point of math that when you're presented with a problem, you figure out how to solve it?
What you're presenting is not a problem, it's just gibberish. For example "Does anyone know how to factor (x-y) in a general way?" makes no sense because (x-y) is already factored.

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>>7779454
Alright smartass

x and y are numbers
numbers have prime factors
So x=P, where P is some permutation of prime numbers
Similarly, y=Q, where Q is a (presumably different) permutation of potentially different prime numbers
When you do X-Y, the result is equal to a new number, Z
Z=R, where R is some permutation of prime numbers

What I'm trying to find out is... hold up let me phrase this as clearly as possible.

Let's say X=7*11*13, which happens to be 1001, but we care more about how it's 7*11*13.
And then we'll have Y=5*7*19, which is some other number
So what happens when I take the difference of X and Y? I get Z. Z has factors. How many factors? What are they? That's what I want to know.

Basically, if you know what P is, and you know what Q is, can you determine what R is without actually evaluating Z?

What you normally do (and what a calculator does) is evaluate these numbers. Multiply all of X's prime factors together, the product is X. Same thing for Y. Then, you do arithmetic with the products, and you get Z. Then, if you want, you can do trial divisions on Z to get its prime factorization.

I want to go from prime-factored numbers directly to prime-factored numbers, without those other steps. And (hopefully) without using transcendental numbers.

>>7779258
The easiest way is to apply the theory of semi-graphs of anabelioids, Frobenioids, the étale theta function and log-shells. You can fix initial [math]\Theta[/math]-data over Z and then construct a [math]\Theta^{\pm ell}[/math] NF-Hodge theater associated to the data, and then look at the Hodge theater as a miniature model of conventional scheme theory where the two underlying combinatorial dimensions of Z (i.e. the additive and multiplicative structures of the ring Z) has been 'disentangled'. Proceeding from here should be fairly trivial.

>>7779516
Yeah I tried that
The problem is that the combinatorial dimensions of Z are infinite, because P, Q and even R are permutations of {P}, which happens to contain infinitely many elements to... permutate?

And there's no restriction as to whether P, Q, or R can or can't have non-unique elements.

Any ideas?

>>7779499

Is.. is this bait?

>>7779542
you wish, nigger
I'm over here trying to circumvent the order of operations

OP, if you are actually interested in mathematics you should take some courses so you can actually learn how to use mathematics without sounding like a crank. I assume you are "working on" something like the Collatz conjecture. The simple fact is that, without any formal education whatsoever, any techniques you come up with will be far too elementary to deal with the problems you face. In particular, it's very hard to say anything general about how certain prime factorizations are related to one another.

For example, an analogue of the Riemann hypothesis has been proved over finite fields. This required the extremely detailed development of scheme theory and etale cohomology; Alexander Grothendieck had to spend nearly his entire career to come up with these techniques, and I can guarantee you are no Alexander Grothendieck. Another example: the abc conjecture is much more closely related to the questions about prime factorizations you are concerned with. Maybe you already know, but Shinichi Mochizuki published a purported proof of abc in 2012. His work, which took more than a decade to complete, totals over 500 pages, and even the most advanced mathematicians working today don't understand what's going on there yet.

My point is: the most talented mathematicians in the world have struggled with Collatz, or lonely runner, or Riemann, or whatever basic yet elusive problem you could come up with. What makes you think that anything you do hasn't already been tried by hundreds of people more knowledgeable than you? And the fact that you think that *you* are the one who can solve these problems speaks volumes about your unwarranted self-importance and delusional ideas about math.

>>7779499
And what on earth makes you think you will need to use transcendental numbers? Do you have the slightest clue what this vague "use transcendental numbers" actually entails?
>permutation
You keep using this word. I don't think you know what this word means.

>>7779537
>Yeah I tried that
Get back to work Mochi-san

>>7779499
>So x=P, where P is some permutation of prime numbers
You mean combination.

>Basically, if you know what P is, and you know what Q is, can you determine what R is without actually evaluating Z?
No, of course not. The only thing you can tell is that if P and Q share a prime, then R contains that prime. R will never contain a prime in P or Q that isn't in both.

>>7779562
>don't try to solve a problem because some guy who might be smarter than you couldn't solve it quickly
I don't know how smart I am, and if I thought I could solve this by myself I never would have come to this site.
I thought this was the math board; why did you come here? To shitpost?

>>7779570
If I've got a set that has elements in it, and I group these elements in some kind of a way, it's either a combination or a permutation.
If I had a set that has all the prime numbers in it, and I wanted to group them (in this case to do multiplication), and I didn't want to use the same element more than once, that's a permutation. I don't know what's confusing you. Did you actually study any of this? Or are you just another undergraduate shitposter?
And "use transcendental numbers" means "write an equation with e in it". Or "write an equation with pi in it". The goal is to avoid doing that, if it can be done. And maybe it can't.

brb reading the proof of the abc conjecture

>>7779608
>If I had a set that has all the prime numbers in it, and I wanted to group them (in this case to do multiplication), and I didn't want to use the same element more than once, that's a permutation.
No, no it's not. First of all, this has nothing to do with what you described above because prime factorization does not mean you can only use one of each prime. Second, permutation implies an ordered sequence or set. It doesn't matter how you order the primes because multiplication is independent of order.

What we have here is a stupid person too stupid to see that their stupid.

>>7779608
No, you don't understand. Every serious mathematician has already tried any possible idea you could think of and seen that it can't work. And on top of that they have so many more ideas than you could possibly ever imagine, and they're tried all those too. You are wasting your time; maybe if you actually learned some math you would understand why.

>>7779608
>And "use transcendental numbers" means "write an equation with e in it". Or "write an equation with pi in it".
And why on earth do you think that will help you deal with your problem? Transcendental numbers don't have prime factorizations, you know.

>>7779608
>brb reading the proof of the abc conjecture
That should keep you busy for a few years.

>>7779340
>Mostly because every prime number is of the form (6^n)m+1
?????????????????

>>7779258
Try:
[Root(x)+root(y)] × [root(x) - root(y)]

>>7779659
this seems to be what i needed
Maybe if I keep expanding that expression something interesting will happen

>>7779310
Lol.

OP's question just got 100x more complex.

Guys he's not looking to factor it algebraically. He's looking to take two integers x and y and factor their sum into prime numbers.

Is that right OP?

You're going to need this:
https://en.wikipedia.org/wiki/General_number_field_sieve

You can generalize the Number Field Sieve to suit your purposes and it's going to take a lot of work. If that doesn't work, try the Quadratic Sieve instead and then if you still can't do it, read up on number theory and about Fermat's work in this area and try again. Good luck.

Nice question.

>>7779659
I tried posting that up here: >>7779317

But it didn't like my tags.

>>7779774
Shhh this >>7779753
is the answer OP was looking for.

OP is actually a genius in disguise and on the verge of a breakthrough in number theory. Isn't that right OP?

>>7779776
hes just an idiot that doesnt know what words mean

>>7779742
lmao something very interesting will happen. you will get x-y

Shout out to BIG Sean Don nigga.

>>7779659
>how to factor (x-y)
[math]( \sqrt{x}- \sqrt{y})( \sqrt{x}+ \sqrt{y})[/math]

>>7779272
>>7779274
>>7779295
>>7779308
>>7779317
>>7779320
change your major to Liberal Arts, fgt pls

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>>7780956
>>7780946
/thread

>>7779258
>Pic Related
Do your own work you fucking faggot. (x - y) is already factored completly

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Alright son, go grab some prime numbers and do some math with this. Use the biggest prime factor as your most important one. If you choose a prime expressed to a power, you have to set up the rest of the primes equal to that prime to the power.
What I mean is, if you've got 5^4, 37, and 13, it's probably best to set it up as 5^4, ((5^4)-((5^4)-37)), and ((5^4)-((5^4)-13))

>>7781063
If it was already factored, how did I just factor it?
Checkmate, nigger

>>7780946
I realized this wasn't going to give me integers so I stopped using it.

This is useful to me, because if r>1 and s>1, (x-y) is a multiple of 6. And how close a number is to 6 is important. Because, like I said, all prime numbers are 6m+1 or 6m-1. And you can go look that shit up if you don't believe me, you niggers.

>>7781440
>all prime numbers are 6m+1 or 6m-1
You do know 2 is prime, right ? So is 3 by the way

>>7781781
Do I have to say "for all n>6" to satisfy your autism every time
Who the fuck cares about counting primes less than 6? You can do it on one hand. The interesting shit comes later

>>7781781
>>7779653
OP is surely retarded but the 6k+1, 6k-1 thing is true tho