Discovery-factorization

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FAIZE

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Good morning
working on prime numbers I discovered that to factorize a number resulting from the multiplication of two prime numbers we can facilitate the task by looking in a closer number for the number to factorize
example if you want to factorize the numbers 2305057
there is close to him the number 2301923 by deducting 2305057-2301923 that gives us 3134
dividing it by 2 the result is 1567
1567 is one of the prime numbers other is 1471
rest ' prove this operation
 
FAIZE said:
working on prime numbers I discovered that to factorize a number resulting from the multiplication of two prime numbers we can facilitate the task by looking in a closer number for the number to factorize
example if you want to factorize the numbers 2305057
there is close to him the number 2301923
Click to expand...
On what basis did you pick this "close by" number?

FAIZE said:
by deducting 2305057-2301923 that gives us 3134
dividing it by 2 the result is 1567
Click to expand...
On what basis did you apply these operations?

FAIZE said:
1567 is one of the prime numbers other is 1471
rest ' prove this operation
Click to expand...
What is meant by "rest ' prove this operation"?

Also, please note replies posted elsewhere, such as here.
 
Last edited:
Good morning
here is another
the 4159751 by looking for the closest we found the 4163917
deducing gives us 4166 divided by 2 = 2083
the factorization is done 2083*1999=4163917
I developed this method by following the path of the multiplication and I found that there is a number with a number in common
as here the number 2083 exists among the two numbers 2083*1999---2083*1997
you can give me a number which is the product of 2 prime numbers and which is 8 or 9 or 10 digits I give you the decomposition
by using this method which is easier than the others

choose or search number-NEAR- THIS DONE with a formula in can say algoritme
 
As stapel pointed out, this is not a new result, and the critical point is how ”close” is close enough to be practical, a critical point that the original poster failed to mention. Finally, the original poster neglected to mention that the proposition is not relevant to even products of a pair of primes.

It is almost trivial to prove the basic proposition that if u and v are both odd integers > 1 and w = uv, there exists an integer x such that (w-x)/2 = v.

[math] \text {Given: } p,\ q \in \mathbb Z, \ 1 \le p \le q, \ u = 2p + 1, \ v = 2q + 1 \text { and } w = uv.\\ \therefore \ u, \ v \in \mathbb Z, \ 3 \le u \ v, \ u^2,\ v^2,\ w \in \mathbb Z, \ 9 \le u^2 \le w \le v^2,\\ u^2 = 4p^2 + 4p + 1, \text { and } w = 4pq + 2p + 2q + 1.\\ \text {Let } x = 4pq + 2p - 2q - 1.\\ \therefore x \text { is an odd integer} < w \text { and}\\ \dfrac{w - x}{2} = \dfrac{4pq + 2p + 2q + 1 - (4pq + 2p - 2q - 1)}{2} = \dfrac{2q + 1 + 2q + 1}{2} = 2q + 1 \implies\\ \dfrac{w - x}{2} = v. [/math]
Because that proof applies to the product of any pair of odd numbers, each greater than 2, it necessarily applies to the product of any pair of primes, each greater than 2. What that proof does not do is to say what “close” means.
 
The proof in the previous post does permit determination of the number of tests needed to determine a non-trivial factorization of the product of two odd primes by the given method. Keeping the same notation as in the previous post but adding the constraint that u and v are odd primes, we know that [imath](w - x)/2 = v > 2.[/imath] We also know that w = uv and x are odd numbers and that x < w. The number of odd integers greater than x but less than w is [imath](w - x)/2 - 1 = v - 1.[/imath]

Therefore, we know we can find x with at most v - 1 tests and that x exactly 2v less than w. But a non-trivial factorization can be found in fewer than [imath]\sqrt{w}/2 - 2[/imath] tests, which will be fewer than v - 1. So this is not the fastest method.
 
I adapted my method with python it works well I would like to distribute it
so that many people can benefit from it
methode.jpg
 
the proof is that I can write the prime numbers from 2 to 20000 or more without errors and without knowing them in advance only by helping me with this sequence
 
I suspect what he trying to get at is that he believes that he has an efficient method to find the number that I called x in my posts, which then allows him to factor the product of two odd primes easily. He does not say what the method is, nor does he attempt a proof that his method is more efficient than any other method.

He has constructed an example. How does he know that 4453 is the product of exactly two primes? (Probably because he constructed it by multiplying 73 and 61 in the first place.) Why does he divide 4453 by numbers of the form 3.5 +3(k-1)? Possibly because 73 - 61 = 3, a fact that he would not know unless he knew the factorization in advance.

If my suspicion is correct, this whole thing is numerology dressed up in computers.
 
topsquark said:
Of course there is a "logical" sequence of prime numbers. Any sequence that is well defined is logical in the sense that you can find all of them.

What are you actually trying to say here?

-Dan
Click to expand...
topsquark said:
Of course there is a "logical" sequence of prime numbers. Any sequence that is well defined is logical in the sense that you can find all of them.

What are you actually trying to say here?

-Dan
Click to expand...
-if you pretend i'm not aware of what i'm saying Or am I exaggerating then ,this is not the case
I am well aware and the method is in my hands
-but if you underestimate me that's another thing
Even the prophets were underestimated by the population
 
FAIZE said:
-if you pretend i'm not aware of what i'm saying Or am I exaggerating then ,this is not the case
I am well aware and the method is in my hands
-but if you underestimate me that's another thing
Even the prophets were underestimated by the population
Click to expand...
I am asking "What are you saying?" Your comment is not correct as it stands, though it is extremely vague.

-Dan
 
NotJeffM said:
I suspect what he trying to get at is that he believes that he has an efficient method to find the number that I called x in my posts, which then allows him to factor the product of two odd primes easily. He does not say what the method is, nor does he attempt a proof that his method is more efficient than any other method.

He has constructed an example. How does he know that 4453 is the product of exactly two primes? (Probably because he constructed it by multiplying 73 and 61 in the first place.) Why does he divide 4453 by numbers of the form 3.5 +3(k-1)? Possibly because 73 - 61 = 3, a fact that he would not know unless he knew the factorization in advance.

If my suspicion is correct, this whole thing is numerology dressed up in computers.
Click to expand...
with a 14-digit number
is this numerology?
you excuse me
14.jpg
 
What I am saying is that

(1) You have never stated what your “method“ is;

(2) You did not give a proof of the trivial propositions that if w is an odd composite number = uv, then there exists
positive integer x = w - 2v and that x can be found in 2v - 1 tests;

(3) Your examples are not compelling because they may have been contrived with the purpose of demonstrating whatever it is you may be vaguely asserting; and

(4) Your apparent assertion that your method is faster than other methods is without any effort at proof and is testable by no one because you have not disclosed your method (see # 1);

In short, you have yet to say anything of mathematical importance.

And no, I do not excuse you for inflicting this upon us.
 
He posted his method on MHF. He's simply using Excel to do the calculation to find his seed number, he is not using an algorithm.

-Dan
 
topsquark said:
He posted his method on MHF. He's simply using Excel to do the calculation to find his seed number, he is not using an algorithm.

-Dan
Click to expand...
Thank you Dan.

That youtube video he posted is acutely painful to watch. It is not even an efficiently designed excel program. It requires a lot of human observation. He could use excel to find his integer quotient.

It has, however, given me a problem that I hope does not burrow its way into what is left of my brain. A program, even one as clumsy as this, embodies an algorithm. He has shown that his algorithm will identify the factorization of some products of pairs of odd primes. He is not checking all odd integers as divisors. in fact, he is checking as divisors rational numbers of the form

[math]3.5 + 3k, \text { where } k \in \mathbb Z_{\ge 0}.[/math]
It is not intuitively clear to me why this algorithm finds a factor of the product of any pair of primes. I hope I manage to avoid jumping down this probably ridiculous rabbit hole of which ones it can find.
 
[math] w = uv, \text { where } u \text { and } v \text { are both odd primes.}\\ \text {Algorithm: (1) Find } z \text { such that } z \in \mathbb Z^+, \ z > 2, \text { and } z = \dfrac{w}{3.5 + 3k}, \text { where } k \in \mathbb Z_{\ge 0};\\ \text {(2) Calculate } v = \dfrac{w - z}{2}; \text { and}\\ \text {(3) Calculate } u = \dfrac{w}{v}. [/math]
I previously proved the relatively trivial result that

[math] \exists \text { integer } x = w - 2v \implies v = \dfrac{w - x}{2}.[/math]
Consider 33 = 3 * 11, a product of odd primes.

9 < 33/3.5 < 10

5 < 33/6.5 < 6

3 < 33/9.5 < 4

33/12.5 < 3

Thus, the so-called algorithm will not necessarily find x because x must be an integer.

Now because there necessarily is such an odd integer x = w - 2v, you can find x by calculating successively

[math]\dfrac{w}{\dfrac{\dfrac{w - (w - 2k)}{2}} = \dfrac{2w}{2k} = \dfrac{uv}{k}.[/math]

As soon as k is incremented to u, we get [imath]\dfrac{uv}{u} = v.[/imath]

In short, incrementing by 1, finds v in exactly u steps, which simultaneously identifies u as the other factor. (I made an error in calculating the number of steps required in previous steps.) But this is less efficient that simply dividing w by 2k + 1 until u is found.

So, we have an algorithm that does not work reliably as specified, and that, if correctly specified, is not efficient,
 
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