# Integer Factorization in PT (Polynomial Time)

### Integer Factorization in PT (Polynomial Time)

Can we achieve integer factorization in polynomial time?

'Integer Factorization'

https://en.wikipedia.org/wiki/Integer_factorization.

Hmm. Can we combine tools (continued fraction factorization algorithm, general number field sieve algorithm, etc.) to achieve integer factorization in polynomial time?

Dave.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

We are seeking proof/algorithm or disproof that integer factorization is achievable in polynomial time.

'Randomness can be a useful tool for solving problems.'

https://www.math10.com/forum/viewtopic.php?f=1&t=8855&sid=ee4551598cc1fa9c3dc5a0b2f292b994&start=140;

'Number Theory and Cryptography',

https://www.math10.com/forum/viewtopic.php?f=63&t=8634.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

The Problem: We are given a large positive integer, I, which is a product of two unknown odd primes, p > q.

What are p and q?

The Solution Formulation:

We generate a system of two equations with three unknowns since $$p - q = 2 \lambda$$ for some unknown positive integer, $$2 \lambda$$:

1. $$p * q = I$$;

2. $$p - q = 2 \lambda$$ such that $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$.

The Question: Can we achieve integer factorization of I in polynomial time?

The Answer: Yes! We can affirmatively achieve integer factorization of I in polynomial time.

Dave,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Dave wrote:The Problem: We are given a large positive integer, I, which is a product of two unknown odd primes, p > q.

What are p and q?

The Solution Formulation:

We generate a system of two equations with three unknowns since $$p - q = 2 \lambda$$ for some unknown positive integer, $$2 \lambda$$:

1. $$p * q = I$$;

2. $$p - q = 2 \lambda$$ such that $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$.

The Question: Can we achieve integer factorization of I in polynomial time?

The Answer: Yes! We can affirmatively achieve integer factorization of I in polynomial time.

Dave,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!

"Simple seeks simplest (best) solution."

Moreover,

1a. $$q = -\lambda + \sqrt{I + \lambda^{2}}$$ where $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$;

1b. $$p = \frac{I}{q}$$.

Dave.

Go Blue!
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Oops! $$(log(p * q))^{2}$$ is the wrong formula! The $$(log(p * q))^{2}$$ works for the largest gap between consecutive primes less than p * q.

Therefore, we could try $$(log(p * q))^{3}$$ or larger.

Dave.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Dave wrote:Oops! $$(log(p * q))^{2}$$ is the wrong formula! The $$(log(p * q))^{2}$$ works for the largest gap between consecutive primes less than p * q.

Therefore, we could try $$(log(p * q))^{3}$$ or larger.

Dave.

Correction! $$2 \le 2 \lambda < \frac{\sqrt{I}}{2}$$ is probably closer to the truth.

And we do not know if we can achieve integer factorization in polynomial time.

Dave
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Dave wrote:Oops! $$(log(p * q))^{2}$$ is the wrong formula! The $$(log(p * q))^{2}$$ works for the largest gap between consecutive primes less than p * q.

Correction! $$2 \le 2 \lambda < \frac{\sqrt{I}}{2}$$ is probably closer to the truth.

Is integer factorization achievable in polynomial time? WE DO NOT KNOW!

Simple Dave

"We will know!" -- David Hilbert, a great mathematician.
Attachments
Silhouette of Sherlock Holmes.jpg (5.13 KiB) Viewed 957 times
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

where $$\lambda$$ is a positive integer such that $$1 \le \lambda < \frac{\sqrt{I}}{4}$$.

Given the integral value, $$I = p * q$$, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

Dave.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Dave wrote:Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

Update:

where $$\lambda$$ is a positive integer such that $$1 \le \lambda < \frac{\sqrt{I}}{2}$$.

Given the integral value, $$I = p * q$$, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

FYI: Newton's Method may not work, but it is worth a try. We must beware of the global minimum, 0, versus many local minima in the open interval, (0, 1), oscillations, and the failure analysis associated with Newton's Method.

However, Newton's Method could be a valuable complementary tool for other tools (continued fraction factorization algorithm, general number field sieve algorithm, etc.) in regards to solving our problem.

'Failure Analysis associated with Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method#Failure_analysis.

Dave.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Dave wrote:Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

where $$\lambda$$ is a positive integer such that $$1 \le \lambda \le \frac{p - 3}{2}$$. (Update)

Given the integral value, $$I = p * q$$ for odd primes, p > q, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

### Re: Integer Factorization in PT (Polynomial Time)

STOP!

Integer factorization in poynomial time is classified research. Please do not develop it further. You have done enough!
Attachments
Classified!.jpg (6.48 KiB) Viewed 862 times
Classification of Research.jpg (122.1 KiB) Viewed 862 times
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Update!

Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = 0$$ if $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

Our analysis is wrong! Where did we go astray?

Maybe it's time to retire... We are discouraged at this point.

And we shall focus our attention elsewhere.

We apologize for the flawed analysis. But all is not lost... And there's always hope.

And we expect no feedback from our readers (mathematicians). What a predicament!

Dave.

P.S. Yes, it's time to retire and focus our energies elsewhere. Goodbye!
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Guest wrote:$$\sum_{k=1}^{\infty }\frac{cos(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \infty$$ if $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

The answer is the very important Harmonic Series! And that's all folks!
Guest

### Re: Integer Factorization in PT (Polynomial Time)

Guest wrote:Update!

Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = 0$$ if $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

Our analysis is wrong! Where did we go astray?

Maybe it's time to retire... We are discouraged at this point.

And we shall focus our attention elsewhere.

We apologize for the flawed analysis. But all is not lost... And there's always hope.

And we expect no feedback from our readers (mathematicians). What a predicament!

Dave.

P.S. Yes, it's time to retire and focus our energies elsewhere. Goodbye!

Hey Dave,