by Guest » Wed Jul 22, 2020 10:51 pm
\lambda
Dave wrote:We are given the integer,
I (a 1000-digit integer: please see attachment below), which is a product two unknown primes,
p (a 500-digit integer) and
q (a 500-digit integer).
What is [tex]\sqrt{I}[/tex] ?
What are
p and
q?
Source Link:
https://www.wolframalpha.com/input/?i=NextPrimeNumber%5BRandomReal%5B%7B1%2C10%7D%5D+*10%5E499%5D+*+NextPrimeNumber%5BRandomReal%5B%7B1%2C10%7D%5D+*10%5E499%5D.
Hmm. Can we combine tools (continued fraction factorization, general number field sieve algorithm, etc.) to achieve integer factorization of
I (see attachment below) in polynomial time?
Dave.
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 [tex]p - q = 2 \lambda[/tex] for some unknown positive integer, [tex]2 \lambda[/tex]:
1. [tex]p * q = I[/tex];
2. [tex]p - q = 2 \lambda[/tex] such that [tex]1 \le \lambda < \frac{log^{2} (pq)}{2}[/tex].
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. [tex]q = -\lambda + \sqrt{I + \lambda^{2}}[/tex] where [tex]1 \le \lambda < \frac{log^{2} (pq)}{2}[/tex];
1b. [tex]p = \frac{I}{q}[/tex].
For our particular problem, we have [tex]1 \le \lambda < 2,650,949[/tex].
Dave.
Go Blue!\lambda[quote="Dave"]We are given the integer, [b]I [/b] (a 1000-digit integer: please see attachment below), which is a product two unknown primes, [b]p[/b] (a 500-digit integer) and [b]q[/b] (a 500-digit integer).
What is [tex]\sqrt{I}[/tex] ?
What are [b]p[/b] and [b]q[/b]?
Source Link:
[url]https://www.wolframalpha.com/input/?i=NextPrimeNumber%5BRandomReal%5B%7B1%2C10%7D%5D+*10%5E499%5D+*+NextPrimeNumber%5BRandomReal%5B%7B1%2C10%7D%5D+*10%5E499%5D[/url].
Hmm. Can we combine tools (continued fraction factorization, general number field sieve algorithm, etc.) to achieve integer factorization of [b]I[/b] (see attachment below) in polynomial time?
Dave.
[b][u]The Problem[/u][/b]: We are given a large positive integer, [b]I[/b], which is a product of two unknown odd primes,[b] p > q[/b].
What are [b]p[/b] and [b]q[/b]?
[b][u]The Solution Formulation[/u][/b]:
We generate a system of two equations with three unknowns since [tex]p - q = 2 \lambda[/tex] for some unknown positive integer, [tex]2 \lambda[/tex]:
1. [tex]p * q = I[/tex];
2. [tex]p - q = 2 \lambda[/tex] such that [tex]1 \le \lambda < \frac{log^{2} (pq)}{2}[/tex].
[b][u]The Question[/u][/b]: Can we achieve integer factorization of [b]I[/b] in polynomial time?
[b][u]The Answer[/u]: Yes! We can affirmatively achieve integer factorization of I in polynomial time[/b].
Dave,
[url]https://www.researchgate.net/profile/David_Cole29[/url].
[color=#0000BF]Go Blue![/color] :D[/quote]
[size=85]"[i]Simple seeks simplest (best) solution[/i]."[/size] :)
Moreover,
1a. [tex]q = -\lambda + \sqrt{I + \lambda^{2}}[/tex] where [tex]1 \le \lambda < \frac{log^{2} (pq)}{2}[/tex];
1b. [tex]p = \frac{I}{q}[/tex].
For our particular problem, we have [tex]1 \le \lambda < 2,650,949[/tex].
Dave.
[color=#0000BF]Go Blue![/color]
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