Dave wrote:An Update: ([tex]\sqrt{I}[/tex])
The Minimum Square Greater Than I = p * q:
Generally, our integer factorization problem, I = p * q, is an integer optimization problem.
We seek the minimum positive integer, [tex]1 \le\lambda < \sqrt{I}[/tex], such that
[tex]\sqrt{I + \lambda^{2}}[/tex] is a positive integer.
Remark: [tex]p - q = 2\lambda[/tex] where p and q are unknown odd primes such that p > q and such that I = p * q where I is known.
Dave.
Guest wrote:An Update: ( [tex]\sqrt{I} / 2[/tex] )
The Minimum Square Greater Than I = p * q:
Generally, our integer factorization problem, I = p * q, is an integer optimization problem.
We seek the unique positive integer, [tex]1 \le\lambda < \sqrt{I} / 2[/tex], such that
[tex]\sqrt{I + \lambda^{2}}[/tex] is a positive integer.
Remark: [tex]p - q = 2\lambda[/tex] where p and q are unknown odd primes such that p > q and such that I = p * q where I is known.
Dave.
Does fast convergence (polynomial time) from [tex]\sqrt{I} / 2[/tex], exclusively, to [tex]\lambda[/tex] exists? Why? How?
Does fast convergence (polynomial time) from 1, inclusively, to [tex]\lambda[/tex] exists? Why? How?
Guest wrote:Now suppose we have access to a worldwide network of one billion personal computers for solving our integer factorization problem.
We program computer one to search randomly for a solution in the range:
[tex]1 \le \lambda\ < 3.09 * 10^{489}[/tex];
We program computer two to search randomly for a solution in the range:
[tex]3.09 * 10^{489} \le \lambda < 2 * 3.09 * 10^{489}[/tex];
We program computer three to search randomly for a solution in the range:
[tex]2 * 3.09 * 10^{489} \le \delta < 3 * 3.09 * 10^{489}[/tex];
...
...
...
We program the billionth computer to search randomly for a solution in the range:
[tex](10^{9} - 1) * 3.09 * 10^{489} \le \lambda < 3.09 * 10^{498}[/tex].
Guest wrote:An Update for [tex]I = p * q = 8.671... * 10^{999}[/tex]:Guest wrote:Now suppose we have access to a worldwide network of one billion personal computers for solving our integer factorization problem.
We program computer one to search randomly for a solution in the range:
[tex]1 \le \lambda\ < 3.09 * 10^{489}[/tex];
We program computer two to search randomly for a solution in the range:
[tex]3.09 * 10^{489} \le \lambda < 2 * 3.09 * 10^{489}[/tex];
We program computer three to search randomly for a solution in the range:
[tex]2 * 3.09 * 10^{489} \le \lambda < 3 * 3.09 * 10^{489}[/tex];
...
...
...
We program the billionth computer to search randomly for a solution in the range:
[tex](10^{9} - 1) * 3.09 * 10^{489} \le \lambda < 3.09 * 10^{498}[/tex].
Guest wrote:Guest wrote:An Update: ( [tex]\sqrt{I} / 2[/tex] )
The Appropriate Square Greater Than I = p * q:
Generally, our integer factorization problem, I = p * q, is an integer optimization problem.
We seek the unique positive integer, [tex]1 \le\lambda < \sqrt{I} / 2[/tex], such that
[tex]\sqrt{I + \lambda^{2}}[/tex] is a positive integer.
Remark: [tex]p - q = 2\lambda[/tex] where p and q are unknown odd primes such that p > q and such that I = p * q where I is known.
Dave.
Does fast convergence (polynomial time) from [tex]\sqrt{I} / 2[/tex], exclusively, to [tex]\lambda[/tex] exists? Why? How?
Does fast convergence (polynomial time) from 1, inclusively, to [tex]\lambda[/tex] exists? Why? How?
Let's ponder the following sum of two functions, [tex]n(\lambda)[/tex] and [tex]∆n(\lambda)[/tex]:
[tex]n(\lambda) + ∆n(\lambda) = \sqrt{I + \lambda^{2}}[/tex] for some positive integers, [tex]n(\lambda)[/tex] and [tex]\lambda[/tex], where [tex]0 \le ∆n(\lambda) < 1[/tex].
Remark: [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] and I = p * q where I is known.
Key Ideas: Power Series Expansion, Convergence, Rate of Convergence, etc.
Relevant Reference Link:
'Power series',
https://en.wikipedia.org/wiki/Power_series.
Guest wrote:Let's ponder the following sum of two functions, [tex]n(\lambda)[/tex] and [tex]∆n(\lambda)[/tex]:
[tex]n(\lambda) + ∆n(\lambda) = \sqrt{I + \lambda^{2}}[/tex] for some positive integers, [tex]n(\lambda)[/tex] and [tex]\lambda[/tex], where [tex]0 \le ∆n(\lambda) < 1[/tex].
Remark: [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] and I = p * q where I is known.
Key Ideas: Power Series Expansion, Convergence, Rate of Convergence, etc.
Relevant Reference Link:
'Power series',
https://en.wikipedia.org/wiki/Power_series.
Can we explicitly identify [tex]n(\lambda)[/tex] in the expression,[tex]\sqrt{I + \lambda^{2}}[/tex] ?
And can we explicitly identify [tex]∆n(\lambda)[/tex] in the expression, [tex]\sqrt{I + \lambda^{2}}[/tex] ?
Guest wrote:Remark: We are seeking either a proof/algorithm or a disproof that our integer factorization problem can be solved in polynomial time.
Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
Guest wrote:Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
The general continued fraction of [tex]\sqrt{I + \lambda^{2}}[/tex] offers great promise. So between the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] and its general continued fraction representation, we hope to solve [tex]\triangle n(\lambda) = 0[/tex]...
Good Luck!
Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
The generalized continued fraction of [tex]\sqrt{I + \lambda^{2}}[/tex] offers great promise. So between the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] and its generalized continued fraction representation, we hope to solve [tex]\triangle n(\lambda) = 0[/tex]...
Good Luck!
Guest wrote:Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
The generalized continued fraction of [tex]\sqrt{I + \lambda^{2}}[/tex] offers great promise. So between the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] and its generalized continued fraction representation, we hope to solve [tex]\triangle n(\lambda) = 0[/tex]...
Good Luck!
A Tentative Update: [tex]\beta(I) * (\frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}) \le \lambda < \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}[/tex] where [tex]0 < \beta(I) \le 1[/tex].
Remark: We want to construct a convergent sequence to solve [tex]\lambda[/tex].
Guest wrote:\geGuest wrote:Dave wrote:FYI: 'Continued fraction',
https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents.
The generalized continued fraction of [tex]\sqrt{I + \lambda^{2}}[/tex] offers great promise. So between the power series expansion of [tex]\sqrt{I + \lambda^{2}}[/tex] and its generalized continued fraction representation, we hope to solve [tex]\triangle n(\lambda) = 0[/tex]...
Good Luck!
A Tentative Update: [tex]\beta(I) * (\frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}) \le \lambda < \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}[/tex] where [tex]0 < \beta(I) \le 1[/tex].
Remark: We want to construct a convergent sequence that solves [tex]\lambda[/tex].
Depending on I, only one of the following two cases is true!
Case 1: [tex]\lambda > \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex] when [tex]\beta(I) = 1[/tex].
Case 2: [tex]\lambda = \beta(I) * (\frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2})[/tex] when [tex]0 < \beta(I) \le 1[/tex].
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