Randomness can be a useful tool for solving problems.

[Guest]

Suppose Case 1 is true, but [tex]\lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex].

Case 1.1: [tex]\beta_{11}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}) \le \lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex] when [tex]0 < \beta_{11}(I) \le 1[/tex].

Now only one of the two following cases is true!

Case 1.1A: [tex]\lambda >\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}[/tex] when [tex]\beta_{11}(I) = 1[/tex].


Case 1.1B: [tex]\lambda = \beta_{11}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2})[/tex] when [tex]0 < \beta_{11}(I) \le 1[/tex].


Remarks: The development process is ongoing, and it is quickly becoming convoluted too. We stress caution and accuracy.

[Guest]

Oops! We choose the wrong case! Please consider case two instead. We have made corrections below.

Suppose Case 2 is true, but [tex]\lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex].

Case 2.1: [tex]\beta_{21}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}) \le \lambda < \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}[/tex] when [tex]0 < \beta_{21}(I) \le 1[/tex].

Now only one of the two following cases is true!

Case 2.1A: [tex]\lambda >\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2}[/tex] when [tex]\beta_{21}(I) = 1[/tex].


Case 2.1B: [tex]\lambda = \beta_{21}(I) * (\frac{1 + \frac{1 + \sqrt{ \frac{(\sqrt{2} - 1) * I}{2}}}{2}}{2})[/tex] when [tex]0 < \beta_{21}(I) \le 1[/tex].


Remarks: The development process is ongoing, and it is quickly becoming convoluted too. We stress caution and accuracy.

[Guest]

Remarks: In either case, one or two, we must be mindful of how close we are achieving a true solution for [tex]\lambda[/tex]. Do our efforts make sense?

[Guest]

Remarks: We must make sure there's no gap in our programming. All cases must be fully developed and verified. We must test and do more testing. Math is hard work!

[Guest]

Remarks: We must make sure there's no gap in our programming. All cases must be fully developed and verified. We must test and do more testing. Math is hard work!

For each case for [tex]\lambda[/tex], is [tex]I + \lambda^{2}[/tex] becoming a perfect square?

And we expect the two factors of I to be prime.

[Guest]

[/b]

Remarks: We must make sure there's no gap in our programming. All cases must be fully developed and verified. We must test and do more testing. Math is hard work!

For each case of [tex]\lambda[/tex], is [tex]I + \lambda^{2}[/tex] becoming a perfect square?

And we expect the two factors of I to be prime.

We recall: [tex]n(\lambda) + \triangle n(\lambda) = \sqrt{I + \lambda^{2}}[/tex].

Important Question: Does the error, [tex]0 \le \triangle n(\lambda) < 1[/tex], associated with each relevant case of [tex]\lambda[/tex], helps us to decide how to proceed with our programming?

[Guest]

Remarks: We are confident that we can crack our integer factorization problem. But there's a great deal of work we must accomplish...

[Guest]

Remark: Some randomness/guessing may help our programming.

Remark: We are hopeful that the integer factorization problem can be solved in polynomial time.

[Guest]

Question: Can we define the function, [tex]0 < \beta(I) \le 1[/tex], more explicity in terms of I?

A "Redundant" Question: For some [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex], the perfect square, [tex]I + \lambda^{2}[/tex], guarantees that the factors, [tex]\sqrt{I + \lambda^{2}} \pm \lambda[/tex], of I are prime?
We had assumed the known value, I, is a product of unknown primes.

[Guest]

"Point us (mathematicians) in the right direction, and we will find the right solutions to our problems."

[Guest]

Relevant Reference Link:

'Congruence of Squares',

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

[Guest]

Relevant Reference Link:

'General number field sieve',

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

[Guest]

Remarks: We have existing technology to factor large composites (product of two large primes, each with 250 decimal digits). What we are seeking is an edge/breakthrough that will guarantee integer factorization in polynomial time.

[Guest]

Relevant Reference Link:

'General number field sieve',

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

FYI: 'An integrated parallel GNFS algorithm for integer factorization based on Linbox Montgomery block Lanczos method over GF(2)',

https://www.sciencedirect.com/science/article/pii/S0898122110000428.

[Guest]

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].

Relevant Reference Link:

'Continued fraction factorization',

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

[Guest]

We spoke very briefly on how integer factorization and integer optimization are related/connected in a past post.

Please refer to the attachment below for more details.

[Guest]

Remark: We have enough information to solve our problem either affirmatively or negatively.

Good luck! Go Blue!

[Guest]

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] ?

[tex]n(\lambda) = \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex];

[tex]\triangle n(\lambda) = \sqrt{I + \lambda^{2}} - \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex].

[Guest]

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] ?

[tex]n(\lambda) = \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex];

[tex]\triangle n(\lambda) = \sqrt{I + \lambda^{2}} - \left \lfloor{\sqrt{I + \lambda^{2}}} \right \rfloor[/tex].

Remark: The discrete graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I + \lambda^{2}}}{2}[/tex] is quite interesting and valuable!

[Guest]

Updated Remark: The discrete graph of [tex]\triangle n(\lambda)[/tex] for [tex]1 \le \lambda < \frac{\sqrt{I }}{2}[/tex] is quite interesting and valuable!