Fermat's factorization method: Difference between revisions

Revision as of 04:09, 13 June 2025

Revision as of 04:12, 13 June 2025

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For the last term, we use the chain rule <math>\left(f\left(g\right)\right)' = f'\left(g\right)g'</math> and the power rule <math>x^n = nx^{n-1}</math>

<math>\frac{d}{dd}\sqrt{2\sqrt{N}d + d^2} = \frac{\left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}}{2} \frac{d}{dd}2\sqrt{N}d + d^2 = \frac{\left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}}{2} \left(2\sqrt{N} + 2d\right) = \left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N} + d\right)</math>SubstutingSubstitute the known derivate formulas

<math>\frac{d}{dd}\sqrt{N} +\frac{d}{dd}d\left(1+\frac{1}{l}\right) - \frac{d}{dd}\sqrt{2\sqrt{N}d + d^2} = \left(1+\frac{1}{l}\right) - \left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N} + d\right) </math>

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=== Cost ===

SubstudeSubstitute our optimal <math>d</math> into <math>C\left( d , N,l\right) </math>

<math>\sqrt{N} +d\left(1+\frac{1}{l}\right) - \sqrt{2\sqrt{N}d + d^2} \to \sqrt{N} +\left(-\sqrt{N} + \sqrt{N}\frac{l+1}{\sqrt{2l+1}}\right)\left(1+\frac{1}{l}\right) - \sqrt{2\sqrt{N}\left(-\sqrt{N} + \sqrt{N}\frac{l+1}{\sqrt{2l+1}}\right) + \left(-\sqrt{N} + \sqrt{N}\frac{l+1}{\sqrt{2l+1}}\right)^2} </math>

Simplifying the monster of aan equation, we get <math>\frac{\sqrt{N}\left(\sqrt{2l+1}-1\right)}{l}</math>.

=== Facts ===

* If <math>l = 1</math> that means no sieving, <math>a_{\mathrm{max}} = \frac{2\sqrt{N}}{3}</math> and the cost becomes <math>\sqrt{N}\left(\sqrt{3}-1\right)

</math>, which is still better than pure trial divisondivision or pure fermatFermat

* The <math>a_{\mathrm{max}} - \sqrt{a_{\mathrm{max}}^2 - N}</math> which is the trial division bound becomes <math>\frac{\sqrt{N}}{\sqrt{2l+1}}</math> when subsututing the optimal.

=== Example ===

Using the same <math>N=2345678917</math>, if theresthere's no sieving then you should choose <math>a_{\mathrm{max}}</math> around 55924.69838392813, the reasonable choice <math>a_{\mathrm{max}} = 55000</math> is not that far off from the optimal with a bound of 28937, but the optimal choice gets a bound of 27962. If we are seivingsieving modulo 20, then <math>l = \frac{1}{\frac{4}{20}} = 5</math> and you should choose around <math>\sqrt{2345678917}\frac{6}{\sqrt{11}} \approx 87617.1636423325</math> and thsthis should intuitvlyintuitively make sense. If the fermatFermat part costs less, the spend more time in the fermatFermat part to lower <math>a_{\mathrm{max}} - \sqrt{a_{\mathrm{max}}^2 - N}</math>

==Multiplier improvement==