Fermat's factorization method: Difference between revisions

Revision as of 04:12, 13 June 2025

Revision as of 04:22, 13 June 2025

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{{Short description|MathematicsFactorization equationmethod based on the difference of two squares}}{{For|Fermat's method on determining extreme values|Interior extremum theorem}}{{Refimprove|date=February 2022}}

'''Fermat's factorization method''', named after [[Pierre de Fermat]], is based on the representation of an [[even and odd numbers|odd]] [[integer]] as the [[difference of two squares]]:

:<math>N = a^2 - b^2.</math>

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An optimal <math>a_{\mathrm{max}}</math> can be computed using derivative methods.

The cost of executing Fermat’s method from <math>\sqrt{N}</math> up to <math>a_{\mathrm{max}}</math> is roughly proportional to a constant we will call <math>d</math>. Using sevingsieving we can reduce it by some constant we call <math>l</math>. In the combined method the trial division bound becomes <math>a_{\mathrm{max}} - \sqrt{a_{\mathrm{max}}^2 - N}</math>. Writing <math>a_{\mathrm{max}} = \sqrt{N} + d</math>, one gets:

<math>a^2_{\mathrm{max}} - N = (\sqrt{N} + d)^2 - N = \sqrt{N}^2 + 2\sqrt{N}d + d^2 - N = 2\sqrt{N}d + d^2</math>

subsututingSubstitute the new formula, we get

<math>a_{\mathrm{max}} - \sqrt{a_{\mathrm{max}}^2 - N} \to a_{\mathrm{max}} - \sqrt{2\sqrt{N}d + d^2} </math>

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

Because <math>\sqrt{N}</math> dosentdoesn't depend on <math>d</math> we can drop that derivative term

for the <math>d\left(1+\frac{1}{l}\right)</math> term, use the multiple rule <math>\left(fg\right)'=f'g+g'f</math>:

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<math>\left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N} + d\right) = 0 \to \frac{\sqrt{N} + d}{\sqrt{2\sqrt{N}d + d^2}} = 1+\frac{1}{l}</math>

Square both sudessides to remvoeremove the root then cross multiply

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

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<math>\left(1+\frac{1}{l}\right)^2-1 = 1^2 + 2\times1\times\frac{1}{l}+\left(\frac{1}{l}\right)^2 - 1 = \frac{2}{l}+\frac{1}{l^2} = \frac{2l+1}{l^2} </math>

So, the equation is now

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