Fermat's factorization method: Difference between revisions

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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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It is not necessary to compute all the square-roots of <math>a^2-N</math>, nor even examine all the values for {{mvar|a}}. Squares are always congruent to 0, 1, 4, 5, 9, 16 [[Modular arithmetic|modulo]] 20, because these are the [[quadratic residue]]s of 20. The values repeat with each increase of {{mvar|a}} by 10. In this example, N is 17 mod 20, so subtracting 17 mod 20 (or adding 3), <math>a^2-N</math> produces 3, 4, 7, 8, 12, and 19 modulo 20 for these values. It is apparent that only the 4 from this list can be a square. Thus, <math>a^2</math> must be 1 mod 20, which means that {{mvar|a}} is 1, 9, 11 or 19 mod 20; it will produce a <math>b^2</math> which ends in 4 mod 20 and, if square, {{mvar|b}} will end in 2 or 8 mod 10.

This can be performed with any modulus. Using the same <math>N=2345678917</math>,

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=== Premise ===

An optimal <math>a_{\mathrm{max}}</math> can be computed using derivative methods. This derivation assumes you didint use any seiving.

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 sieving 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>

The goal is to choose a <math>a_{\mathrm{max}}</math> such that <math>C\left(d,N,l\right) = \frac{d}{l} + \left(a_{\mathrm{max}} - \sqrt{2\sqrt{N}d + d^2}\right) \to d\frac{1}{l}+ \left(\sqrt{N}+d\right) - \sqrt{2\sqrt{N}d + d^2} = \sqrt{N} +2dd\left(1+\frac{1}{l}\right) - \sqrt{2\sqrt{N}d + d^2}</math> is minimized.

=== Finding the Optimum ===

[[Derivative|Differentiate]] <math>C\left( d , N,l\right) </math> with respect to <math>d</math>. Due to the linearity of derivatives

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

Notice theBecause <math>\sqrt{N}</math> term dosentdoesn't depend on <math>d</math> sowe itscan derivativedrop respectthat toderivative d is 0term

Forfor the <math>2d </math> term we can use the constant multiple rule to get <math>\frac{d}{dd} 2d \to 2 left(1+\frac{d1}{ddl} d \right)</math> butterm, noticeuse the derivativemultiple ofrule <math>d</math> is just 1 so <math>2 \frac{d}{dd} d left(fg\to 2 right)'=f'g+g'f</math>.:

<math>C\left(-d\sqrt{N} left(1+ \frac{2\sqrt{N}1}{\sqrt{3l}},N\right)\right)' = \sqrt{N} +2d'\left(-\sqrt{N} 1+ \frac{2\sqrt{N}1}{\sqrt{3}l}\right) - \sqrt{2\sqrt{N}+\left(-\sqrt{N} 1+ \frac{2\sqrt{N}1}{\sqrt{3}l}\right)'d += \left(-\sqrt{N} 1+ \frac{2\sqrt{N}1}{\sqrt{3}l}\right)^2}</math>Simplyfing

For the last term <math>\sqrt{2\sqrt{N}d + d^2} </math> we use the chain rule:

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

Use linearity on the term to get <math>\frac{d}{dd}\sqrt{2\sqrt{N}d + d^2} = \tofrac{\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}{dd2}}}{2} \left(2\sqrt{N} + 2d\right) = \left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N} + d\right)</math>Substitute 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>

For the first term, use the constant multiple rule and the power rule to get:

To find the minimum, notice at the minimum the derivative vanishes, so st the derivative to 0

<math>\frac{d}{dd}2\sqrt{N}d \to 2\sqrt{N}\frac{d}{dd}d \to 2\sqrt{N} </math>

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

For the second term use the power rule: <math>\frac{d}{dd}d^2 = 2d </math>

Square both sides to remove the root then cross multiply

So subsututing known derivatives we get

<math>\left(\frac{\sqrt{N} + d}{dd}\sqrt{2\sqrt{N}d + d^2}}\right)^2 = \left(1+\frac{d1}{ddl}d\right)^2 \to \frac{\left(\sqrt{N} + d\right)^2}{2\sqrt{N}d + 2dd^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>

Expand LHS

<math>\frac{\left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}}{2}\frac{d}{dd}2\sqrt{N}d + d^2 \to \frac{\left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}}{2}2\left(\sqrt{N}+d\right) = \left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N}+d\right)</math><math>\frac{d}{dd} \sqrt{N} + \frac{d}{dd} 2d - \frac{d}{dd} \sqrt{2\sqrt{N}d + d^2} = \frac{d}{dd} \sqrt{N} + \frac{d}{dd} 2d - \frac{d}{dd} \sqrt{2\sqrt{N}d + d^2} = 2 - \left(2\sqrt{N}d + d^2\right)^{-\frac{1}{2}}\left(\sqrt{N}+d\right) </math>Setting this to 0 gives you the optimal d

<math>\left(\sqrt{N} + d\right)^2 -= \left(2\sqrt{N}d + d^2\right)^{-\left(1+\frac{1}{2}l}\left(\sqrt{N}+d\right) = 0^2 \to \frac{\sqrt{N} +d}{\sqrt{ 2\sqrt{N}d+d^2}} = 2 \to left(2\sqrt{N}+d =+ d^2\sqrt{2right)\sqrtleft(1+\frac{N1}d+d{l}\right)^2} </math>

Bring the right side to the left, Factor the common factor, Then, bring the second term to the right-hand side

Eliminating the square root

<math>N + 2\sqrt{N}d+d^2 = 2 \sqrt{left(2\sqrt{N}d + d^2} \to right)\left(1+\sqrtfrac{N1}{l}+d\right)^2 \to N = 4\left(2\sqrt{N}d+d^2\right)\left[\left(1+\frac{1}{l}\right)^2-1\right] </math>

Simplify the bracket

Simplifying

<math> \left(1+\sqrtfrac{N1}{l}+d\right)^2-1 = 41^2 + 2\times1\times\frac{1}{l}+\left(2\sqrtfrac{N1}{l}d+d^2\right)^2 \to- 1 = \sqrtfrac{N}^2}{l}+2\sqrtfrac{N1}d+d{l^2} = 8\sqrtfrac{N}d2l+2d1}{l^2} </math>

So, the equation is now

Subtract RHS from both sides and combining like terms

<math> \sqrt{N}^2+2\sqrt{N}d+d^2 = 8\sqrt{N}d+2d^2 \to N+left(2\sqrt{N}d+d^2 - 8\sqrtright)\frac{N2l+1}d+2d{l^2 = 0} \to N-6\sqrt{N}d-3dNl^2 = 0 \to 6left(2\sqrt{N}d+3dd^2-N = 0\right)\left(2l+1\right) </math>Apply the quadratic formula

To apply the quadratic formula to solve <math>d</math> we have to rewrite the equation to a quadratic. Write the right side as:

<math>d = \frac{-2\sqrt{N}\pm\sqrt{\left(2N\right)^2-4\times1\times\left(-\frac{N}{3}\right)}}{2} \to -\sqrt{N} \pm \frac{2\sqrt{N}}{\sqrt{3}}</math>

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

Since <math> d > 0 </math> take the positive solution.

Since <math>2l+1 \neq 0 </math>, you could divide through by it to get

Now get the optimal <math>a_{\mathrm{max}}</math> from the optimal <math>d</math>

<math>a_{\mathrm{max}}d^2 =+ 2\sqrt{N} + d \to \sqrt{N} + -\sqrt{N} + \frac{Nl^2\sqrt{N}}{\sqrt{3}2l+1} = \frac{2\sqrt{N}}{\sqrt{3}}0 </math>

This is the quadratic equation we been looking for, we can now apply the quardratic formula:

So under then permise of no seiving, optimally you should chose <math> \frac{2\sqrt{N}}{\sqrt{3}} </math>

<math>d = \frac{-2\sqrt{N}\pm\sqrt{\left(2N2\sqrt{N}\right)^2-4\times1\times\left(-\frac{NNl^2}{32l+1}\right)}}{2} \to -\sqrt{N} \pm \frac{2\sqrt{N}\frac{l+1}{\sqrt{32l+1}} </math>

Since <math> d > 0 </math> we take the positive solution. Since <math>a_{\mathrm{max}} = \sqrt{N} + d</math> one gets:

<math>a_{\fracmathrm{dmax}{dd}2 = \sqrt{N} + d \to 2\sqrt{N} + -\sqrt{N} + \sqrt{N}\frac{dl+1}{dd\sqrt{2l+1}}d \to= 2\sqrt{N} \frac{l+1}{\sqrt{2l+1}}</math>

=== Cost ===

SubstuteSubstitute our optimal <math> -\sqrt{N} + \frac{2\sqrt{N}}{\sqrt{3}} d</math> forinto <math>C\left( d , N,l\right) </math> we get

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

=== Facts ===

<math>C\left(-\sqrt{N} + \frac{2\sqrt{N}}{\sqrt{3}},N\right) = \sqrt{N} +2\left(-\sqrt{N} + \frac{2\sqrt{N}}{\sqrt{3}}\right) - \sqrt{2\sqrt{N}\left(-\sqrt{N} + \frac{2\sqrt{N}}{\sqrt{3}}\right) + \left(-\sqrt{N} + \frac{2\sqrt{N}}{\sqrt{3}}\right)^2}</math>Simplyfing

* If <math>\sqrt{N}l +2\left(-\sqrt{N}= +1</math> \frac{2\sqrt{N}}{\sqrt{3}}\right)that -means \sqrtno sieving, <math>a_{2\sqrtmathrm{Nmax}\left(-\sqrt{N} += \frac{2\sqrt{N}}{\sqrt{3}}\right)</math> +and \left(-\sqrt{N}the +cost \frac{2\sqrt{N}}{\sqrt{3}}\right)^2} \tobecomes <math>\sqrt{N}\left(\sqrt{3}-1\right)</math>

</math>, which is still better than pure trial division or pure Fermat

* 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 ===

So using Fermat + Trial divison we get a cost of <math>\sqrt{N}\left(\sqrt{3}-1\right)</math> which is a imporvement over plain trial division

Using the same <math>N=2345678917</math>, if there'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 sieving 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 this should intuitively make sense. If the Fermat part costs less, the spend more time in the Fermat part to lower <math>a_{\mathrm{max}} - \sqrt{a_{\mathrm{max}}^2 - N}</math>

==Multiplier improvement==