- Fermat's factorization method
is based on the representation of an oddFermat 's factorization methodinteger as thedifference of two squares : :$N\; =\; a^2\; -\; b^2.$ That difference isalgebra ically factorable as $(a+b)(a-b)$; if neither factor equals one, it is a proper factorization of "N".Each odd number has such a representation. Indeed, if $N=cd$ is a factorization of "N", then :$N\; =\; [(c+d)/2]\; ^2\; -\; [(c-d)/2]\; ^2.$ Since "N" is odd, then "c" and "d" are also odd, so those halves are integers. (A multiple of four is also a difference of squares: let "c" and "d" be even.)

In its simplest form, Fermat's method might be even slower than trial division (worst case). Nonetheless, the combination of trial division and Fermat's is more effective than either.

**The basic method**One tries various values of "a", hoping that $a^2-N\; =\; b^2$ is a square.

:FermatFactor(N): // N should be odd::A ← ceil(sqrt(N))::Bsq ← A*A - N::while Bsq isn't a square::::A ← A + 1:::Bsq ← A*A - N // equivalently: Bsq ← Bsq + 2*A + 1::endwhile::return A - sqrt(Bsq) // or A + sqrt(Bsq)

For example, to factor $N=5959$, one computes

The third try produces a square. $A=80$, $B=21$, and the factors are $A-B\; =\; 59$, and $A+B\; =\; 101$.A: 78 79 80 Bsq: 125 282 441 Suppose N has more than two prime factors. That procedure first finds the factorization with the least values of "a" and "b". That is, $a+b$ is the smallest factor ≥ the square-root of "N". And so $a-b\; =\; N/(a+b)$ is the largest factor ≤ root-"N". If the procedure finds $N=1*N$, that shows that "N" is prime.

For $N=cd$, let "c" be the largest subroot factor. $a\; =\; (c+d)/2$, so the number of steps is approximately $(c+d)/2\; -\; sqrt\; N\; =\; (sqrt\; d\; -\; sqrt\; c)^2\; /\; 2\; =\; (sqrt\; N\; -\; c)^2\; /\; 2c$.

If "N" is prime (so that $c=1$), one needs $O(N)$ steps! This is a bad way to prove primality. But if "N" has a factor close to its square-root, the method works quickly. More precisely, if c differs less than $\{left(4N\; ight)\}^\{1/4\}$ from $sqrt\; N$ the method requires only one step. Note, that this is independent of the size of N.

**Fermat's and trial division**Let's try to factor the prime number N=2345678917, but also compute B and A-B throughout. Going up from $sqrt\{N\}$, we can tabulate:

A: 48433 48434 48435 48436 B ^{2}:76572 173439 270308 367179 B: 276.7 416.5 519.9 605.9 A-B: 48156.3 48017.5 47915.1 47830.1 In practice, one wouldn't bother with that last row, until "B" is an integer. But observe that if "N" had a subroot factor above $A-B=47830.1$, Fermat's method would have found it already.

Trial division would normally try up to 48432; but after only four Fermat steps, we need only divide up to 47830, to find a factor or prove primality.

This all suggests a combined factoring method. Choose some bound $c\; >\; sqrt\{N\}$; use Fermat for factors between $sqrt\{N\}$ and $c$. This gives a bound for trial division which is $c\; -\; sqrt\{c^2\; -\; N\}$. In the above example, with $c\; =\; 48436$ the bound for trial division is 47830. A reasonable choice could be $c\; =\; 55000$ giving a bound of 28937.

In this regard, Fermat's method gives diminishing returns. One would surely stop before this point:

A: 60001 60002 B ^{2}:1254441084 1254561087 B: 35418.1 35419.8 A-B: 24582.9 24582.2 **ieve improvement**One needn't compute all the square-roots of $a^2-N$, nor even examine all the values for $a$. Examine the tableau for $N=2345678917$:

A: 48433 48434 48435 48436 Bsq: 76572 173439 270308 367179 B: 276.7 416.5 519.9 605.9 One can quickly tell that none of these values of Bsq are squares. Squares end with 0, 1, 4, 5, 9, or 16 modulo 20. The values repeat with each increase of $a$ by 10. For this example $a^2-N$ 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, $a^2$ must be 1 mod 20, which means that $a$ is 1 or 9 mod 10; it will produce a Bsq which ends in 4 mod 20, and if Bsq is a square, $b$ will end in 2 or 8 mod 10.

This can be performed with any modulus. Using the same $N=2345678917$,

One generally chooses a power of a different prime for each modulus.modulo 16: Squares are 0, 1, 4, or 9 N mod 16 is 5 so $a^2$ can only be 9 and $a$ must be 3 or 5 modulo 16 modulo 9: Squares are 0, 1, 4, or 7 N mod 9 is 7 so $a^2$ can only be 7 and $a$ must be 4 or 5 modulo 9 Given a sequence of "a"-values (start, end, and step) and a modulus, one can proceed thus:

:FermatSieve(N, Astart, Aend, Astep, Modulus)::A ← Astart::do Modulus times::::Bsq ← A*A - N:::if Bsq is a square, modulo Modulus:::::FermatSieve(N, A, Aend, Astep * Modulus, NextModulus):::endif:::A ← A + Astep::enddo

But one stops the recursion, when few "a"-values remain; that is, when (Aend-Astart)/Astep is small. Also, because "a"'s step-size is constant, one can compute successive Bsq's with additions.

**Multiplier improvement**Fermat's method works best when there is a factor near the square-root of "N". Perhaps one can arrange for that to happen.

If one knows the approximate ratio of two factors ($d/c$), then one can pick a rational number $v/u$ near that value. $Nuv\; =\; cv\; *\; du$, and the factors are roughly equal: Fermat's, applied to "Nuv", will find them quickly. Then $gcd(N,cv)=c$ and $gcd(N,du)=d$. (Unless "c" divides "u" or "d" divides "v".)

Generally, one does not know the ratio, but one can try various $u/v$ values, and try to factor each resulting "Nuv". R. Lehman devised a systematic way to do this, so that Fermat's plus trial-division can factor N in $O(N^\{1/3\})$ time.See R. Lehman, "Factoring Large Integers", "Mathematics of Computation", 28:637-646, 1974.

**Other improvements**The fundamental ideas of Fermat's factorization method are the basis of the

quadratic sieve andgeneral number field sieve , the best-known algorithms for factoring "worst-case" largesemiprimes . The primary improvement that quadratic sieve makes over Fermat's factorization method is that instead of simply finding a square in the sequence of "a"^{2}−"n", it finds a subset of elements of this sequence whose "product" is a square, and it does this in a highly efficient manner. The end result is the same: a difference of square mod "n" that, if nontrivial, can be used to factor "n".See also J. McKee, " [

*http://www.ams.org/mcom/1999-68-228/S0025-5718-99-01133-3/home.html Speeding Fermat's factoring method*] ", "Mathematics of Computation", 68:1729-1737 (1999).**External links*** [

*http://www.patrickkonsor.com/code/ Java Implementation Of Fermat's method and trial division*]

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