Fermat's factorization method

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

${\displaystyle N=a^{2}-b^{2}.}$

That difference is algebraically factorable as ${\displaystyle (a+b)(a-b)}$; if neither factor equals one, it is a proper factorization of N.

Each odd number has such a representation. Indeed, if ${\displaystyle N=cd}$ is a factorization of N, then

${\displaystyle N=\left({\frac {c+d}{2}}\right)^{2}-\left({\frac {c-d}{2}}\right)^{2}}$

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.

Basic method[]

One tries various values of a, hoping that ${\displaystyle a^{2}-N=b^{2}}$, a square.

FermatFactor(N): // N should be odd
    a ← ceiling(sqrt(N))
    b2 ← a*a - N
    repeat until b2 is a square:
        a ← a + 1
        b2 ← a*a - N 
     // equivalently: 
     // b2 ← b2 + 2*a + 1 
     // a ← a + 1
    return a - sqrt(b2) // or a + sqrt(b2)

For example, to factor ${\displaystyle N=5959}$, the first try for a is the square root of 5959 rounded up to the next integer, which is 78. Then, ${\displaystyle b^{2}=78^{2}-5959=125}$. Since 125 is not a square, a second try is made by increasing the value of a by 1. The second attempt also fails, because 282 is again not a square.

Try:	1	2	3
a	78	79	80
b2	125	282	441
b	11.18	16.79	21

The third try produces the perfect square of 441. So, ${\displaystyle a=80}$, ${\displaystyle b=21}$, and the factors of 5959 are ${\displaystyle a-b=59}$ and ${\displaystyle a+b=101}$.

Suppose N has more than two prime factors. That procedure first finds the factorization with the least values of a and b. That is, ${\displaystyle a+b}$ is the smallest factor ≥ the square-root of N, and so ${\displaystyle a-b=N/(a+b)}$ is the largest factor ≤ root-N. If the procedure finds ${\displaystyle N=1\cdot N}$, that shows that N is prime.

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

If N is prime (so that ${\displaystyle d=1}$), one needs ${\displaystyle O(N)}$ 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 ${\displaystyle {\left(4N\right)}^{1/4}}$ from ${\displaystyle {\sqrt {N}}}$, the method requires only one step; this is independent of the size of N.^{[citation needed]}

Fermat's and trial division[]

Consider trying to factor the prime number N = 2345678917, but also compute b and a − b throughout. Going up from ${\displaystyle {\sqrt {N}}}$, we can tabulate:

a	48,433	48,434	48,435	48,436
b2	76,572	173,439	270,308	367,179
b	276.7	416.5	519.9	605.9
a − b	48,156.3	48,017.5	47,915.1	47,830.1

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 ${\displaystyle a-b=47830.1}$, Fermat's method would have found it already.

Trial division would normally try up to 48,432; 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 ${\displaystyle c>{\sqrt {N}}}$; use Fermat's method for factors between ${\displaystyle {\sqrt {N}}}$ and ${\displaystyle c}$. This gives a bound for trial division which is ${\displaystyle c-{\sqrt {c^{2}-N}}}$. In the above example, with ${\displaystyle c=48436}$ the bound for trial division is 47830. A reasonable choice could be ${\displaystyle c=55000}$ giving a bound of 28937.

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

a	60,001	60,002
b2	1,254,441,084	1,254,561,087
b	35,418.1	35,419.8
a − b	24,582.9	24,582.2

Sieve improvement[]

When considering the table for ${\displaystyle N=2345678917}$, one can quickly tell that none of the values of ${\displaystyle b^{2}}$ are squares:

a	48,433	48,434	48,435	48,436
b2	76,572	173,439	270,308	367,179
b	276.7	416.5	519.9	605.9

It is not necessary to compute all the square-roots of ${\displaystyle a^{2}-N}$, nor even examine all the values for a. Squares are always congruent to 0, 1, 4, 5, 9, 16 modulo 20. The values repeat with each increase of a by 10. In this example, N is 17 mod 20, so subtracting 17 mod 20 (or adding 3), ${\displaystyle a^{2}-N}$ 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, ${\displaystyle a^{2}}$ must be 1 mod 20, which means that a is 1, 9, 11 or 19 mod 20; it will produce a ${\displaystyle b^{2}}$ which ends in 4 mod 20 and, if square, b will end in 2 or 8 mod 10.

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

modulo 16:	Squares are	0, 1, 4, or 9
	N mod 16 is	5
	so ${\displaystyle a^{2}}$ can only be	9
	and a must be	3 or 5 or 11 or 13 modulo 16
modulo 9:	Squares are	0, 1, 4, or 7
	N mod 9 is	7
	so ${\displaystyle a^{2}}$ can only be	7
	and a must be	4 or 5 modulo 9

One generally chooses a power of a different prime for each modulus.

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:
        b2 ← a*a - N
        if b2 is a square, modulo modulus:
            FermatSieve(N, a, aend, astep * modulus, NextModulus)
        endif
        a ← a + astep
    enddo

But the recursion is stopped 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 b2's with additions.

A further modular improvement can be made by applying the division algorithm as an affine transformation, that is ${\displaystyle X=\beta x+a}$, ${\displaystyle Y=\beta y+b}$, ${\displaystyle N=\beta z+r}$, over any integer ring ${\displaystyle \mathbb {Z} _{\beta }}$ where ${\displaystyle \beta <X}$. After a small amount of algebra, one can conclude that ${\displaystyle \lfloor N/\beta ^{2}\rfloor -s=xy}$ and ${\displaystyle \lfloor ab/\beta \rfloor =t}$ where s and t are identical to determining the carries one does in multiplying the divisors over base ${\displaystyle \beta }$.^{[citation needed]}

Multiplier improvement[]

Fermat's method works best when there is a factor near the square-root of N.

If the approximate ratio of two factors (${\displaystyle d/c}$) is known, then the rational number ${\displaystyle v/u}$ can be picked near that value. For example, if ${\displaystyle Y/X\approx q}$, then ${\displaystyle X\approx {\sqrt {N/q}}}$ is a good estimate for the smaller of a divisor pair. ${\displaystyle Nuv=cv\cdot du}$, and the factors are roughly equal: Fermat's, applied to Nuv, will find them quickly. Then ${\displaystyle \gcd(N,cv)=c}$ and ${\displaystyle \gcd(N,du)=d}$. (Unless c divides u or d divides v.) A further generalization of this approach assumes that ${\displaystyle Y^{n}/X^{m}\approx q}$, meaning that ${\displaystyle X\approx N^{m/(m+n)}/q^{1/(m+n)}}$.

Generally, if the ratio is not known, various ${\displaystyle u/v}$ values can be tried, 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 ${\displaystyle O(N^{1/3})}$ time.^[1]

Other improvements[]

The fundamental ideas of Fermat's factorization method are the basis of the quadratic sieve and general number field sieve, the best-known algorithms for factoring large semiprimes, which are the "worst-case". The primary improvement that quadratic sieve makes over Fermat's factorization method is that instead of simply finding a square in the sequence of ${\displaystyle a^{2}-n}$, 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[]

• Completing the square
• Factorization of polynomials
• Factor theorem
• FOIL rule
• Monoid factorisation
• Pascal's triangle
• Prime factor
• Factorization
• Euler's factorization method
• Integer factorization
• Program synthesis
• Table of Gaussian integer factorizations
• Unique factorization

Notes[]

References[]

• Fermat (1894), Oeuvres de Fermat, 2, p. 256
• McKee, J (1999). "Speeding Fermat's factoring method". Mathematics of Computation (68): 1729–1737.

External links[]

• Fermat's factorization running time, at blogspot.in
• Fermat's Factorization Online Calculator, at windowspros.ru