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In mathematics, the rational sieve is a general algorithm for factoring integers into prime factors. It is a special case of the general number field sieve. While it is less efficient than the general algorithm, it is conceptually simpler. It serves as a helpful first step in understanding how the general number field sieve works.

Method

Suppose we are trying to factor the composite number n. We choose a bound B, and identify the factor base (which we will call P), the set of all primes less than or equal to B. Next, we search for positive integers z such that both z and z+n are B-smooth — i.e. all of their prime factors are in P. We can therefore write, for suitable exponents $a_{i}$ ,

$z=\prod _{p_{i}\in P}p_{i}^{a_{i}}$

and likewise, for suitable $b_{i}$ , we have

$z+n=\prod _{p_{i}\in P}p_{i}^{b_{i}}$ .

But $z$ and $z+n$ are congruent modulo $n$ , and so each such integer z that we find yields a multiplicative relation (mod n) among the elements of P, i.e.

\prod _{p_{i}\in P}p_{i}^{a_{i}}\equiv \prod _{p_{i}\in P}p_{i}^{b_{i}}{\pmod {n}}

(where the a_i and b_i are nonnegative integers.)

When we have generated enough of these relations (it's generally sufficient that the number of relations be a few more than the size of P), we can use the methods of linear algebra to multiply together these various relations in such a way that the exponents of the primes are all even. This will give us a congruence of squares of the form a²≡b² (mod n), which can be turned into a factorization of n = gcd(a-b,n)×gcd(a+b,n). This factorization might turn out to be trivial (i.e. n=n×1), in which case we have to try again with a different combination of relations; but with luck we will get a nontrivial pair of factors of n, and the algorithm will terminate.

Example

We will factor the integer n = 187 using the rational sieve. We'll arbitrarily try the value B=7, giving the factor base P = {2,3,5,7}. The first step is to test n for divisibility by each of the members of P; clearly if n is divisible by one of these primes, then we are finished already. However, 187 is not divisible by 2, 3, 5, or 7. Next, we search for suitable values of z; the first few are 2, 5, 9, and 56. The four suitable values of z give four multiplicative relations (mod 187):

2^{1}3^{0}5^{0}7^{0}=2\equiv 189=2^{0}3^{3}5^{0}7^{1}

1

2^{0}3^{0}5^{1}7^{0}=5\equiv 192=2^{6}3^{1}5^{0}7^{0}

2

2^{0}3^{2}5^{0}7^{0}=9\equiv 196=2^{2}3^{0}5^{0}7^{2}

3

2^{3}3^{0}5^{0}7^{1}=56\equiv 243=2^{0}3^{5}5^{0}7^{0}

4

There are now several essentially different ways to combine these and end up with even exponents. For example,

(1)×(4): After multiplying these and canceling out the common factor of 7 (which we can do since 7, being a member of P, has already been determined to be coprime with n^[1]), this reduces to 2⁴ ≡ 3⁸ (mod n), or 4² ≡ 81² (mod n). The resulting factorization is 187 = gcd(81-4,187) × gcd(81+4,187) = 11×17.

Alternatively, equation (3) is in the proper form already:

(3): This says 3² ≡ 14² (mod n), which gives the factorization 187 = gcd(14-3,187) × gcd(14+3,187) = 11×17.

Limitations of the algorithm

Like the general number field sieve, the rational sieve cannot factor numbers of the form p^m, where p is a prime and m is an integer. This is not a huge problem, though—such numbers are statistically rare, and moreover there is a simple and fast process to check whether a given number is of this form. Probably the most elegant method is to check whether $\lfloor n^{1/b}\rfloor ^{b}=n$ holds for any $1<b\leq \log _{2}(n)$ using an integer version of Newton's method for the root extraction.^[2]

The biggest problem is finding a sufficient number of z such that both z and z+n are B-smooth. For any given B, the proportion of numbers that are B-smooth decreases rapidly with the size of the number. So if n is large (say, a hundred digits), it will be difficult or impossible to find enough z for the algorithm to work. The advantage of the general number field sieve is that one only needs to search for smooth numbers of order $\exp(C(\log n)^{2/3}(\log \log n)^{1/3})$ for some C > 0, rather than of order n as required here.^[3]

References

A. K. Lenstra, H. W. Lenstra, Jr., M. S. Manasse, and J. M. Pollard, The Factorization of the Ninth Fermat Number, Math. Comp. 61 (1993), 319-349. Available at [2].
A. K. Lenstra, H. W. Lenstra, Jr. (eds.) The Development of the Number Field Sieve, Lecture Notes in Mathematics 1554, Springer-Verlag, New York, 1993.

Footnotes

^ Note that common factors cannot in general be canceled in a congruence, but they can in this case, since the primes of the factor base are all required to be coprime to n, as mentioned above. See modular multiplicative inverse.
^ R. Crandall and J. Papadopoulos, On the implementation of AKS-class primality tests, available at [1]
^ A. K. Lenstra, H. W. Lenstra, Jr., M. S. Manasse, and J. M. Pollard, The Factorization of the Ninth Fermat Number, Math. Comp. 61 (1993), p. 328