Williams's p + 1 algorithm

Choose some integer A greater than 2 which characterizes the Lucas sequence:

${\displaystyle V_{0}=2,V_{1}=A,V_{j}=AV_{j-1}-V_{j-2}}$

where all operations are performed modulo N.

Then any odd prime p divides ${\displaystyle \gcd(N,V_{M}-2)}$ whenever M is a multiple of ${\displaystyle p-(D/p)}$, where ${\displaystyle D=A^{2}-4}$ and ${\displaystyle (D/p)}$ is the Jacobi symbol.

We require that ${\displaystyle (D/p)=-1}$, that is, D should be a quadratic non-residue modulo p. But as we don't know p beforehand, more than one value of A may be required before finding a solution. If ${\displaystyle (D/p)=+1}$, this algorithm degenerates into a slow version of Pollard's p − 1 algorithm.

So, for different values of M we calculate ${\displaystyle \gcd(N,V_{M}-2)}$, and when the result is not equal to 1 or to N, we have found a non-trivial factor of N.

The values of M used are successive factorials, and ${\displaystyle V_{M}}$ is the M-th value of the sequence characterized by ${\displaystyle V_{M-1}}$.

To find the M-th element V of the sequence characterized by B, we proceed in a manner similar to left-to-right exponentiation:

x := B           
y := (B ^ 2 − 2) mod N     
for each bit of M to the right of the most significant bit do
    if the bit is 1 then
        x := (x × y − B) mod N 
        y := (y ^ 2 − 2) mod N 
    else
        y := (x × y − B) mod N 
        x := (x ^ 2 − 2) mod N 
V := x

With N=112729 and A=5, successive values of ${\displaystyle V_{M}}$ are:

V₁ of seq(5) = V_1! of seq(5) = 5

V₂ of seq(5) = V_2! of seq(5) = 23

V₃ of seq(23) = V_3! of seq(5) = 12098

V₄ of seq(12098) = V_4! of seq(5) = 87680

V₅ of seq(87680) = V_5! of seq(5) = 53242

V₆ of seq(53242) = V_6! of seq(5) = 27666

V₇ of seq(27666) = V_7! of seq(5) = 110229.

At this point, gcd(110229-2,112729) = 139, so 139 is a non-trivial factor of 112729. Notice that p+1 = 140 = 2² × 5 × 7. The number 7! is the lowest factorial which is multiple of 140, so the proper factor 139 is found in this step.

Using another initial value, say A = 9, we get:

V₁ of seq(9) = V_1! of seq(9) = 9

V₂ of seq(9) = V_2! of seq(9) = 79

V₃ of seq(79) = V_3! of seq(9) = 41886

V₄ of seq(41886) = V_4! of seq(9) = 79378

V₅ of seq(79378) = V_5! of seq(9) = 1934

V₆ of seq(1934) = V_6! of seq(9) = 10582

V₇ of seq(10582) = V_7! of seq(9) = 84241

V₈ of seq(84241) = V_8! of seq(9) = 93973

V₉ of seq(93973) = V_9! of seq(9) = 91645.

At this point gcd(91645-2,112729) = 811, so 811 is a non-trivial factor of 112729. Notice that p−1 = 810 = 2 × 5 × 3⁴. The number 9! is the lowest factorial which is multiple of 810, so the proper factor 811 is found in this step. The factor 139 is not found this time because p−1 = 138 = 2 × 3 × 23 which is not a divisor of 9!

As can be seen in these examples we do not know in advance whether the prime that will be found has a smooth p+1 or p−1.

Based on Pollard's p − 1 and Williams's p+1 factoring algorithms, Eric Bach and Jeffrey Shallit developed techniques to factor n efficiently provided that it has a prime factor p such that any k^th cyclotomic polynomial Φ_k(p) is smooth.[1] The first few cyclotomic polynomials are given by the sequence Φ₁(p) = p−1, Φ₂(p) = p+1, Φ₃(p) = p²+p+1, and Φ₄(p) = p²+1.

• Williams, H. C. (1982), "A p+1 method of factoring", Mathematics of Computation, 39 (159): 225–234, doi:10.2307/2007633, MR 0658227

• P Plus 1 Factorization Method, MersenneWiki.