Prime Numbers

6.1 The quadratic sieve factorization method 277
where
c2 =(a 2 2 − a1)b 2 2 − 2a2b1b2 + b 2 1 +2b0b2,
c1 =(a1a2 − a0)b 2 2 − 2a1b1b2 +2b0b1,
c0 = a0a2b 2 2 − 2a0b1b2 + b 2 0.
Since b0,b1,b2 are free variables, perhaps they can be chosen so that they are
small integers and that c2 =0.Indeed,theycan.Let
b2 =2, b1 =2b, b0 = a1 − a 2 2 +2a2b − b 2 ,
where b is an arbitrary integer. With these choices of b0,b1,b2 we have
where
x(b) 2 ≡ y(b) (modn), (6.6)
x(b) =2m 2 +2bm + a1 − a 2 2 +2a2b − b 2 ,
y(b) = 4a1a2 − 4a0 − 4a1 +4a 2 2 b +8a2b 2 − 4b 3 m
+4a0a2 − 8a0b + a1 − a 2 2 +2a2b − b 2 2 .
Theproposalistoletbrun through small numbers, use a sieve to search for
smooth values of y(b), and then use a matrix of exponent vectors to find a
subset of the congruences (6.6) to construct two congruent squares mod n
that then may be tried for factoring n. Ifa0,a1,a2, andbare all O(nɛ ), where
0 ≤ ɛ