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Factorization of n = 87463 with the
Quadratic Sieve
To find a factor base consider the values of
(n p
(^ p^ 2 3^5 7 11 13 17 19^ 23 29^31 n p
We thus select the factor base 2 , 3 , 13 , 17 , 19 , 29. Solutions for x^2 ≡ n (mod p) are: p 2 3 13 17 19 29 x 1 1, 2 5, 8 7, 10 5, 14 12, 17
We now start sieving, using a sieving interval of length 2 · 30 around b
nc = 295. For the values of x for which x^2 − n splits completely, the exponent vector modulo 2 is:
x −1 2 3 13 17 19 29 265 1 1 1 0 1 0 0 278 1 0 1 1 0 0 1 269 0 0 0 0 1 0 0 299 0 1 1 0 1 1 0 307 0 1 0 1 0 0 1 316 0 0 0 0 1 0 0
We now solve (the matrix is transposed as we solve Av = 0 and not vA = 0 ):
· v = 0
modulo 2. One solution is
v = (1, 1 , 1 , 0 , 1 , 0)
We thus take the 1st, 2nd, 3rd and the 4th x-value and get
x = 265 · 278 · 296 · 307 = 6694540240 ≡ 34757 (mod n)
y =
(265^2 − n) · (278^2 − n) · (296^2 − n) · (307^2 − n) = 2 · 34 · 132 · 17 · 29 = 13497354 ≡ 28052 (mod n)
This yields the gcds:
gcd(x − y, n) = 149, gcd(x + y, n) = 587
which give a factorization
