A prime p is said to be safe prime if (p-1)/2 is also a prime. Safe primes are heavily used in cryptography. In order to generate a random prime of 512 bits, I use
random_prime(2^512-1, false, 2^511)
How to randomly generate a safe prime of given length?
You can use a rejection method: try a new random prime until you get a safe one:
sage: def rdp(nbits=512):
....: while True:
....: p = random_prime(2^nbits-1, false, 2^(nbits-1))
....: if ZZ((p-1)/2).is_prime():
....: return p
....:
sage: rdp()
9075892275451579482861175762932106084253268411540277039314929550059578374557168263115660009649748089090540343402740578371326513035684052420493647284687893
For other number of bits:
sage: rdp(200)
1011992113066581800000651861935373321407012277464235251049677
Thanks for your snippet, but a typo/bug! Must be: "ZZ((p-1)/2).is_prime()". Not plus. Minus.
A much faster method is to generate a random integer p, to try dividing both p and (p-1)/2 by the first 100000 primes (just generate a table with those primes, using any suitable method) and reject p as soon as it is found that either p or (p-1)/2 has a divisor. If this is successful, then use the Miller-Rabin test on p and on (p-2)/2, using the first 10 primes as witnesses.
This is much faster because most p value will be rejected quickly, while first finding a prime p is much work and this work is useless because most of the time you can quickly find that (p-1)/2 is NOT prime. Better evaluate p and (p-1)/2 together, not focusing on one of them while ignoring the other one for a long time.
Asked: 2018-12-24 00:32:36 +0100
Seen: 6,985 times
Last updated: Apr 30 '22
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