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Sep 30, 2019 at 0:20 comment added Charles @VS. Suppose you have a semiprime that's 1 mod 4. (If it was 3 mod 4 you'd know it wasn't a sum of squares already.) If you learn that it's a sum of two squares, the prime factors are both 1 mod 4 (or equal). If you learn it's not the sum of two squares, the prime factors are both 3 mod 4. Either way you go from knowing their last bit to knowing their last 2 bits.
Sep 30, 2019 at 0:09 comment added VS. @Charles I mean I want to know 'If it was easy, then it would give away the second-lowest bit in both prime factors of such numbers'. Also what is the blog link?
Sep 29, 2019 at 23:38 comment added Charles @VS. No method is known or believed to exist. Tao has a post on his blog about this.
Sep 29, 2019 at 21:17 comment added VS. @Charles How do you get second-lowest bit?
Mar 13, 2011 at 3:50 comment added Charles See also Tao's answer mathoverflow.net/questions/3820/… .
Mar 13, 2011 at 3:45 comment added Charles ?? You would CRT all the known residues together so that for one of the primes dividing n, p = r mod m. Then use Coppersmith to find numbers in that residue class which have a nontrivial gcd with n, recovering the factor. Isn't that a standard use of the algorithm? Of course you'd need a lot of information, but Coppersmith makes "a lot" a much smaller number than if the only available algorithm was testing numbers of the form r, r+m, r+2m, ... sequentially. (Obviously, if only a very small amount of information is known, general-purpose factorization is faster.)
Mar 12, 2011 at 20:12 comment added H A Helfgott Please explain - how would you use Coppersmith's algorithm? What information would you be assuming, precisely?
Mar 9, 2011 at 19:28 history edited Charles CC BY-SA 2.5
Coppersmith's algorithm
Mar 9, 2011 at 19:20 history answered Charles CC BY-SA 2.5