The "universal" RSA keys

This all starts with an earlier observation in IRC, and the discussion thereupon :

asciilifeform check it out, all the nontrivial factors found have very interesting binary symmetries.
key 4.1 : 17742509903907 = 1000000100011 0000000000000000000 1000000100011
key 4.2 : 4294967297 = 100000000000000000000000000000001
key 5.1 : 7301444404 = 1101100 11001100 11001100 1100110100
key 5.2 : 270582939711 = 111111 00000000000000000000000000 111111
key 6.1 : like key 4.2.
key 6.2 : absent
key 7.1 : 98784247831 = 1011100000000000000000000000000010111
key 7.2 : 30064771079 = 11100000000000000000000000000000111
key 8.1 : 12884901891 = 1100000000000000000000000000000011
key 8.2 : 21474836485 = 10100000000000000000000000000000101
key 9.1 : like key 4.2.
key 9.2 : absent
key 10.1 : 4294967297 = like key 4.2.
key 10.2 : like key 8.1

Leaving aside for the moment all other considerations, let's look at the symmetric factors. Known so far :

100000000000000000000000000000001,
	which is 1 on both ends of a string of 31 0s (33 bits total).
1100000000000000000000000000000011,
	which is 11 on both ends of a string of 30 0s (34 bits total).
10100000000000000000000000000000101,
	which is 101 on both ends of a string of 29 0s (35 bits total).
11100000000000000000000000000000111,
	which is 111 on both ends of a string of 29 0s (35 bits total).
1011100000000000000000000000000010111,
	which is 10111 on both ends of a string of 27 0s (37 bits total).
11111100000000000000000000000000111111,
	which is 111111 on both ends of a string of 28 0s (38 bits total).
100000010001100000000000000000001000000100011
	which is 1000000100011 on both ends of a string of 19 0s
	(45 bits total).

One way to describe this set would be to say that a 0-padded 32 bit representation of a numeric value is appended to the end of that numeric valueⁱ. The numbers in question are

• 1₂ = 1₁₀ ;
• 11₂ = 3₁₀ ;
• 101₂ = 5₁₀ ;
• 111₂ = 7₁₀ ;
• 10111₂ = 23₁₀ ;
• 111111₂ = 63₁₀ (=3² * 7) ;
• 1000000100011₂ = 4131₁₀ (=3⁵ * 17).

The first five of these Elements 2 through 5 are prime, and the first four an exhaustive list of primes (excepting of course 2), but both these circumstances appear coincidental in light of the larger values encountered.

It seems at this point altogether reasonable to take the following steps :

1. Take all odd numbers up to 8192 (14 bit integers) and add the 32 bit zero padded number in question to create 64 bit values.
2. Multiply these values together, first to last, second to penultimate etc
3. Repeat the previous step 7 times to obtain 512 different 4096 bit "moduli".
4. Multiply the smallest by 2 to make sure we've introduced that factor in the pool as well.
5. Transform these into keys and introduce them into Phuctor

This would potentially be useful because Phuctor is only capable to identify shared divisors - up until the moment we found the 2nd key divisible by 3 it was unaware the first one existed. Consequently, the explanation as to why 111 and 10111 are present but 10101 and 11101 are not not may well be that they are, but only appear once. (And for that matter even 110ⁱⁱ could be in there for all we know!)

If these "universal" RSA keys actually open any further keys in the database, it'd be the first time since the original results were announced last Sunday that anyone proposed a useful, testable hypothesis - a sorer indictment of the soi dissant cryptographical community resident on the Internet being scarcely conceivable.

1. In a process muchly reminiscent of the diddled exponents discussed earlier. [↩]
2. 110 is distinct from 11 here - 11 exists and appears as 1100000000000000000000000000000011₂ = 4294967299₁₀, whereas 110 would appear as 11000000000000000000000000000000110₂ = 25769803782₁₀ [↩]