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Old 2014-08-12, 00:41   #6
Rich
 
Apr 2011

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Sorry, I must be confused because n = 601 * 1801 * 6151 = 6657848551 seems to work. This is the same order of magnitude as the other solution quoted so it's unlikely to have been missed. It's less than 1010 and certainly less than 1020 that Erick Wong is supposed to have checked to.
I claim that O = 1025 = 210 + 1 = 52 * 41
21025 = 1 mod n
21025 / 5 = 5533233944 mod n and 21025 / 41 = 33554432 (which seem to have more than their fair share of digit pairs 33, 44 and 55)
This proves that the order of 2 is 1025 as required.
Finally n = 1025 * 6495462 + 1 so O | n-1 as required.

The next solution seems to be 13857901601 = 6151 * 2252951 = 1025 * 13519904 + 1
21025 / 5 = 12903300634 mod 13857901601 (hmm.. more digit pairs)
while
21025 = 1 mod 13857901601
This proves the order of 2 is 1025 again while n-1 is of the required form.

I believe there are a total of 4079 solutions with O=1025 of which an impressive one (but by no means the largest) is n = 19858924506932274923192217121413840253555986701099656443876494287833804013531009915252291156116144835199447717419022262305584364955410801

Indeed shouldn't any product of the primes where 2 has the required order qualify as a composite of the required form? As the orders get larger, the number of primes with the required order increase as well.

Contrary to what Bob claims, I believe that when we start talking about numbers n where log(log(n)) is not small then these sorts of numbers are actually quite common.

Last fiddled with by Rich on 2014-08-12 at 00:52 Reason: Adding material
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