[math-fun] Wieferich numbers?
Wow. Nice. It appears that the base 2 6.7x10^15 search https://cs.uwaterloo.ca/journals/JIS/VOL14/Klyve/klyve3.html omitted pseudoprimes. A moronic Mma search thru 5.7G is so far fruitless. Has no one tried this before? --rwg rcs>If we vary the base, we can arrange for Wieferich pseudoprimes: Start with the known base 2 psp 341: 2^340 = 1 (mod 341) while 341 = 11*31 is composite. But 2^340 = 34783 (mod 341^2), so the "tens digit" is 102, [34782/341]. We muck around with the tens digit: (2+341K)^340 (mod 341^2) = 2^340 + 340 * 2^339 * 341K + stuff * 341^2. The coefficient of 341K is effectively -1/2, so Set K=204 to zero out the tens digit from 2^340. 69566^340 = 1 (mod 341^2) and 69566^341 = 69566 (mod 341^2). Rich ----- Quoting Bill Gosper <billgosper@gmail.com>: Is it obvious that if n∈N and 2==Mod[2^n,n^2] then n is (a Wieferich) prime? Why don't we just call A001220 Wieferich numbers? --rwg
On 4/30/2013 8:06 PM, Bill Gosper wrote:
Wow. Nice. It appears that the base 2 6.7x10^15 search https://cs.uwaterloo.ca/journals/JIS/VOL14/Klyve/klyve3.html omitted pseudoprimes. A moronic Mma search thru 5.7G is so far fruitless. Has no one tried this before? --rwg
Any prime divisor of a "Wieferich pseudoprime" (i.e., a composite n such that 2^n == 2 (mod n^2)) must itself be a Wieferich prime. A proof can be found here (of all places): http://forums.xkcd.com/viewtopic.php?t=23397&p=699486
Is it obvious that if n∈N and 2==Mod[2^n,n^2] then n is (a Wieferich) prime? Why don't we just call A001220 Wieferich numbers? --rwg
The term "Wieferich number" has been used for n such that 2^phi(n) == 1 (mod n^2); see A077816. -- Fred W. Helenius fredh@ix.netcom.com
participants (2)
-
Bill Gosper -
Fred W. Helenius