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"To show that a number is a Riesel number, one must find a 'covering set'. A covering set is a set of small prime numbers that will divide any member of a sequence, so called because it is said to 'cover' that sequence."
There are at least three necessary conditions for that statement:
- there's no such thing as a nonconstructive proof that a number is Riesel
- one can define "small"
- every Riesel number even has a finite covering set - can't there be one whose sequence of smallest factors is unbounded?
Does anybody have a proof of any of these statements? -- Smjg 16:54, 14 September 2005 (UTC)
- I suppose "small" here would simply be finite. Depending on the exact definition of covering set used, I could see an argument for them, but if the definition is that broad (lacking a proof of your first and third points) then I can't see any justification for talking about them, because they could be too unwieldy to work with. I've changed the wording in the article to reflect the possibility of nonconstructive proofs; anyone with proofs for your first or third points should add them to the article. CRGreathouse 22:50, 25 May 2006 (UTC)
- The burden of proof is on us for implying that every Riesel number can be shown with this method, if it's true that numbers don't have a finite covering set when the sequence of smallest factors is unbounded. If we can't find proof that it will work for all the numbers, we should only discuss methods that will. The existing material is unsourced, after all, it doesn't need immunity. ᛭ LokiClock (talk) 13:18, 30 December 2011 (UTC)
It would be good to have the reference to Riesel's original work. Two web sites suggest H. Riesel (1956). "Naagra stora primtal". Elementa. 39: 258–260. but I can't verify that (it isn't in ZMATH). Richard Pinch (talk) 06:39, 3 August 2008 (UTC)
(my) Riesel's theorem (as it should be named): ... let R= k*2^n -1, n >2 is a natural number and k <= 2^n -1. If for some base 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime; (similar to Proth's theorem, only gcd(b, 3)= 1 is necessary!) ... proof: if 'm' is from the set of natural numbers, then every odd prime divisor 'r' of a^(2^m) +1 implies that q == +1(mod a^(m+1)) [concluded from generalized Fermat-number 'proofs' by Proth, and with after my examination of Proth's theorem]. ... now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^(2^(n-1)) == +1 (mod q) implies that q == +1 (mod 2^n). ... thus, if 'S' is composite, 'S' will be the product of at least two primes each of which may have a maximum value of (2^n -1), and it follows that... ... k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n) +1; implies that -2 <= -2*(2^n) and dividing by 2^n and multiplying by -1, we have.. 1 > 2^n, contradiction! ... hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +1 (mod R), then 'R' is prime.
... I didn't violate any copyrights; it's MY modification that a high school student could verify. Thanks, Bill; my e-mail is email@example.com ...