mersenneforum.org Possible P-1 Entry Example Error
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 2013-03-04, 03:07 #1 Jayder     "Jared" Dec 2012 2×139 Posts Possible P-1 Entry Example Error Been trying to figure out P-1 (I've mostly got it now) and I think there's an error with the example. 237876521^29 does not equal 171331425 (mod 2^29-1), but rather is 337474461 (mod 2^29-1). Further: when you take the gcd of either the old incorrect remainder or this new one, you get a gcd of 1. A fail for P-1, I believe. I'm guessing the bound needs to be risen? This may belong in the Discussion part of the wiki, but I don't have a wiki account yet and I wasn't sure if it would be noticed.
 2013-03-04, 03:32 #2 LaurV Romulan Interpreter     Jun 2011 Thailand 20D616 Posts Try figuring out more :D Code: gp > a=Mod(237876521,1<<29-1)^29 %1 = Mod(171331425, 536870911) gp > gcd(%-1,1<<29-1) %2 = Mod(486737, 536870911) gp >
 2013-03-04, 04:10 #3 Jayder     "Jared" Dec 2012 4268 Posts I'm afraid I don't get your point. The way that is written makes no sense to me. Are you showing me why I'm wrong, or why I'm right? Sorry if I made a mistake.
2013-03-04, 04:20   #4
Batalov

"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

11·19·43 Posts

Quote:
 Originally Posted by Jayder 237876521^29 does not equal 171331425 (mod 2^29-1), but rather is 337474461 (mod 2^29-1).
He was demonstrating that yes, 237876521^29 does indeed equal 171331425 (mod 2^29-1), and that your calculation was wrong.

You can use multiple tools to do that; it is far from being tricky because 2^29-1 (536870911) is a small number.
One is shown above (Pari/GP).
Another way is using dc:
Code:
> dc
237876521 29 ^    2 29 ^ 1 - % p
171331425
Yet another way is to write it in C (Visual basic, C#, etc etc), using "long long" or a similar type and calculating 237876521^29 by brute force: multiply 237876521 by 237876521 28 times, taking %536870911 every time. Can you try that?

 2013-03-04, 04:32 #5 Jayder     "Jared" Dec 2012 2×139 Posts I'm very sorry, the calculator I was using doesn't return that result. Too large a number I suppose. Thought I double checked in wolframalpha but I guess I forgot to. But then, what about the gcd? gcd(1713314245-1,2^29-1) in wolfram alpha returns 1. I am using windows so cannot use dc (or do not know how). I would use calc but the instructions provided for compiling(?) are not clear enough/too much trouble. I am a layman. I can read pseudocode and that's about it. But I am trying to learn. Last fiddled with by Jayder on 2013-03-04 at 04:42
 2013-03-04, 04:41 #6 Batalov     "Serge" Mar 2008 Phi(3,3^1118781+1)/3 11·19·43 Posts That's because you have a typo.
 2013-03-04, 04:45 #7 Jayder     "Jared" Dec 2012 2×139 Posts Yeah, I just figured that out. Only it's not my typo, it's in the wiki. As I do not have an account I cannot edit it myself. Thanks for clearing this up for me. Last fiddled with by Jayder on 2013-03-04 at 04:48
 2013-03-04, 04:52 #8 LaurV Romulan Interpreter     Jun 2011 Thailand 2·32·467 Posts My point was not only about calculus. You did not get the method either, as you missed subtracting 1 from the "old result" before taking the gcd (see your "further" part), threfore missed the factor. This is part of the method, not of the calculus. P-1 is very simple: From Fermat theorem, one has the fact that [TEX]b^{p-1}=1[/TEX] (mod p) for any prime p. Therefore, raising both sides at any power k, you get [TEX]b^{k*(p-1)}=1^k=1[/TEX] (mod p), or [TEX]b^{k*(p-1)}-1=0[/TEX] (mod p). Assuming you want to factor a big number n, which has a factor p, you might get lucky and find a multiple of p-1 very fast, if that p-1 has nothing but small factors. Then, taking the gcd of n with b[SUP]k*(p-1)[/SUP][COLOR=Red][B]-1 [/B][/COLOR]may reveal that factor. Edit: scrap this! Indeed there was a typo on wiki page. I just corrected it, by eliminating additional "4". Thanks for signaling it. Last fiddled with by LaurV on 2013-03-04 at 04:58

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