Talk:Pollard's p - 1 algorithm

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a^(p-1) = 1 mod (p) for a in U*(Z/nZ) is meaningless and is not the correct statement of fermat's little theorem.

I don't think you really understand the meaning of the notation. For starters mod is an equivalence relation on Z and

a^(p-1) = 1 mod (p) means

[a^(p-1) -1] / p = 0

But when you say that a is in U*(Z/nZ) then what does it mean to divide an element of U*(Z/nZ) by the integer p?

Fermat's little theorem does say that for all a!

a^(p-1) = 1 mod (p)

I don't know what you're trying to say ,but it's not fermat's little theorem and it's mathematically ungramatical.

First of all Fermat's little theorem isn't stated in its verbatim form for a reason. However it's a short hop from FLT to the statement in the article, and the statement in the article is correct.

The notation YOU are using isn't standard, although you may think it correct. a^(p-1) = 1 mod (p) is not valid notation. To indicate the equivalence you need to use ≡ (note: three lines) and (mod p) (note: parentheses surrounding the whole thing). To indicate a computation you would simply use = (note: two lines) and mod p (note: no parentheses).

Lastly, Fermat's little theorem doesn't say that for all a that "a^(p-1) = 1 mod (p)", there is a condition that a has to be coprime to p. For instance, if you picked a and p to be the same value then the equation is wrong (as I mentioned in the summary when I reverted your change before). CryptoDerk 01:45, Oct 22, 2004 (UTC)

Well it doesn't really matter, say whatever you like. Nobody is going to rely on this entry for anything.

[edit] Prime factors?

"It is possible that all the prime factors of n are divisible by small primes, at which point the Pollard p-1 algorithm gives you n again."

If a prime factor is divisible by a small prime.... how is it a prime factor? —Preceding unsigned comment added by 129.97.22.119 (talk) 23:53, 11 February 2008 (UTC)

I think the mistake is between P being a prime factor and P-1 being divisible by a small prime. The quoted statement is an error. It needs to be worded differently. Robomojo (talk) 09:14, 31 May 2008 (UTC)