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.