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 Subject: "Modular arithmetic" Previous Topic | Next Topic
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Monty
Member since Jul-27-04
May-12-07, 10:25 PM (EST)    "Modular arithmetic"

 The definition of mod is, I understand, if a = b mod n, then (a-b) is divisible by n, or a-b=kn, with k an integer I'm applying this to Fermat's little theorem: so if a^(p-1) = 1 mod p, then a^(p-1)-1 = kp multiplying both sides by a, we get a^p - a = kap = cpSo a^p - a = cp and a^(p-1)-1 = kp would seem to be equivalent.HOWEVER, they seem NOT equivalent if a=p a=p=2, 2^2-2=2=1*2, but 2^1-1=1 not a product of 2 a=p=3, 3^3-3=27-3=24=8*3, but 3^2-1=9-1=8 not a product of 3Why is this?It would seem that I'm somehow dividing by a-p, which is of course NG if a=p. But I don't see where.... ??? Monty

alexb Charter Member
2009 posts
May-13-07, 08:04 AM (EST)    1. "RE: Modular arithmetic"
In response to message #0

 If a is divisible by p then a = 0 (mod p). So, in this case, multipliying by a is multiplying by 0. Monty guest
May-15-07, 12:29 PM (EST)

2. "RE: Modular arithmetic"
In response to message #1

 I see what you're saying, but still don't understand. From the definition of mod I got an equation which said a^(p-1)-1=kp, that is it equals a product of p. I then multiplied by a and got a^p-a=cp, so it's also a product of p. If a=0 (mod p), then a=fp -- a is a product of p. So my second equation is fp^p-fp=cp. Let's see, I can factor out p, so I get fp^(p-1)-f=c. Now the function y=x^(x-1)+k is a little irregular (when x is negative....), but otherwise it's an ok function.So I'm still missing the point. alexb Charter Member
2009 posts
May-15-07, 12:39 PM (EST)    3. "RE: Modular arithmetic"
In response to message #2

 >I see what you're saying, but still don't understand. From >the definition of mod I got an equation which said >a^(p-1)-1=kp, This is only true if a is coprime with p.>I then >multiplied by a and got a^p-a=cp, This is true for any a, for if p|a the fact is trivial.>so it's also a product of p. If a=0 (mod p), then a=fp -- a is a product of p. So my >second equation is fp^p-fp=cp. This is trivially true, yes. >Let's see, I can factor out >p, so I get fp^(p-1)-f=c. Now the function y=x^(x-1)+k is a >little irregular (when x is negative....), but otherwise >it's an ok function. >>So I'm still missing the point. I am not sure what point it is, so I am going to check your original message.

alexb Charter Member
2009 posts
May-15-07, 12:42 PM (EST)    4. "RE: Modular arithmetic"
In response to message #0

 >The definition of mod is, I understand, > if a = b mod n, then (a-b) is divisible by n, or a-b=kn, >with k an integer > I'm applying this to Fermat's little theorem: > so if a^(p-1) = 1 mod p, then a^(p-1)-1 = kp > multiplying both sides by a, we get > a^p - a = kap = cp >So a^p - a = cp and a^(p-1)-1 = kp would seem to be >equivalent. In a sense, yes, but you must be cautious as to the meaning of equivalence.a^(p-1)-1 = kp holds only for a coprime with p, whereasa^p - a = cp holds for any a,

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