Right. I'm kinda getting a bit mixed up on whether my question is actually true (i.e, is it an 'iff' question or just an 'if' question)...
The counter example given was 561 but the problem as I understand it is...
If for all , then is prime.
But this is not true for 561.
mod 561. Hence 561 does not satisfy the above... What am I missing here? It holds true for may primes that I have tested using Maple leading me to believe that the part I'm stuck on can be proved.
Ok look. I get what iff means but 561 is not a counter example.
For 561 to be a counter example then for every , mod 561.
But after running it through in Maple you can show that for many values for a
mod 561. (don't know the Latex for not equivalent)
Hence 561 is not prime.
Ok folks, I'm back.
Hopefully with an understanding.
It seems perhaps the proof wasn't so complicated as I first thought...
=> Fermat's little theorem. (since all a in this example are coprime to p if p is prime)
<= If , all a are coprime to p, hence p has no divisors and so is prime.
Is that all it is?