.I was wondering if someone might be able to look over my proof's involving Fermat's Theorem.
When n = 2p, where p is an odd prime, then a^(n-1)= a (mod n) for any integer a.
So let n=2p
We have a^(2p-1)
This last step is legal only if
How?? Perhaps you meant ? Otherwise it's wrong.
Then from Fermat's Theorem, a^p=a mod p and thus a^(2p-1)=a(mod2p)
Don't get it: you had ,and applying flt, you get . How
does this translate into modulo 2p?
For n = 195 = 3 * 5 * 13, we have a^(n-2) = a (mod n) for any integer a.
Then we have a^193
Since 193 is a prime, we have a^193=a mod 193 by Fermat's Theorem and thus a^193=a (mod195)