Proving Fermat's Theorem (1): a^(n-1) = a (mod n)

[kathrynag]

(1). Prove the following statements.
(a). When n = 2p, where p is an odd prime, then a^(n-1) = a (mod n) for any integer a.
(b). For n = 195 = 3 * 5 *13, we have a^(n-2) = a (mod n) for any integer a


If I am correct Fermat's Theorem comes into play.
a)n=2p
a^(2p-1)
a^(2p)*(1/a)
a^2(a^p)*(1/a)
a^p=a for any prime
a^2*(a)*(1/a)
a^(2-1)*a
a*a
a^2
a mod n


b)a^(195-2)
a^(193)
193 is a prime so by Fermat's Theorem, a^p=a mod p
So a^193=a mod(193)
and thus a^193=a mod (195)

I feel like I did something wrong on both problems...

 

[mighty2000]

Your response:

Your use of Fermat's Theorem is correct and your steps are accurate. However, there is a minor mistake in the first problem. It should be a^p = a (mod p) instead of a^2 = a (mod p). Other than that, your proofs are correct and demonstrate the application of Fermat's Theorem in these scenarios. Good job!