Originally Posted by

**mathematic** 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)

a^(2p)*1/a

This last step is legal only if $\displaystyle a\neq 0\!\!\pmod n$

a^2a^p*1/a

a^2*1/aa^p

How?? Perhaps you meant $\displaystyle a^2\cdot\frac{1}{a}\cdot a^p$? Otherwise it's wrong.

a^(2-1)a^p

aa^p

Then from Fermat's Theorem, a^p=a mod p and thus a^(2p-1)=a(mod2p)

Don't get it: you had $\displaystyle aa^p$ ,and applying flt, you get $\displaystyle aa^p=a^2\!\!\pmod p$ . How

does this translate into modulo 2p?

Tonio

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)