Euler's theorem

Now we can state Euler's generalization.

Let $ n>1$ be an integer. Recall that $ \phi(n)$ is then number of positive integers relatively prime to $ n$ .

Theorem 1.8.4 (Euler's Theorem)   If $ n>1$ is an integer and $ gcd(a,n)=1$ then $ a^{\phi(n)}\equiv 1 ({\rm mod} n)$ .

proof: The proof of Fermat's Little Theorem above can be modified somewhat to work in this more general case as well. Exercise: Try to do this. $ \Box$

We know that if a number $ p$ is prime then $ a^p\equiv  a ({\bf mod} p)$ . An integer $ n$ for which $ a^n\equiv  a ({\bf mod} n)$ is called a Fermat pseudoprime to base $ a$ . Roughly speaking, if $ n$ is a Fermat pseudoprime to many different bases then $ n$ is ``probably'' a prime 1.7.

Example 1.8.5   We have $ \phi(10)=4$ (see Example 1.4.10 above), so by Euler's theorem, $ 1001^4 \equiv 1 ({\rm mod} 10)$ . We can check this by hand without a computer: by the binomial expansion $ 1001^4=(1000+1)^4=1000^4+4\cdot 1000^3 +6\cdot 1000^2 +4\cdot 1000 + 1
=1004006004001\equiv 1 ({\rm mod} 10)$ .



david joyner 2008-04-20