# Generalized Fermat's Little Theorem

Fermat's Little Theorem says that if p is prime and if p and a are coprime then p divides (a^p-a). Now if n is a positive integer is there a function of n f(n) such that if a and n are coprime then p divides (a^f(n)-a)? I have worked on some examples and from the fact that f(p)=p-1 p a prime I think that f(n) is the number of positive integers less than n and coprime to it, but how can it be proved? Note the number of positive integers less than n and coprime to it for the kth power of a prime are (p^(k-1))(p-1).

Note by Samuel Queen
5 years, 7 months ago

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Actually Euler's totient function serves as a generalisation to Fermat's Little theorem as this theorem is often referred to as.Since totient function for any prime number p equals (p-1), Fermat's theorem gives us a special case of Euler's theorem

- 5 years, 7 months ago

Your hunch is correct provided a typo in your statement is corrected. The statement should read $n \text{ divides }a^{f(n)}-1$ or $f(n) = 1 + \text{Number of numbers relatively prime to n and less than n}.$ The fact that, if $$f(n)$$ is the number of relatively prime numbers less than $$n$$ and if $$\gcd(a,n)=1$$, then $$n$$ divides $$a^{f(n)}-1$$, is known as Euler's theorem. Calvin has a wonderful blog post on this. The function $$f(n)$$ is called the Euler's totient function and is usually denoted as $$\phi(n)$$.

- 5 years, 7 months ago