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).

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TopNewestActually 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 – Jaydutt Kulkarni · 4 years, 5 months ago

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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 \(\phi(n)\). – Marvis Narasakibma · 4 years, 6 months ago

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