Find the biggest integer number $k$ such that $k$ divides $n^{ 55 }-n$, where $n$ is any integer number.
This question was a challenge from my teacher, here's my attempt :
- because $\left( n \right) \left( n-1 \right) \left( { n }^{ 53 }+{ n }^{ 52 }+{ n }^{ 51 }+...+n+1 \right) \equiv 0 \pmod k $, therefore $n$ could be $2$,
but my teacher told me that there is a number which is bigger than $2$.
I've tried to solve it many times, but unfortunately I didn't get any solution. So I hope that you can help me to approach this problem.
Hint $\ $ By the Theorem below we deduce $\,k\mid n^{\large 55}-n\,$ for all $n$ iff $\,k\,$ is a product of distinct primes $p$ such that $\,p-1\mid 54$, i.e. $\,p =2,3,7,19.\,$ Thus the largest such $k$ is their product $= 798.$
Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$
$\qquad\rm n\:|\:a^{\large e}-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\:\Rightarrow\: p\!-\!1\:|\:e\!-\!1$
Proof $\ (\Leftarrow)\ \ $ Since a squarefree natural divides another iff all its prime factors do, we need only show $\rm\:p\:|\:a^{\large e}\!-\!a\:$ for each prime $\rm\:p\:|\:n,\:$ or, that $\rm\:a \not\equiv 0\:\Rightarrow\: a^{\large e-1} \equiv 1\pmod p,\:$ which, since $\rm\:p\!-\!1\:|\:e\!-\!1,\:$ follows from $\rm\:a \not\equiv 0\:$ $\Rightarrow$ $\rm\: a^{\large p-1} \equiv 1 \pmod p,\:$ by little Fermat.
$(\Rightarrow)\ \ $ See this answer.