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.

  • 2
    $\begingroup$ Did you really need to enclose all your text in dollar signs? $\endgroup$ – Sean Roberson Jan 27 '17 at 18:43
  • 2
    $\begingroup$ Why man, why. I can't even edit it without considerable effort due to all the backslashes.. $\endgroup$ – s.harp Jan 27 '17 at 18:55
  • $\begingroup$ Life is hard... $\endgroup$ – TheGeekGreek Jan 27 '17 at 18:56
  • 1
    $\begingroup$ Why don't you try to similar things that worked here? $\endgroup$ – Jyrki Lahtonen Jan 27 '17 at 18:59
  • 1
    $\begingroup$ @Jyrki But there is no need to factor any large numbers. It suffices to factor $54$ - see my answer. $\endgroup$ – Bill Dubuque Jan 27 '17 at 19:38

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.