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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.

Jyrki Lahtonen
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    Did you really need to enclose all your text in dollar signs? – Sean Roberson Jan 27 '17 at 18:43
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    Why man, why. I can't even edit it without considerable effort due to all the backslashes.. – s.harp Jan 27 '17 at 18:55
  • Life is hard... – TheGeekGreek Jan 27 '17 at 18:56
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    Why don't you try to similar things that worked here? – Jyrki Lahtonen Jan 27 '17 at 18:59
  • For example by Little Fermat we know that $19\mid n^{19}-n$ for all $n$. Do you see how that implies that $19\mid n^{55}-n$ for all $n$ also? Same with $3$ and $7$? – Jyrki Lahtonen Jan 27 '17 at 19:01
  • To aid the search: $$2^{55}-2=2\cdot3^4\cdot7\cdot19\cdot73\cdot87211\cdot262657.$$ Check the highest power of $3$ dividing $3^{55}-3$ and use what you know about primitive roots to deal with the larger primes. – Jyrki Lahtonen Jan 27 '17 at 19:29
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    @Jyrki But there is no need to factor any large numbers. It suffices to factor $54$ - see my answer. – Bill Dubuque Jan 27 '17 at 19:38
  • @Bill Agreed. That was admittedly a Sonnhardish way of going about it. Mind you, I was mildly worried about divisibility by prime powers. Your other answer saves the day :-) – Jyrki Lahtonen Jan 27 '17 at 20:30
  • @Jyrki Oh no, it's contagious? That way can be useful when less is known, e.g. if we have instead of $,n^5-n,$ some arbitrary sequence $,f(n),$ then we can take the gcd of the first few values to attempt to gain some insight. – Bill Dubuque Jan 27 '17 at 20:36

1 Answers1

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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.

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Bill Dubuque
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