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Find n such that $n^5 - n$ is divisible by $120$. (*Here $n$ belongs to the natural number set)

Seems a trivial problem, but the answer mentioned in the text book seemed incomplete to me so I'm here just to ensure whether approach to this problem is correct or not.

So my approach is as follows: We know that $n^5 - n$ can be written as $n^5-n = n(n-1)(n+1)(n^2 + 1)$ Now since it's product of three consecutive natural numbers so it's divisible by $3! = 6$. Now using Fermat's Little theorem we can prove that $n^5= n \mod5$ hence $n^5 - n$ is divisible by $5$ as well. So $n^5 - n$ is divisible by $30$. Now to ensure their is another factor of $4$ in the term we need to ensure that $n^5 - n$ is divisible by $8$. I found that all odd natural numbers and multiples of $8$ makes $n^5 - n$ divisible by $8$, hence $n^5 - n$ is divisible by $120$. But in the text book they have only mentioned odd natural numbers as answer, which seemed incomplete to me.

Thanking everyone in advance:P

Bill Dubuque
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Caratheodory_Enthusiast
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  • Your book may have made a mistake. – Anurag A Nov 22 '19 at 21:43
  • Clearly the textbook is wrong, as you can verify that $8^5-8$ is divisible by 120. – arkeet Nov 22 '19 at 21:45
  • I just wanted to confirm. Thanks – Caratheodory_Enthusiast Nov 22 '19 at 21:50
  • Or $0^5-0$, for that matter! –  Nov 22 '19 at 21:53
  • 0 is a vague example considering n is a natural number – Caratheodory_Enthusiast Nov 22 '19 at 21:54
  • Another thought: Divisibility of $n^5-n$ by 8 depends only on the value of $n$ mod 8 (for any integer n), and you can check it's true iff $n \equiv 0,1,3,5,7 \pmod8$. – arkeet Nov 22 '19 at 22:00
  • @arkeet This is clearer using Euler's theorem - see my answer. – Bill Dubuque Mar 21 '20 at 15:56
  • Well obviously multiples of $120$ are divisibile but $120$ so the answers can't all be odd. – fleablood Mar 21 '20 at 15:58
  • The factor $n$ is one parity and the factors $n+1,n-1,n^2+1$ are the other. So if $n$ is odd there are $3$ distinct even factors and it is divisible by $8$. If $n$ is even there is only one even factor so it will not be divisible by $8$ unless $n$ is a multiple of $8$ which is, of course, possible. So $n^5-n$ is divisible by $8$ precisely when either $n$ is odd or $n$ is a multiple of $8$. So the book got too focused in the "if $n$ is even then then $n$ is the only even factor so it's not divisible by $8$" to forget to hypothesis check the obvious "unless $n$ itself is a multiple of $8$". – fleablood Mar 21 '20 at 16:03

3 Answers3

1

You're right and the textbook is wrong.

For odd $n$, $n-1$ and $n+1$ are even, and one of them is divisible by $4$; hence the product is divisible by $8$.

For even $n$, neither $n-1$, nor $n+1$, nor $n^2+1$ is even, so factors of two only come from $n$. Thus, for even $n$, $n^5-n$ is divisible by $8$ (and thus by $120$) if and only if $n$ itself is divisible by $8$.

joriki
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0

We need $8$ to divide $$n(n^4-1)$$

As $n(n^3)+(-1)(n^4-1)=1,(n,n^4-1)=1$

So, either $8|n$

Or $n$ must be odd and

$$8|(n^4-1)\iff2\mid\dfrac{n^2-1}2\cdot\dfrac{n^2+1}2$$ which is true for all odd $n$ as the later is the product of two consecutive integers

lab bhattacharjee
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0

You are correct. It's case $p^k = 2^3,\,$ and $\, f = \phi(p^k) = 4\,$ below

Lemma $ $ If $\,p\,$ is prime and $\,\phi(p^k)\mid f>0\,$ then $\,p^k\mid n(n^f-1)\iff p^k\mid n\,$ or $\,(p,n)=1$

Proof $\,\ (n,\,n^f-1) = 1\,$ so $\,p^k\mid n(n^f-1)\iff p^k\mid n\,$ or $\,p^k\mid n^f-1,\,$ and the latter holds iff $\,(p,n)=1,\,$ since then $\,\phi\mid f,\, n^\phi\equiv 1\Rightarrow\, n^f\equiv 1\pmod{\!p^k}\,$ by Euler & modular order reduction.

Remark $ $ Stronger, we can replace Euler phi by Carmichael lambda (= half of phi for $2^k,\ k>2)$

See also this generalization of Euler and Fermat.

Bill Dubuque
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