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I first tried to answer this using proof by induction, however my problem got more complicated when I got to the induction step. Is there another way of solving this problem?

Martin Sleziak
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Jnyeboah93
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    It might be easier to prove that $n^5-n$ is divisible by each of $2$, $3$ and $5$ separately. – hrkrshnn Oct 08 '14 at 15:18
  • If you lack any other inspiration, the binomial theorem might help. You could also just check all the numbers between 0 and 29. – Joel Oct 08 '14 at 15:18
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    Possible duplicate of http://math.stackexchange.com/questions/132210/how-to-prove-n5-n-is-divisible-by-30-without-reduction. – lhf Oct 08 '14 at 16:16
  • http://math.stackexchange.com/questions/132210/how-to-prove-n5-n-is-divisible-by-30-without-reduction, http://math.stackexchange.com/questions/174558/divisibility-of-composite-numbers, http://math.stackexchange.com/questions/451220/question-show-that-n5-n-is-divisible-by-30-for-all-natural-n, http://math.stackexchange.com/questions/963877/can-anyone-explain-how-to-show-that-n5-n-%E2%89%A10-mod-30-for-every-n-in-ma – Martin Sleziak Oct 08 '14 at 16:35
  • If you are interested specifically in a proof by induction: http://math.stackexchange.com/questions/350675/prove-n5-n-is-divisible-by-5-by-induction, http://math.stackexchange.com/questions/529757/showing-30-divides-n5-n-for-all-n-in-bbb-n-using-induction – Martin Sleziak Oct 08 '14 at 16:36

4 Answers4

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Note that $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1).$$

First, $n(n-1)$ is even because these are two consective numbers.

Second, $(n-1)n(n+1)$ is divisible by $3$ because these are three consective numbers.

Third, one of $n,n-1,n+1,n^2+1$ is divisible by $5$. In mod $5$, if $n\equiv 0$, then it's obvious. If $n\equiv 1$, then $n-1\equiv 0$. If $n\equiv 2\ \text{or}\ 3$, then $n^2+1\equiv 0$. If $n\equiv 4$, then $n+1\equiv 0$.

Hence, $n^5-n$ is divisible by $2\times 3\times 5=30$.

mathlove
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  • Thanks. I understand how $n^{5} -n \equiv\space 0\space mod\space 2$ , $n^{5} -n \equiv\space0\space mod\space 3$ and $n^{5} -n \equiv\space 0\space mod\space 5$ from your explanation. But how does that prove that $n^{5} -n \equiv\space 0\space mod\space 30?$ – Jnyeboah93 Oct 08 '14 at 15:41
  • @Jnyeboah93: Note that $2,3,5$ are prime numbers. So, let $A=n^5-n$, if $A\equiv 0\pmod 2,A\equiv 0\pmod 3,A\equiv 0\pmod 5$, then we can say $A\equiv 0\pmod{2\times 3\times 5=30}$. – mathlove Oct 08 '14 at 15:45
  • Oh thanks, has this theorem got a name? – Jnyeboah93 Oct 08 '14 at 15:46
  • @Jnyeboah93: well, not sure. sorry:) – mathlove Oct 08 '14 at 15:48
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HINT

By Fermat's little theorem $$n^5\equiv n \pmod 5\\n^3\equiv n\pmod 3\\n^2\equiv n\pmod 2$$

AgentS
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Verify the equation for 0,1,2,...,29. If it is valid for these numbers, the equation is valid for all $n \in \mathbb{N}$.

user175914
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$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)$$

$$=\underbrace{n(n-1)(n+1)(n-2)(n+2)}_{\text{The product of }5 \text{ consecutive integers}}+5\cdot\underbrace{n(n-1)(n+1)}_{\text{The product of }3 \text{ consecutive integers}}$$

See The product of n consecutive integers is divisible by n factorial

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lab bhattacharjee
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