I started to think about this problem and then factored $n^5 - n$ to $(n^2 - 1)(n^2 + 1)(n)$, and later to $(n-1)(n)(n+1)(n^2 + 1)$. I know that $(n-1)(n)(n+1)$ is divisible by $6$, but it is not that case $5$ divides $n^2 + 1$ for any integer $n$, so i can´t use the multiplication property. Can anyone help me finish this proof?
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Also: How to prove $n^5 - n$ is divisible by 30 without reduction – all found instantly with Approach0 – Martin R May 01 '18 at 16:17
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$5$ doesn't have to divide $n^2+1$. $5$ could divide $n$, or $n-1$ or $n+1$. Can you prove that $5$ must divide one of those $4$ values? Hint: How can $5$ not divide any of the $n, n+1, n-1$? If $5$ divides non of those three, what possible remainder is there if you divide $n$ by $5$? – fleablood May 01 '18 at 16:31
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Hint:
$$(n^2-1)(n^2+1)n=(n^2-1)(n^2-4+5)n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{product of } 5\text{ consecutive integers}}+5n(n^2-1)$$
lab bhattacharjee
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@MateusBuarque, Welcome! See also: https://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial – lab bhattacharjee May 01 '18 at 16:15
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Oops, i made a mistake, i wanted to prove if it for 30, but this strategy still works. – Mateus Buarque May 01 '18 at 16:18
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@MateusBuarque, That's also done here using the link I have just shared – lab bhattacharjee May 01 '18 at 16:20
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One of the other answers requires that you know something - the other is very clever. A hint for a proof that doesn't require either non-trivial theorems or cleverness: $n=5k+j$, where $j\in\{0,1,2,3,4\}$.
David C. Ullrich
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