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Yeah it looks like a basic, really elementary question, but i'm having hard time with it.

First i tried to show that it's divisible by 9 $$(n-1)^3n^3(n+1)^3 = ((n+1)(n-1))^3n^3 = (n^2-1)^3n^3 = (n^3-n)^3$$ and using eulers theorem we know that $$[n^{\varphi(9)} \equiv 1 (mod \ 9)] = [n^6 \equiv 1 (mod \ 9)]$$ My doubt : can we do that? Cause $n$ and $9$ have to be coprime. Is it right direction? I'd love some help on this, cause i never did tasks which asks for proving divisiblity of some polynomial. Cheers!

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    $\begingroup$ This thing is not necessarily divisible by $7$: See $n=3$. $\endgroup$
    – J. J.
    May 14, 2014 at 12:02
  • $\begingroup$ in the title you have $(n-1)^n$ but in the question you have $(n-1)^3$ $\endgroup$ May 14, 2014 at 12:05
  • $\begingroup$ There is an easier way to show $9$: one of $(n-1), n, (n+1)$ is divisible by $3$, and you're cubing all of them in your product (so it's actually divisible by $27$). $\endgroup$
    – Arthur
    May 14, 2014 at 12:06
  • $\begingroup$ @Alessandro thanks, it was a typo in the title, sorry! Actually i'm solving a task that this polynomial is divisble by 504. Would it mean it can't be true, because it's not divisible by 7? $\endgroup$ May 14, 2014 at 12:08

1 Answer 1

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It's divisible by $9$ and in fact $27$, but not necessarily by $7$. To see it is divisible by $27$, use the fact that that one of $n$, $n - 1$, $n - 2$ is divisible by $3$.

Your way

Expand out what you have: $$ (n^3 - n)^3 = n^{9} - 3 n^{7} + 3n^5 - n^3 $$ If $n$ is divisible by $3$, you are done. Otherwise, as you notice, we have $n^6 \equiv 1 \pmod 9$. This implies $n^9 \equiv n^3$, so the above is $$ \equiv n^3 - 3n^7 + 3n^5 - n^3 = -3(n^7 - n^5) \pmod 9 $$

Now since you have one factor of $3$ for sure, you consider $n^7 - n^5$ modulo $3$. Euler's theorem gives $n^{\varphi(3)} = n^2 \equiv 1 \pmod 3$, so $n^7 \equiv n^5$. Therefore $n^7 - n^5 \equiv 0 \pmod 3$, and the expression in question is $$ \equiv 0 \pmod 9. $$

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  • $\begingroup$ Oh... Yes, you are right! But could you help me with my idea? Is it possible to do this like that? I'd love to learn something new :-) $\endgroup$ May 14, 2014 at 12:10
  • $\begingroup$ @Chris I've added things to my answer. Does that help? $\endgroup$ May 14, 2014 at 12:21

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