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Let n be a natural number, show that at least one of $(n^5)-1$, $(n^5)$ and $(n^5)+1$ is divisible by $11$.

I started by considering two cases: $n$ is even. $n=2k$, so the three numbers become $32(k^5)-1$, $32k^5$, and $32(k^5)+1$ In mod $11$, it is equal to $-(k^5)-1$, $-k^5$, and $-(k^5)+1$

If $n$ is odd, $n^5=32k^5+80k^4+80k^3+40k^2+10k+1$ in mod $11$ it is $-k^5+3k^4+3k^3+7k^2-k+1$

And I want show that one of them eventually become $0$ in mod $11$, but it doesn't seem to help...

Henry Lee
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Y.Z.
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    Why don't you evaluate all the fifth powers modulo $;11;$ ? That ends the problem at once... – DonAntonio Dec 13 '18 at 00:35
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    Hint: $n^{11}\equiv n\pmod{11}$ by fermat's little theorem. If $n$ were divisible by $11$ then we are done. Otherwise then we have $(n^5)^2 \equiv 1\pmod{11}$ by a bit of factoring and dividing each side by $n$. – JMoravitz Dec 13 '18 at 00:37
  • I am not with closing this question. – Maged Saeed Dec 13 '18 at 07:37

1 Answers1

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If $11\mid n$, then $11\mid n^5$.

Otherwise, $11\mid n^{10}-1$ by Fermat's little theorem. And since $n^{10}-1=(n^5-1)(n^5+1)$, it follows from this that $11$ divides one of the numbers $n^5-1$ or $n^5+1$.

José Carlos Santos
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