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Use Fermat's Little Theorem to prove that

$11|(9n^{23}-5n^{13}+7n^3)$

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    Ahh yes, I don't know how I didn't see it. $9n^{23}$ is congruent to $9n^3$ and so on for all the others. Thanks Harald!2012-11-11

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$n^{23}\equiv n^{13}\equiv n^3\pmod{11}$ so we get that $ 9n^{23}-5n^{13}+7n^3\equiv11n^3\equiv0\pmod{11} $ Thus, $11|(9n^{23}-5n^{13}+7n^3)$.

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    ok, sorry for the noise.2012-11-11
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If modular arithmetic is unfamiliar then one may use that $\rm\:11\mid \color{#C00}{n-n^{11}}\:$ by little Fermat, hence

$\rm 11\mid7n^3\! - 5n^{13}\! + 9 n^{23}\, =\ (7n^2\! + 2n^{12}) (\color{#C00}{n-n^{11}}) + 11 n^{23}$