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Problem:

Show that for every odd integer, $n^3 - n^2 - 5n +5$ is divisible by 8. using modulo arithmetic

I'm not sure how to use the modulo arithmetic concept to answer this question.

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$n^3-n^2-5n+5=n^2(n-1)-5(n-1)=(n-1)(n^2-5)=(2p)(4p^2+4p+1-5)=(2p)(4p^2+4p-4)=8p(p^2+p-1)$

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$$n^3-n^2-5n+5=(n^2-5)n-(n^2-5)=(n^2-5)(n-1)=(n^2-1-4)(n-1)=(n+1)(n-1)^2-4(n-1)$$

If $n$ is odd, then $n+1$ and $n-1$ are both even, so $8\mid(n+1)(n-1)^2$ and $8\mid4(n-1)$.

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  • $\begingroup$ Interesting factorisation. $\endgroup$ – zwim May 23 '17 at 1:09
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You could factor this into $n^2(n-1)-5(n-1)=(n^2-5)(n-1)$ now check the four cases when $$n\equiv 1\pmod{8}\\n\equiv 3\pmod{8}\\n\equiv 5\pmod{8}\\n\equiv 7\pmod{8}$$

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