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I have a problem here that I have no idea how to go about solving. It states: Let $n∈Z$ with $n>1$.

(a) If $n=2k$ for some odd integer $k$, prove that $k^3≡k \pmod{2n}$.

(b) If $n=2k$ for some even integer $K$, prove that $k^2≡0 \pmod n$.

I only have knowledge of proofs using mathematical induction. However, I have the section of the book open but I'm still not getting any ideas of how to solve the problem.

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For part b, as $k$ is even integer so you can choose $k=2t$ for some integer $t$. Then what is $k^2$? It is $16t^2$ and that is congruent to $0$ mod n as $n=4t$ as $4t |16t^2$.

Now for part a) as $k$ is odd, choose $k=2t+1$ , then $k^3=8t^3+12t^2+6t+1$ and thus $k^3-k=8t^3+12t^2+4t$ which is divisible by $2n=8t+4$. Right, long divide it, if you can't see. quotient will be $t^2+t$.

I hope it helps..!

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Hint: For a) take $k=2m+1$ and look at how $k^3$ looks like.

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  • $\begingroup$ How does the help solve the problem; I'm lost? $\endgroup$ – dreamin Nov 23 '14 at 17:45

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