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So I've noticed that if $n \in \{4k\},$ with $k \in \mathbb{N}$ then $(\frac{n}{2})^2 \equiv 0$ mod $n$.

Example $2^2 \equiv 0$ mod $4$, $4^2 \equiv 0$ mod $8$, $8^2 \equiv 0$ mod $16$, and so on...

I wanna go about proving this but I'm not entirely sure how to finish it.

Proof

Let $n=4k, k \in \mathbb{N}$. We want to show that $(\frac{n}{2})^2\equiv 0$ mod $n$.

$(\frac{n}{2})^2\equiv 0$ mod $n$

$(\frac{4k}{2})^2\equiv 0$ mod $n$

$(2k)^2\equiv 0$ mod $n$

$4k^2\equiv n$ mod $n$

$4k^2\equiv 4k$ mod $n$

$k^2\equiv k$ mod $n$

I've been playing around with this a little bit, but don't know if I'm headed in the right direction. Any tips or hints?

ANSWER: (Thanks to @Onir)

$4k^2\equiv 0$ mod $n$

$4k^2\equiv 0$ mod $4k$

$4k*k \equiv 0$ mod $4k$

Thus, $(\frac{4k}{2})^2$ is a multiple of $4k$.

Bill Dubuque
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SunRoad
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1 Answers1

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We want to prove $(\frac{4k}{2})^2$ is a multiple of $4k$.

$(\frac{4k}{2})^2 = (2k)^2 = 4k^2 = (4k)k$, so indeed it is a multiple of $4k$.

Asinomás
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