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I'm working in $\mathbb{Z}_n$ under multiplication (not addition) for some time now and I'm beginning to notice something:

For $n \in \mathbb{N}$ where $n$ is even and $\frac{n}{2}$ is odd, it seems to be always true that $\frac{n}{2}$ is idempotent, i.e. $(\frac{n}{2})^2 \equiv \frac{n}{2}$ mod $n$ when $n$ satisfies the aforementioned definition.

So I wanna try to prove this but I don't know how to approach it. My first thought was induction since this iterates across $n \in \mathbb{N}$ satisfying my definition of $n$.

Proof

Let $n \in \mathbb{N}$ such that $n$ is even and $\frac{n}{2} = 2k+1$ for some other $k \in \mathbb{N} \cup \{0\}$ (Note that $k$ could equal $0)$

Define $P(n)$ as the statement:$(\frac{n}{2})^2 \equiv \frac{n}{2}$ mod $n$

Our basis case would be $P(2): (\frac{2}{2})^2=(1)^2 \equiv 1$ mod $2 \equiv \frac{2}{2}$ mod $2$ So $P(2)$ holds as true.

Assume $P(i)$ is true for some $i \in \mathbb{N}$ where $i = 2(2k+1)$.

Inductively, then we have:

$$P(i+1): (\frac{i+1}{2})^2 = ?? $$

Then I get stuck here and don't know how to proceed. Obviously, I can't do much once I put $k$'s into the equation because $k$ can be any number. All we know about $k$ is that $k \leq n$ by how we defined $n$.

Grigor Hakobyan
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  • @Jorge Sorry that's a typo. I meant to square it, Hold on. It's fixed now – Grigor Hakobyan Apr 02 '21 at 19:53
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    I don't think it makes a lot of sense to use induction because the congruence mod $2n$ isn't related to the congruence $\bmod 2n+2$ (other than because they both tell you the parity of the numbers). – Asinomás Apr 02 '21 at 19:59
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    Nice observation. A word about strategy: induction on $n$ is not likely to work when what you are interested in is arithmetic modulo $n$. That's because $\mathbb{Z_n}$ and $\mathbb{Z_{n+1}}$ are not related in any useful arithmetic way. – Ethan Bolker Apr 02 '21 at 20:01
  • I see. Do you have a method other than induction that would be more appropriate here? – Grigor Hakobyan Apr 02 '21 at 20:04
  • @EmilyBurkenhamen As the answers show, use divisibilty arguments involving $n$ directly. – Ethan Bolker Apr 02 '21 at 21:23
  • I added an answer explaining how it;s a special case of a property that characterizes idempotents in $\Bbb Z_n.\ \ $ – Bill Dubuque Apr 02 '21 at 21:44

3 Answers3

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You want to prove $k^2\equiv k \bmod 2k$ when $k$ is odd, this is equivalent to proving $k^2-k = k(k-1)$ is a multiple of $2k$. Since $k$ divides $k$ and $2$ divides $k-1$ we have $2k$ divides $k(k-1)$.

What we used here is that if $d$ divides $a$ and $e$ divides $b$ then $de$ divides $ab$. This can be proved by writing $a$ as $dk$ and $b$ as $el$ and then writing $ab$ as $(ed)(kl)$.

Asinomás
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In any $\mathbb{Z}_n,$ computing idempotents is equivalent to solving the congruence $$x^2 \equiv x \pmod{n}$$ or $x(x-1)=ny$ for some $y\in\mathbb{Z}$.

In your case, $n=2(2m+1),$ so, choose $x=2m+1$ and $y=m.$

Shaun
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Bumblebee
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A factor $\,k\,$ of modulus $\,n=jk\,$ is idempotent $\!\bmod n\!\iff\! \color{#c00}{k\equiv 1}\,$ mod its cofactor $\,j,\,$ by

$$\ jk\mid (k\!-\!1)k\iff j\mid \color{#c00}{k\!-\!1}.\ \text{ OP is case } \,j=2,\ k = n/2\qquad$$

Notice that here $\,k\,$ is coprime to $\,j.\,$ This is a characteristic property of idempotents, namely generally idempotents in $\:\Bbb Z_{ n}\:$ correspond to factorizations of $\:n\:$ into two coprime factors. For if $\:e^2 = e\in\Bbb Z_n\:$ then $\:n\mid (e\!-\!1)e\:$ thus $\:n = jk,\ j\:|\:e\!-\!1,\ k\mid e,\:$ so $\:(j,k)= 1\:$ by $\:(e\!-\!1,e) = 1.\:$ Conversely if $\:n = jk\:$ for $\:(j,k)= 1,\:$ then by CRT, $\:\Bbb Z_n\cong \Bbb Z_j\times \Bbb Z_k\:$ which has nontrivial idempotents $\:(0,1),\,(1,0).\:$

Bill Dubuque
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