${x+1 \choose 2} = \frac{(x+1)x}2$ is a bijection on $\Bbb Z_m\!\!\iff m = 2^k$

Question

Given the set $S_m=\{[\frac{k(k+1)}{2}]_m\}$ for $k\in\mathbb{N}$. Where $[x]_m$ denotes the equivalence class (modm).

I am trying to show that: $S_m=\mathbb{Z}_m$ if and only if $m=2^s$ for some $s \in\mathbb{N}$.

A hint is also given that for each prime $p>2$ there is a $b\in\mathbb{Z}_p$ such that the equation $x^2\equiv b\pmod{p}$ has no solutions.

I've tried using the condition for what m has to equal in order for the identity to hold but I'm quite lost in how to proceed really, any help would be greatly appreciated.

Answer 1

Case 1: $\ m=2^s$

For $k$ and $l$ residues modulo $2^{s+1}$ suppose $$\frac{k(k+1)}{2}\equiv \frac{l(l+1)}{2}\pmod {2^s}.$$

Then $k(k+1)\equiv l(l+1)\pmod {2^{s+1}}$

i.e. $(k-l)(k+l+1)\equiv 0\pmod {2^{s+1}}$.

Now one of the two factors on the LHS is odd and so the other factor has to be divisible by $2^{s+1}$. So equality occurs if and only if $k+l+1\equiv 0\pmod {2^{s+1}}$. Hence everyone of the $2^{s+1}$ values for $\frac{k(k+1)}{2}$ is taken precisely twice and so, by the pigeon-hole principle, $\frac{k(k+1)}{2}$ takes every value modulo $2^s$.

Case 2: If $m$ has an odd prime factor $p$ then, following the given hint, let $b$ be an element such that $x^2\equiv b\pmod p$ cannot be solved.

Suppose that $\frac{k(k+1)}{2}\equiv \frac{b-1}{8}\pmod p$ could be solved. Then $(2k+1)^2\equiv b \pmod p$, a contradiction.

Answer 2

For each positive integer $m$, let $f_m:Z_m\to Z_m$ be given by $$ f(x) = \Bigl( {\small{\frac{x^2+x}{2}}} \Bigr) {\,\text{mod}\,} m $$ Claim:$\;f_m$ is injective if and only if $m$ is a power of $2$.

Proof:

First assume $m$ is not a power of $2$.

We want to show that $f_m$ is not injective.

Suppose instead that $f_m$ is injective.

Let $n > 1$ be an odd divisor of $m$.

Then $2$ is a unit in $Z_n$.

Since $f_m$ is injective, $f_m$ is surjective, hence $f_n$ is surjective.

Since $f_n$ is surjective, $f_n$ is injective, contrary to $$ f_n(n-1) = (2^{-1})(n^2-n) = 0 = f_n(0) $$ hence $f_m$ is not injective, as was to be shown.

Next assume $m$ is a power of $2$.

We want to show that $f_m$ is injective.

The injectivity of $f_1$ is immediate, so assume $m > 1$.

Since $m$ is a power of $2$, so is $2m$.

It follows that all odd integers are invertible mod $2m$.

Now suppose $f_m(x)=f_m(y)$ for some $x,y\in\{0,...,m-1\}$ with $x > y$.

Our goal is to derive a contradiction.

From $f_m(x)=f_m(y)$ we get \begin{align*} & {\small{\frac{x^2+x}{2}}}\equiv{\small{\frac{y^2+y}{2}}}\;(\text{mod}\;m) \\[4pt] \implies\;& x^2+x\equiv y^2+y\;(\text{mod}\;2m) \\[4pt] \implies\;& (x^2+x)-(y^2+y)\equiv 0\;(\text{mod}\;2m) \\[4pt] \implies\;& (x-y)(x+y+1)\equiv 0\;(\text{mod}\;2m) \\[4pt] \end{align*} Consider two cases . . .

Case $(1)$:$\;x-y$ is even.

Then $x+y+1$ is odd, so $x+y+1$ is invertible mod $2m$.

Also we have $1\le x-y\le m-1$, so $x-y$ is not divisible by $2m$.

But then \begin{align*} & (x-y)(x+y+1)\equiv 0\;(\text{mod}\;2m) \\[4pt] \implies\;& x-y\equiv 0\;(\text{mod}\;2m) \\[4pt] \end{align*} contradiction.

Case $(2)$:$\;x-y$ is odd.

Then $x-y$ is invertible mod $2m$.

Also we have $2\le x+y+1\le 2m-2$, so $x+y+1$ is not divisible by $2m$.

But then \begin{align*} & (x-y)(x+y+1)\equiv 0\;(\text{mod}\;2m) \\[4pt] \implies\;& x+y+1\equiv 0\;(\text{mod}\;2m) \\[4pt] \end{align*} contradiction.

Thus both cases yield a contradiction.

It follows that $f_m$ is injective.

This completes the proof.