Is open adressing with prime steps bijective?
The following proof only uses basic modular arithmetic, and the fact that the least common multiple $\mathrm{lcm}(a,b)$ is equal to $a\cdot b$, if $\gcd(a,b)=1$.

Injectivity:

Assume there are two different $a,b\in \{0,...,m-1\}$, $a&gt;b$, such that $f(a)=f(b)$. This means that

$$
a\cdot w \mod m = b\cdot w \mod m
$$
and
$$
a\cdot w=b\cdot w+k\cdot m
$$
for some integer $k&gt;0$. We reorder to
$$
(a-b)\cdot w=k\cdot m
$$
Both sides of the equation are a multiple of $w$ and $m$. However, since $w$ and $m$ don't share any divisor, any common multiple is larger than $w\cdot m$. The left hand side of the equation clearly contradicts this fact. Hence, there are no such choices for $a$ and $b$, and $f$ is injective.

Surjectivity

This can be argued by counting. The function $f$ is total on the set $\{0,...,m-1\}$, and since $f$ is injective as shown above, $f$ is also surjective.