Name of Modulus Theorems

Question

Let p and q be two positive integers with r being their greatest common factor. Consider the set E such that

$E = \{0,\:\:r, \:\:2r,\:\:3r,\:\: ...,\:\: (q-r)\}$

Let a_n be a sequence defined for integers n within the interval [0, q-1] such that

$a_n = (np) \bmod q$

For every integer, k, within set, E, there is a corresponding n(k) such that

$a_{n(k)} = k$

I am able to prove this theorem from scratch, but I would like to know if there are any existing theorems within Discrete Mathematics (or another branch of math) that is able to help prove this theorem quicker. Do such theorems exists? And if so, I would also like to ask if there is a resource that presents a large number of them.

Answer 1

If we unwind your definitions then your claim is equivalent to the following very well-known result about the solution of a general linear congruence, replacing your $\,n\,$ by $\,x,\,$ and where $\frac{a}b := ab^{-1}$

$${\rm if}\ \ \ r := (p,q)\mid k\ \ \ {\rm then}\ \ \ p\,x \equiv k\!\!\!\pmod{\!q}\!\iff\! x\equiv \dfrac{k/r}{p/r}\!\!\!\pmod{\!q/r}\qquad\qquad$$

You can find a proof of this and further discussion this answer, and in most number theory textbooks.

Remark $ $ The congruence has a solution $\,x\in\Bbb Z\iff r\mid k.\,$ Above shows $(\Leftarrow)$. For the converse $(\Rightarrow)\ $ if $\,x_0\,$ is a solution then for some $\,j\!:\, $ $\, p\,x_0+q\,j = k\,$ so $\,r\mid p,q\,\Rightarrow\, r\mid p\,x_0+q\,j = k.\,$ Or, by congruences $\,\bmod r\!:\,\ k\equiv px\equiv 0\,$ by $\,p\equiv 0\ $ (the congruence persists $\!\bmod r\,$ by $\,r\mid q)$