I am self studying analytic number theory from Tom M apostol and couldn't think about how apostol proves a result related to complete residue systems.
The original theorem in which deduction appears is ->
I clearly understand the arguments involving Hecke operators but I couldn't deduce the result which apostol proves in the end which is only related to elementary number theory .
Result to be deduced-- if b and $\beta $ runs through a complete residue system mod d and $\delta $ respectively, then prove that linear combination $\alpha$b+$\beta$d runs through a complete residue system mod d$\delta$ .
Here $\alpha$ >0 and d >0 and both belongs to integers. Also $\alpha$ $\delta$=m and ad=n. m and n are co-prime and as d and $\delta$ run through positive divisors of n and my respectively, d$\delta$ runs through positive divisors of mn as (m, n) = 1 .
Can somebody please explainhow to deduce this result?
Let $n,m\in \mathbb N$ with $(n,m) = 1$. Fix $a,d,\alpha,\delta\in \mathbb N$ with $ad = n$ and $\alpha \delta = m$. Since $n$ and $m$ are coprime, we deduce $(a,\delta) = (\alpha,d) = 1$.
The claim is that \begin{equation} \label{eq:1} \tag{1} \{ \alpha b + \beta d \,|\, 0\le b < d\text{ and } 0\le \beta < \delta\} \end{equation} is a complete residue system modulo $d\delta$. Since it consists of $d\delta$ elements, it suffices to show that they are pairwise inequivalent modulo $d\delta$.
Assume $\alpha b + \beta d \equiv \alpha b' + \beta' d \mod{d\delta}$ for some $0\le b, b' < d$ and $0\le \beta, \beta' <\delta$.
Reducing modulo $d$ yields $\alpha b \equiv \alpha b'\mod d$. Now, $\alpha$ is invertible modulo $d$, since $(\alpha, d) = 1$. Therefore, $b\equiv b' \mod{d}$ and hence $b=b'$.
We are left with $\beta d \equiv \beta' d\mod{d\delta}$. Since $d\neq 0$ this is equivalent to $\beta \equiv \beta' \mod{\delta}$. Hence $\beta = \beta'$.
This shows that the elements of \eqref{eq:1} are pairwise inequivalent modulo $d\delta$.