Exercise 3.4.3 (Integers represented by $mf$)

Question

If $\mathscr{C}$ is any class of integers, finite or infinite, let $m\mathscr{C}$ denote the class obtained by multiplying each integer of $\mathscr{C}$ by the ineger $m$. Prove that if $\mathscr{C}$ is the class of integers represented by any form $f$, then $m\mathscr{C}$ is the class of integers represented by $mf$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
By definition,
$$\mathscr{C} = \{c \in \Z \mid \exists (x,y) \in \Z^2,\ c = f(x,y),$$
and 
$$m \mathscr{C} = \{n \in \Z \mid \exists c \in \mathscr{C}, \ n = mc\}.$$
Put 
$$\mathscr{S} = \{n \in \Z \mid \exists (x,y) \in \Z^2,\ n =(m f)(x,y)\}$$
the set of integers represented by $mf$.
\begin{itemize}
\item If $n \in m \mathscr{C}$. Then $n = mc$, where $c = f(x,y)$ for some integers $x,y$. Thus $n = (mf)(x,y)$, and so $n \in \mathscr{S}$.

\item Conversely, if $n \in \mathscr{S}$, then $n = mf(x,y)$ for some integers $x,y$, thus $n = m c$, where $c = f(x,y) \in \mathscr{C}$, therefore $n \in m \mathscr{C}$.
\end{itemize}
This proves $\mathscr{S} = m \mathscr{C}$, and so $m\mathscr{C}$ is the class of integers represented by $mf$.

\end{proof}

Exercise 3.4.3 (Integers represented by $mf$)

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Exercise 3.4.3 (Integers represented by $mf$)

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