Exercise 1.6

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $a, b, c \in \Z$. Show that the equation $ax + by = c$ has solutions in integers iff $(a, b) |c$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
 Let $d = a \wedge b$. 
 
 $\bullet$ If $a x + by = c, x,y \in \Z$, as $d \mid a, d \mid b$, $d \mid ax + by = c$.
 
 $\bullet$ Conversely, if $d \mid c$, then $c = d c',\ c' \in \Z$. 
 
 From Prop. 1.3.2., $d\Z = a \Z + b\Z$, so $d = a u + b v,\ u,v \in \Z$, and $c = dc' = a (c'u) + b (c'v) =a x + b y$, where $x = c' u, y = c' v$ are integers.
 
 Conclusion : 
 
 $$\exists (x,y) \in \Z 	imes \Z, \ ax + by = c \iff a\wedge b \mid c.$$
\end{proof}

Exercise 1.6

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Exercise 1.6

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