the book can be downloaded from here (the solution is on the pages 23-28).
The stattement of the problem can be written as a logical equivalence
$(1)$ $17 \; \mid 2x+3y \; \; \Leftrightarrow \; 17 \; \mid 9x+5y.$
We will give a list of equivalences, "completely" in some sense, from which we can deduce (almost) any of such a relations.
LEMA For any integers $x$, $y$ the following equivalences hold:
$17 \; \mid \; I(x,y)=2x+3y \; \Leftrightarrow \; 17 \; \mid \;II(x,y)=x-7x \; \Leftrightarrow \; 17 \; \mid \; III(x,y)=3x-4y \; \Leftrightarrow $
$\Leftrightarrow \;17 \; \mid \; IV(x,y)=5x-y \; \Leftarrow \; 17 \; \mid V(x,y)=7x+2y \; \Leftarrow \; 17 \; \mid \; VI(x,y)=8x-5y.$
Proof. We have $17 \; \mid 2x+3y \; \Leftrightarrow \; \widehat{2}x+\widehat{3}y=\widehat{0}$, where $\widehat{m}$ mean the congruence class modulo $17$ determined by $m$; their set is $\mathbb{Z}/17\mathbb{Z}$ (actually it is the field $GF(17)$). If we multiply last equation by $\widehat{9}$ and we take into account that, in $GF(17)$,
$\widehat{9} \cdot \widehat{2}=\widehat{1}, \; \widehat{9} \cdot \widehat{3}=\widehat{10}=-\widehat{7}$
we obtain $\widehat{1}x-\widehat{7}y=\widehat{0} \; \Leftrightarrow \; 17 \; \mid \; x-7y$.In the same way, starting with $\widehat{2}x+\widehat{3}y=\widehat{0}$ and multiplyng it by $\widehat{10}$, $\widehat{11}$, $\widehat{12}$, $\widehat{4}$ respectively, we get the remained equivalences.
$\blacksquare$
Apparently we didn't find the desired result. It comes if we were to multiply $\widehat{2}x+\widehat{3y}=\widehat{0}$ by $\widehat{13}$ and taking into account that
$\widehat{13} \cdot \widehat{2}=\widehat{9}$ and $\widehat{13} \cdot \widehat{3}=\widehat{5}$
but patience $\cdot$ $\cdot$ $\cdot$ The word "completely" means that no other divisibility relation that can be obtained from $17 \mid 2x+3y$ it must be able to be expressed by ones of the forms $I-VI$ (and apliying other simple divisibility properties). From example$$17 \; \mid I(x,y) \; \stackrel{LEMA} \Leftrightarrow 17 \; \mid VI(x,y) \Leftrightarrow 17 \; \mid 17x-VI(x,y)=9x+5y.$$
Finally, we conclude anothers equivalences that we will use in a similar problem:
$7\; \mid x+2y \; \Leftrightarrow \; 7 \; \mid 2x-3y \; \Leftrightarrow \; 7 \; \mid 3x-y;$
$7\; \mid 3x+y \; \Leftrightarrow \; 7\; \mid x-2y \; \Leftrightarrow \; 7 \; \mid 2x+3y.$
