<s>[INST] Prove that $17$ divides $xy-12x+15y$ 

if $x,y$ are integers and $17$ divides both the expression $x^2-2xy+y^2-5x+7y$ and $x^2-3xy+2y^2+x-y$, then prove that $17$ divides $xy-12x+15y$


My attempt:
I tried to factories both expression and the result obtained is as followed
$x^2-2xy+y^2-5x+7y=(x-y)(x-y+5)+2y.....(1)$
$x^2-3xy+2y^2+x-y=(x-y)(x-2y+1) ...... (2)$
From second expression I concluded that $17$ either divides $(x-y)$ or $(x-2y-1)$
Case 1: Let $17$ divides $(x-y)$ then from equation $(1)$ i concluded that $17$ divides $2y$ and hence $17$ divides $x$ so the $17$ divides  $(xy-12x+15y)$
Is my argument correct that $17$ will divide $2y$?
Also what if $17$ does not divides $(x-y)$ instead it divides $(x-2y-1)$?
I want solution without using modulo arithmetic for, what if $17$ does not divides $(x-y)$ instead it divides $(x-2y-1)$.
Any other proof will also help which does not involve modulo arithmetic.
 [/INST] Note: I'm going to introduce some integer variables like $k,n ,m $ etc. They are only for the purpose of showing the remainder of a number  when divided by a particular number. For example if I say $x=17k+3 $ then what you need to focus on is the $3$ ( remainder) and not $k$ ( I'm stating this explicitly because at some places I've written $3(17k+3)=17k+9$ , which actually isn't correct but the $k$ is only introduced to show divisibility). Actually this is the idea behind using modular arithmetic as well.

*

*Yes, your argument that $17$ divides $ 2y$ if it divides $x-y$ is correct. From there you deduced that $17$ divides $y $ and hence also $x$ and therefore your case $1$ is correct.

case 2: $17$ divides $x-2y+1$
Write the first equation as
$x^2-2xy+y^2-5x+7y=(x-2y+1)(x-6)+y^2-5y+6$.
Since $17$ divides $x-2y+1$, it divides $y^2-5y+6$.
$y^2-5y+6=(y-2)(y-3) \implies 17$ divides atleast one of $y-2$ and $y-3$. That means $y=17k+2$ or $y= 17k+3$.
Since $17$ divides $ x-2y+1$, it is equal to $17n$.
Sub case 1: $y=17k+2$
$\implies x-17k-4+1=17n \implies x= 17k+3$
Now we can see that $xy-12x+15y= 17k+6-36+30=17k$, therefore $17$ divides it.
Sub case 2: $y=17k+3$
$\implies x-17k-6+1=17n\implies x=17k+5$
And again $xy-12x+15y=17k+15-60+45=17k$, hence $17$ divides it.
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