If $ a-b \mid ax-by$, then $\gcd(x,y) \ne1$?

Question

So is this true for positive integers $a,b,x,y>1$ with $a>b$ and $x>y$?

Answer 1

Well $a-b|ax - bx$ and $a-b|ax-by$ so $a-b|(ax-bx) - (ax-by)=b(y-x)$ and ... why not?

So we can have $a-b |b$, for example $a=8;b=6$ and $x,y$ can be anything....$y=57, x = 56$ for example. $8-6|8*57-6*58$.

And even if $\gcd(a,b) = 1$ where can still have $a-b|y-x$ with no issue. Let $y = 81$ and $x=73$ and $a=13$ and $b=9$. Why the heck not? $13-9|13*81 -9*73$. Whoa! wait is that true? $13*81-9*73 = 4*81 + 9*81-9*73 =4*81 + 9*4$.... yep, seems to be true.

Answer 2

No, it is not. Say $a=b+1$, then $x,y$ can be anything.

Still no, say $a= 4$ and $b=2$, again $x,y$ can be anything.