Need to generate numbers with no divisible arithmetic progressions This came up in the course of trying to prove something:-
Given that p$\geq 1$, I need (in terms of p) any two numbers $x, y >p$ such that:  $y$ divides $x$ , and there exist no two numbers $0<a+b \leq p$ for which every member of the AP $y- b, y, y+b,\dots$ divides the corresponding member of the AP $x - a, x, x + a \dots$
All numbers are non-negative.
Solution might be trivial.. couldn't think of anything yet
 A: Choose an arbitrary $y>p$. Now choose an arbitrary $c \geq p$. Take $x=cy>p$. We claim that there does not exist (non-negative integers) $a, b$ such that $0<a+b \leq p$ and $(y+ka) \mid (x+kb)$ for $k=-1, 0, 1, \ldots$
Assume on the contrary that such $a, b$ exist. First consider $a \geq 1$.
Consider any $k$ such that $\gcd(k, y)=1$ (There are clearly infinitely many such positive integers $k$). We have 
$$(y+ka) \mid (x+kb)=(cy+kb)$$
$$(y+ka) \mid c(y+ka)$$
$$(y+ka) \mid (cy+kb)-c(y+ka)=k(b-ca)$$
Since $\gcd(k, y)=1$, we have $\gcd(y+ka, k)=1$. Thus $y+ka \mid (b-ca)$ for all positive integers $k$ relatively prime to $y$. If $b-ca \not =0$, then we necessarily have $|b-ca| \geq (y+ka) \geq y+k$ for all positive integers $k$ relatively prime to $y$, which is clearly impossible. 
Thus $b=ca$, but this gives $p \geq a+b=a+ca=a(c+1) \geq a(p+1)>p$, a contradiction. Therefore $a=0$, and so $y \mid x-b$, $y \mid x$. This implies that $y \mid b$. If $b>0$, then $b \geq y>p$, a contradiction. Therefore $b=0$, so $a+b=0$, a contradiction.
Therefore no such $a, b$ exist. For a concrete example, we can take $y=p+1, x=p(p+1)$.
