a elementary number theory problem please help me to find $ a,b \in \Bbb Z $ such that $a|b$ and  $ \forall z \in \Bbb Z $,we have $a+z|b+z$.

If $a = b$ the result is trivial.
So I will assume that $a \ne b$.
If $a$ and $b$ are positive,
we must have $a < b$.
If we choose $z$ such that
$2(a+z) > (b+z)$
(i.e., $z > b-2a$),
then $1 < \frac{b+z}{a+z} < 2$
which contradicts $(a+z)|(b+z)$.
So $a$ and $b$ cannot both be positive.
If $a$ and $b$ are both negative,
by looking at negative $z$
we come to the same conclusion.
So one of them must be positive and
the other negative.
Assume $a < 0$, and let
$c = -a$. Then
$(z-c)|(z+b)$ and $b$ and $c$ are positive integers.
Again by choosing $z$ large enough,
so $z > c$ and $z+b < 2(z-c)$
(or $z > b+2c$),
we get $1 < \frac{z+b}{z-c} < 2$,
a contradiction.
Therefore the only solution is $a = b$.
Note that this works if $a, b$, and $z$
are restricted to the positive integers.
In this case, we cannot set $z = 0$
to get $a|b$.
