The condition for $x , y , z$ and $t$ Suppose for all $a , b \in \mathbb{Z}$  we have   $a^x | b^y \to a^z | b^t$ . What's the condition for $x , y , z$ and $t$ to satisfying that ? I tried to use Division algorithm but it wasn't helpful .
(Note : My guess is $xt \ge yz$ but I'm unable to proof it .)
 A: In this answer, I'm assuming that all variables are natural numbers.
The condition $tx\ge yz$ should work. Consider, $b^y=ka^x$, so $b^{ny+l}=(k^nb^l)a^{nx}$. If we write $z=nx, t=ny+l$, this is equivalent to $\frac{t-l}y=\frac{z}x$, which implies $\frac{t}{y}\ge\frac{z}{x}$ or $tx\ge yz$.
A: Let the highest powers of prime $p$ that divide $a,b$ are $u,v$  respectively.
So, we have $v \ge ux/y$
So, $vt\ge  uxt/y$
It is sufficient that the later $\ge uz$
A: $(a^x)^z$ divides$b^{yz}$
and. $(a^y)^x$ needs to be $b^{xt}$
So, it is sufficient to have $$yz\ge xt$$
A: Necessity ($\implies$): Take $a=2^y, b=2^x$, then $2^{xy}=a^x=b^y$. Then $(2^{xy})^{z/x} = a^z$ divides $b^t = (2^{xy})^{t/y}$ which means that $z/x \leq t/y$.
Sufficiency ($\impliedby$): Suppose $a^x | b^y$ then $a^{xz} | b^{yz} | b^{xt}$. But $(a^z)^x | (b^t)^x$ implies that $(b^t/a^z)^x =: k$ is an integer. Because the $x$-th root $\sqrt[x]{k} = b^t/a^z$ of an integer is either integer or irrational, $b^t/a^z$ is an integer and $a^z | b^t$.
