jmc

Let $a$ be the first term, and let $d$ be the common difference. Then $a_n=a+(n-1)d$ for all $n$. In particular, $a_4=a+3d$ and $a_2=a+d$, so $$\frac{a+3d}{a+d}=3.$$Multiplying both sides by $a+d$, we get $a+3d=3a+3d$, so $a=0$.

Then $$\frac{a_5}{a_3}=\frac{a+4d}{a+2d}=\frac{4d}{2d}=\boxed{2}.$$