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 \[\frac{a_5}{a_3} = \frac{a + 4d}{a + 2d} = \frac{4d}{2d} = \boxed{2}.\]