Perfect powers, divisibility and constrained prime factors

Answer: 7

1. A square has even prime exponents.
2. Supply one additional copy of each prime whose exponent is odd: m=7.

Answer: 14161

1. Remove 72×172 so both remaining exponents are multiples of 3.
2. The smallest divisor is 14161.

Answer: 3

1. A square has even prime exponents.
2. Supply one additional copy of each prime whose exponent is odd: m=3.