The idea is to go into the binary form of the number.

If it's a 1 following by even number of zeros, then it's a power of four.

e.g.

bin(4**0)

Out[27]: '0b1'

bin(4**1)

Out[28]: '0b100'

bin(4**2)

Out[29]: '0b10000'

bin(4**3)

Out[30]: '0b1000000'

Codes:

```
class Solution(object):
def isPowerOfFour(self, num):
"""
:type num: int
:rtype: bool
"""
return bin(num)[3:] == len(bin(num)[3:])*'0' and len(bin(num)[3:])%2==0 if num!=0 else False
```