Given n, find a,b such that a**b = n

power.py

# Generate all the primes (and quite a few non-primes, but meh)
def primeish():
	for n in 2,3,5:
		yield n
	base = 6
	while True:
		yield base+1
		yield base+5
		base = base+6

# Find prime factors of power (e.g. 144 is 2**4 x 3**2)
# If we find a factor with an exponent of 1, we've lost.
# If we find two factors where their exponents won't work
# together nicely (e.g. 2**3 * 7**5), we've lost.
# Otherwise, combine the factors together for our answer.
def isPP(power):
	factors = []
	for p in primeish():
		if p*p > power:
			return None
		if power%p !=0:
			continue
		power, exp = extractPrimes(power, p)
		if exp< 2:
			return None

		if not factors:
			expGCD = exp
		else:
			expGCD = gcd(expGCD, exp)
			if expGCD < 2:
				return None
		factors.append( (p, exp))

		if power == 1:
			break
	combinedP = 1
	# print factors
	for p, exp in factors:
		combinedP = combinedP * p ** (exp/expGCD)
	return combinedP, expGCD

def gcd(x,y):
	while (y):
		x,y = y, x%y
	return x
	
def extractPrimes(n, p):
	exp = 0
	while n>1:
		n2, rem = divmod(n,p)
		if rem:
			break
		else:
			n, exp = n2, exp+1
	return n, exp


print isPP(1800 ** 14)