a^2 + b^2 = c^2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
"""
Use the fact if a^2 + b^2 = c^2, then there exists m and n such that
a = m^2 - n^2, b = 2mn, and c = m^2 + n^2, where m > n.
Since a + b + c = 1000, we have
2m^2 + 2mn = 1000
m^2 + mn = 500
m(m+n) = 500
Now we just have to check the divisors of 500.
"""
import prime
m = -1
n = -1
for m_possible in prime.get_divisors(500):
n_possible = 500//m_possible - m_possible
if n_possible > 0 and m_possible > n_possible:
m = m_possible
n = n_possible
# there is only possibility, so we know we're done
break
print((m*m - n*n)*2*m*n*(m*m+n*n))
def is_amical(n):
sum1 = sum(prime.get_divisors(n)-set([n]))
sum2 = sum(prime.get_divisors(sum1)-set([sum1]))
#print( n, sum1, sum2 )
return n == sum2 and n != sum1
#/bin/python
# http://projecteuler.net/problem=211
'''
For a positive integer n, let 2(n) be the sum of the squares of its divisors. For example,
2(10) = 1 + 4 + 25 + 100 = 130.
Find the sum of all n, 0 n 64,000,000 such that 2(n) is a perfect square.
'''
from prime import get_divisors
def is_perfect_square(n):
root = n**0.5
return root == int(root)
tot = sum((n for n in range(0,60000000) if is_perfect_square(sum(get_divisors(n)))))
#tot = sum((n for n in range(0,600000) if is_perfect_square(sum(get_divisors(n)))))
print(tot)
