#
#Special Pythagorean triplet
#Problem 9
#
#A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
#
#a^2 + b^2 = c^2
#For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
#
#There exists exactly one Pythagorean triplet for which a + b + c = 1000.
#Find the product abc.
#
#Solution:
#

# a^2+b^2=c^2=(1000-a-b)^2= 1000^2-2000(a+b)+(a+b)^2=
# = 1000^2-2000(a+b)+a^2+2ab+b^2 =>
# => 0=1000^2-2000(a+b)+2ab =>
# => 500000=1000(a+b)-ab
# => 500=a+b - ab/1000
# => ab is dividable to 1000=2*2*2*5*5*5
# => a dividable at least to 5 and b to 2
#
#for a in range(5,1000,5):
#    for b in range(2,1000-a,2):
#        i+=1
#        if (a**2+b**2) == (1000-a-b)**2:
#            print( a, b, 1000-a-b, '->', a**2, '+', b**2, '=', a**2+b**2 )
#            print( ' while (1000-a-b)**2 =', (1000-a-b)**2 )
#            print( 'product =', a*b*(1000-a-b) )
#            print( 'product =', (a+b)*(5*10**5+10**3) - (a+b)**2 - 5*10**8 )
#
# a better solution
# => 500000=1000(a+b)-ab
# => (500000-1000a)/(1000-a)=b
# => 1000(500-a)/(1000-a)=b

for a in range(1,500):
    r=1000*(500-a)
    q=1000-a
    b=r//q
    if (a**2+b**2) == (1000-a-b)**2:
        print( a, b, 1000-a-b, '->', a**2, '+', b**2, '=', a**2+b**2 )
        print( ' while (1000-a-b)**2 =', (1000-a-b)**2 )
        print( 'product =', a*b*(1000-a-b) )
        print( 'product =', (a+b)*(5*(10**5)+10**3) - (a+b)**2 - 5*(10**8) )