#!/usr/bin/env python3

########################################################################
#   Solves problem 197 from projectEuler.net.
#   Finds the sum of 2 values of a recursively defined sequence.
#   Copyright (C) 2011  Santiago Alessandri
#
#   This program is free software: you can redistribute it and/or modify
#   it under the terms of the GNU General Public License as published by
#   the Free Software Foundation, either version 3 of the License, or
#   (at your option) any later version.
#
#   This program is distributed in the hope that it will be useful,
#   but WITHOUT ANY WARRANTY; without even the implied warranty of
#   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
#   GNU General Public License for more details.
#
#   You should have received a copy of the GNU General Public License
#   along with this program.  If not, see <http://www.gnu.org/licenses/>.
#
#   You can contact me at san.lt.ss@gmail.com
#   Visit my blog at http://blog.san-ss.com.ar
#   Visit my wiki at http://wiki.san-ss.com.ar
########################################################################

from itertools import count, takewhile

is_square = lambda x: int(x ** 0.5) ** 2 == x

if __name__ == '__main__':
    for a in count(6):
        a_2 = a ** 2
        for f in (f for f in takewhile(lambda f: f < a, count(4)) if is_square(a_2 - f ** 2)):
            f_2 = f ** 2
            c_2 = a_2 - f_2
            setoff = 3 if (f & 1) else 2
            for e in (e for e in takewhile(lambda e: e ** 2 < c_2, count(setoff, 2)) if is_square(c_2 - e ** 2) and is_square(a_2 - e ** 2)):
                e_2 = e ** 2
                b_2 = c_2 - e_2
                d_2 = a_2 - e_2
                z = -(d_2 - c_2) // 2
                y = -(-d_2 - c_2 + 2 * b_2) // 2
                x = (d_2 + c_2) // 2
                print('The result is: (x){0} + (y){1} + (z){2} = {3}'.format(x, y, z, x + y + z))
                exit(0)