Project-Euler/Problem9.py at master · MGNaik/Project-Euler · GitHub

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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sat Apr  5 16:08:30 2025

@author: root
"""
from math import sqrt

def is_natural_num (n):
    if (n % 1 == 0):
        if (n >= 0):
            return True
        else:
            return False
    else:
        return False

def solve_quad_formula (a,b,c):
    if (b**2 - 4*a*c >= 0):
        b1 = (-b + sqrt(b**2 - 4*a*c))/(2*a)
        b2 = (-b - sqrt(b**2 - 4*a*c))/(2*a)
        return [b1,b2]
    else:
        return 1


# a + b + c = 1000
# a^2 + b^2 = c^2
# a < b < c
# Solve for abc

# Find a,b,c
# Idea: Loop through different c's and find reciprocating a and b by solving
# the two equations


c = 1
while True:

    d = 2
    e = -2*(1000-c)
    f = (1000-c)**2 - c**2
    if (solve_quad_formula(d,e,f) != 1):
        [b1, b2] = solve_quad_formula(d,e,f)
    else:
        c = c + 1
        continue
    if (is_natural_num(b1)):
        b = b1
        a1 = 1000-c-b1
        if (is_natural_num(a1)):
            a = a1
            break
    elif (is_natural_num(b2)):
        b = b2
        a2 = 1000-c-b2
        if (is_natural_num(a2)):
            a  = a2
            break
    c = c+1

product = a*b*c

print(product)


# (1000 - c) - b = a

# sub into a^2+b^2=c^2

# ((1000-c)-b)^2 + b^2  = c^2
# (1000-c)^2 -2*(1000-c)*b + b^2 + b^2 = c^2
# 2b^2 - 2*(1000-c)*b + (1000-c)^2-c^2 = 0

# b1 = [ 2*(1000-c) + Sqrt ((2*(1000-c))^2 - (4*2*((1000-c)^2-c^2)))]/4
# b2 = [ 2*(1000-c) - Sqrt ((2*(1000-c))^2 - (4*2*((1000-c)^2-c^2)))]/4


# check b is a natural number else move onto next c


# Calculate a
# a1 = 1000-c-b1
# a2 = 1000-c-b2
# check a is a natural number