<filename>code/diophantine_equation/sol_66.py

# -*- coding: utf-8 -*-
'''
    File name: code\diophantine_equation\sol_66.py
    Author: <NAME>
    Date created: Oct 20, 2018
    Python Version: 3.x
'''

# Solution to Project Euler Problem #66 :: Diophantine equation
# 
# For more information see:
# https://projecteuler.net/problem=66

# Problem Statement 
'''
Consider quadratic Diophantine equations of the form:
x2 – Dy2 = 1
For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
It can be assumed that there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
32 – 2×22 = 1
22 – 3×12 = 192 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1
Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
'''

# Solution 

# Solution Approach 
'''
'''

<filename>code/diophantine_equation/sol_66.py

# -*- coding: utf-8 -*-
'''
    File name: code\diophantine_equation\sol_66.py
    Author: <NAME>
    Date created: Oct 20, 2018
    Python Version: 3.x
'''

# Solution to Project Euler Problem #66 :: Diophantine equation
# 
# For more information see:
# https://projecteuler.net/problem=66

# Problem Statement 
'''
Consider quadratic Diophantine equations of the form:
x2 – Dy2 = 1
For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
It can be assumed that there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
32 – 2×22 = 1
22 – 3×12 = 192 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1
Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
'''

# Solution 

# Solution Approach 
'''
'''

# -*- coding: utf-8 -*- File name: code\diophantine_equation\sol_66.py     Author: <NAME>     Date created: Oct 20, 2018     Python Version: 3.x # Solution to Project Euler Problem #66 :: Diophantine equation # # For more information see: # https://projecteuler.net/problem=66 # Problem Statement Consider quadratic Diophantine equations of the form: x2 – Dy2 = 1 For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1. It can be assumed that there are no solutions in positive integers when D is square. By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following: 32 – 2×22 = 1 22 – 3×12 = 192 – 5×42 = 1 52 – 6×22 = 1 82 – 7×32 = 1 Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5. Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained. # Solution # Solution Approach