sol1.py

"""
Project Euler Problem 234: https://projecteuler.net/problem=234

For any integer n, consider the three functions

f1,n(x,y,z) = x^(n+1) + y^(n+1) - z^(n+1)
f2,n(x,y,z) = (xy + yz + zx)*(x^(n-1) + y^(n-1) - z^(n-1))
f3,n(x,y,z) = xyz*(xn-2 + yn-2 - zn-2)

and their combination

fn(x,y,z) = f1,n(x,y,z) + f2,n(x,y,z) - f3,n(x,y,z)

We call (x,y,z) a golden triple of order k if x, y, and z are all rational numbers
of the form a / b with 0 < a < b ≤ k and there is (at least) one integer n,
so that fn(x,y,z) = 0.

Let s(x,y,z) = x + y + z.
Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples
(x,y,z) of order 35.
All the s(x,y,z) and t must be in reduced form.

Find u + v.


Solution:

By expanding the brackets it is easy to show that
fn(x, y, z) = (x + y + z) * (x^n + y^n - z^n).

Since x,y,z are positive, the requirement fn(x, y, z) = 0 is fulfilled if and
only if x^n + y^n = z^n.

By Fermat's Last Theorem, this means that the absolute value of n can not
exceed 2, i.e. n is in {-2, -1, 0, 1, 2}. We can eliminate n = 0 since then the
equation would reduce to 1 + 1 = 1, for which there are no solutions.

So all we have to do is iterate through the possible numerators and denominators
of x and y, calculate the corresponding z, and check if the corresponding numerator and
denominator are integer and satisfy 0 < z_num < z_den <= 0. We use a set "uniquq_s"
to make sure there are no duplicates, and the fractions.Fraction class to make sure
we get the right numerator and denominator.

Reference:
https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
"""

from __future__ importannotations

fromfractionsimportFraction
frommathimportgcd, sqrt


defis_sq(number: int) ->bool:
"""
 Check if number is a perfect square.

 >>> is_sq(1)
 True
 >>> is_sq(1000001)
 False
 >>> is_sq(1000000)
 True
 """
sq: int=int(number**0.5)
returnnumber==sq*sq


defadd_three(
x_num: int, x_den: int, y_num: int, y_den: int, z_num: int, z_den: int
) ->tuple[int, int]:
"""
 Given the numerators and denominators of three fractions, return the
 numerator and denominator of their sum in lowest form.
 >>> add_three(1, 3, 1, 3, 1, 3)
 (1, 1)
 >>> add_three(2, 5, 4, 11, 12, 3)
 (262, 55)
 """
top: int=x_num*y_den*z_den+y_num*x_den*z_den+z_num*x_den*y_den
bottom: int=x_den*y_den*z_den
hcf: int=gcd(top, bottom)
top//=hcf
bottom//=hcf
returntop, bottom


defsolution(order: int=35) ->int:
"""
 Find the sum of the numerator and denominator of the sum of all s(x,y,z) for
 golden triples (x,y,z) of the given order.

 >>> solution(5)
 296
 >>> solution(10)
 12519
 >>> solution(20)
 19408891927
 """
unique_s: set=set()
hcf: int
total: Fraction=Fraction(0)
fraction_sum: tuple[int, int]

forx_numinrange(1, order+1):
forx_deninrange(x_num+1, order+1):
fory_numinrange(1, order+1):
fory_deninrange(y_num+1, order+1):
# n=1
z_num=x_num*y_den+x_den*y_num
z_den=x_den*y_den
hcf=gcd(z_num, z_den)
z_num//=hcf
z_den//=hcf
if0<z_num<z_den<=order:
fraction_sum=add_three(
x_num, x_den, y_num, y_den, z_num, z_den
 )
unique_s.add(fraction_sum)

# n=2
z_num= (
x_num*x_num*y_den*y_den+x_den*x_den*y_num*y_num
 )
z_den=x_den*x_den*y_den*y_den
ifis_sq(z_num) andis_sq(z_den):
z_num=int(sqrt(z_num))
z_den=int(sqrt(z_den))
hcf=gcd(z_num, z_den)
z_num//=hcf
z_den//=hcf
if0<z_num<z_den<=order:
fraction_sum=add_three(
x_num, x_den, y_num, y_den, z_num, z_den
 )
unique_s.add(fraction_sum)

# n=-1
z_num=x_num*y_num
z_den=x_den*y_num+x_num*y_den
hcf=gcd(z_num, z_den)
z_num//=hcf
z_den//=hcf
if0<z_num<z_den<=order:
fraction_sum=add_three(
x_num, x_den, y_num, y_den, z_num, z_den
 )
unique_s.add(fraction_sum)

# n=2
z_num=x_num*x_num*y_num*y_num
z_den= (
x_den*x_den*y_num*y_num+x_num*x_num*y_den*y_den
 )
ifis_sq(z_num) andis_sq(z_den):
z_num=int(sqrt(z_num))
z_den=int(sqrt(z_den))
hcf=gcd(z_num, z_den)
z_num//=hcf
z_den//=hcf
if0<z_num<z_den<=order:
fraction_sum=add_three(
x_num, x_den, y_num, y_den, z_num, z_den
 )
unique_s.add(fraction_sum)

fornum, deninunique_s:
total+=Fraction(num, den)

returntotal.denominator+total.numerator


if__name__=="__main__":
print(f"{solution() =}")