<reponame>lcsm29/project-euler
# Solution of;
# Project Euler Problem 723: Pythagorean Quadrilaterals
# https://projecteuler.net/problem=723
# 
# A pythagorean triangle with catheti $a$ and $b$ and hypotenuse $c$ is 
# characterized by the well-known equation $a^2+b^2=c^2$. However, this can 
# also be formulated differently:When inscribed into a circle with radius $r$, 
# a triangle with sides $a$, $b$ and $c$ is pythagorean, if and only if 
# $a^2+b^2+c^2=8\, r^2$. Analogously, we call a quadrilateral $ABCD$ with 
# sides $a$, $b$, $c$ and $d$, inscribed in a circle with radius $r$, a 
# pythagorean quadrilateral, if $a^2+b^2+c^2+d^2=8\, r^2$. We further call a 
# pythagorean quadrilateral a pythagorean lattice grid quadrilateral, if all 
# four vertices are lattice grid points with the same distance $r$ from the 
# origin $O$ (which then happens to be the centre of the circumcircle). Let 
# $f(r)$ be the number of different pythagorean lattice grid quadrilaterals 
# for which the radius of the circumcircle is $r$. For example $f(1)=1$, 
# $f(\sqrt 2)=1$, $f(\sqrt 5)=38$ and $f(5)=167$. Two of the pythagorean 
# lattice grid quadrilaterals with $r=\sqrt 5$ are illustrated below:Let 
# $\displaystyle S(n)=\sum_{d \vert n} f(\sqrt d)$. For example, $S(325)=S(5^2 
# \cdot 13)=f(1)+f(\sqrt 5)+f(5)+f(\sqrt 
# {13})+f(\sqrt{65})+f(5\sqrt{13})=2370$ and $S(1105)=S(5\cdot 13 \cdot 
# 17)=5535$. Find $S(1411033124176203125)=S(5^6 \cdot 13^3 \cdot 17^2 \cdot 29 
# \cdot 37 \cdot 41 \cdot 53 \cdot 61)$.
# 
# by lcsm29 http://github.com/lcsm29/project-euler
import timed


def dummy(n):
    pass


if __name__ == '__main__':
    n = 1000
    i = 10000
    prob_id = 723
    timed.caller(dummy, n, i, prob_id)

# Solution of;
# Project Euler Problem 723: Pythagorean Quadrilaterals
# https://projecteuler.net/problem=723
# 
# A pythagorean triangle with catheti $a$ and $b$ and hypotenuse $c$ is 
# characterized by the well-known equation $a^2+b^2=c^2$. However, this can 
# also be formulated differently:When inscribed into a circle with radius $r$, 
# a triangle with sides $a$, $b$ and $c$ is pythagorean, if and only if 
# $a^2+b^2+c^2=8\, r^2$. Analogously, we call a quadrilateral $ABCD$ with 
# sides $a$, $b$, $c$ and $d$, inscribed in a circle with radius $r$, a 
# pythagorean quadrilateral, if $a^2+b^2+c^2+d^2=8\, r^2$. We further call a 
# pythagorean quadrilateral a pythagorean lattice grid quadrilateral, if all 
# four vertices are lattice grid points with the same distance $r$ from the 
# origin $O$ (which then happens to be the centre of the circumcircle). Let 
# $f(r)$ be the number of different pythagorean lattice grid quadrilaterals 
# for which the radius of the circumcircle is $r$. For example $f(1)=1$, 
# $f(\sqrt 2)=1$, $f(\sqrt 5)=38$ and $f(5)=167$. Two of the pythagorean 
# lattice grid quadrilaterals with $r=\sqrt 5$ are illustrated below:Let 
# $\displaystyle S(n)=\sum_{d \vert n} f(\sqrt d)$. For example, $S(325)=S(5^2 
# \cdot 13)=f(1)+f(\sqrt 5)+f(5)+f(\sqrt 
# {13})+f(\sqrt{65})+f(5\sqrt{13})=2370$ and $S(1105)=S(5\cdot 13 \cdot 
# 17)=5535$. Find $S(1411033124176203125)=S(5^6 \cdot 13^3 \cdot 17^2 \cdot 29 
# \cdot 37 \cdot 41 \cdot 53 \cdot 61)$.
# 
# by lcsm29 http://github.com/lcsm29/project-euler
import timed


def dummy(n):
    pass


if __name__ == '__main__':
    n = 1000
    i = 10000
    prob_id = 723
    timed.caller(dummy, n, i, prob_id)
# Solution of; # Project Euler Problem 723: Pythagorean Quadrilaterals # https://projecteuler.net/problem=723 # # A pythagorean triangle with catheti $a$ and $b$ and hypotenuse $c$ is # characterized by the well-known equation $a^2+b^2=c^2$. However, this can # also be formulated differently:When inscribed into a circle with radius $r$, # a triangle with sides $a$, $b$ and $c$ is pythagorean, if and only if # $a^2+b^2+c^2=8\, r^2$. Analogously, we call a quadrilateral $ABCD$ with # sides $a$, $b$, $c$ and $d$, inscribed in a circle with radius $r$, a # pythagorean quadrilateral, if $a^2+b^2+c^2+d^2=8\, r^2$. We further call a # pythagorean quadrilateral a pythagorean lattice grid quadrilateral, if all # four vertices are lattice grid points with the same distance $r$ from the # origin $O$ (which then happens to be the centre of the circumcircle). Let # $f(r)$ be the number of different pythagorean lattice grid quadrilaterals # for which the radius of the circumcircle is $r$. For example $f(1)=1$, # $f(\sqrt 2)=1$, $f(\sqrt 5)=38$ and $f(5)=167$. Two of the pythagorean # lattice grid quadrilaterals with $r=\sqrt 5$ are illustrated below:Let # $\displaystyle S(n)=\sum_{d \vert n} f(\sqrt d)$. For example, $S(325)=S(5^2 # \cdot 13)=f(1)+f(\sqrt 5)+f(5)+f(\sqrt # {13})+f(\sqrt{65})+f(5\sqrt{13})=2370$ and $S(1105)=S(5\cdot 13 \cdot # 17)=5535$. Find $S(1411033124176203125)=S(5^6 \cdot 13^3 \cdot 17^2 \cdot 29 # \cdot 37 \cdot 41 \cdot 53 \cdot 61)$. # # by lcsm29 http://github.com/lcsm29/project-euler