Problem 723
Pythagorean Quadrilaterals
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 $S(1411033124176203125)=S(5^6 \cdot 13^3 \cdot 17^2 \cdot 29 \cdot 37 \cdot 41 \cdot 53 \cdot 61)$.
毕达哥拉斯四边形
若一个毕达哥拉斯三角形也即直角三角形的直角边为$a$和$b$，斜边为$c$，则它们满足著名的勾股定理$a^2+b^2=c^2$。不过，它们同时也满足另外一条公式：
若一个三边长为$a$、$b$、$c$的三角形内接于一个半径为$r$的圆，当且仅当$a^2+b^2+c^2=8\ r^2$时这个三角形是毕达哥拉斯三角形。
类似地，若一个四边长为$a$、$b$、$c$、$d$的四边形$ABCD$内接于一个半径为$r$的圆，且满足$a^2+b^2+c^2+d^2=8\ r^2$，则我们称之为毕达哥拉斯四边形。
进一步地，如果这个毕达哥拉斯四边形的四个顶点都在格点上，且距离原点$O$的距离均为$r$（此时原点恰好也是圆心），则称之为毕达哥拉斯格点四边形。
记$f(r)$为内接于半径为$r$的圆中的不同毕达哥拉斯格点四边形的数目。例如，$f(1)=1$，$f(\sqrt 2)=1$，$f(\sqrt 5)=38$，$f(5)=167$。
如下图所示为两个$r=\sqrt 5$时的毕达哥拉斯格点四边形：
记$\displaystyle S(n)=\sum_{d \vert n} f(\sqrt d)$。例如，$S(325)=S(5^2 \cdot 13)=f(1)+f(\sqrt 5)+f(5)+f(\sqrt {13})+f(\sqrt{65})+f(5\sqrt{13})=2370$，$S(1105)=S(5\cdot 13 \cdot 17)=5535$。
求$S(1411033124176203125)=S(5^6 \cdot 13^3 \cdot 17^2 \cdot 29 \cdot 37 \cdot 41 \cdot 53 \cdot 61)$。