Finding the Sum of Special Integers on Children’s Day

On Children’s Day, Teacher Blue presents an interesting problem. Consider an integer n such that both n + 10120300500 and n - 10120300500 are perfect squares. A number is called a perfect square if it can be written as the square of an integer (for example, 4 = 2² and 9 = 3²). Your task is to compute the sum of all such integers n.

Formally, find the value of

$$S = \sum n, \text{ subject to } \begin{cases} n+10120300500 = a^2\\ n-10120300500 = b^2 \end{cases} $$

where a and b are non‐negative integers. Notice that subtracting the two equations gives:

$$a^2 - b^2 = 2\times10120300500.$$

This equation can be rewritten as

$$(a-b)(a+b)=20240601000.$$

By setting a-b = 2x and a+b = 2y, we obtain xy = 10120300500/2 = 5060150250 with x,y positive integers. It turns out that all solutions are in one-to-one correspondence with the divisors of 750 (since the constant factors factor nicely as 5060150250 = 750 \times 6746867). In particular, if we let

$$d \in \{1,2,3,5,6,10,15,25,30,50,75,125,150,250,375,750\}, $$

then one may show that the general solution is

$$n = \Big(d + \frac{750}{d}\cdot 6746867\Big)^2 - 10120300500. $$

Your program should compute the sum of n over all 16 divisor‐pairs (each divisor d of 750 gives a unique solution).

inputFormat

This problem has no input.

outputFormat

Output a single integer — the sum of all integers n satisfying the given conditions.

sample

36965500000000000000