Detailed mathematical approach

Problem Summary

The implementations start from a reduced arithmetic form of the Pythagorean quadrilateral counting problem. For an integer

$n=\prod_{i=1}^{r} p_i^{e_i},$

the number of admissible quadrilaterals associated with the exact parameter \(n\) depends only on the exponent vector \((e_1,\dots,e_r)\), not on the actual prime values. The quantity required by the problem is then the divisor-summatory version of that exact count, so the solution never enumerates quadrilaterals geometrically.

Once the reduction has been made, everything becomes multiplicative. The whole task is therefore to evaluate a short list of prime-power factors and combine them with fixed coefficients.

Mathematical Approach

Let \(F(n)\) denote the exact-count function for one fixed arithmetic parameter \(n\), and let

$S(n)=\sum_{d\mid n} F(d)$

be the divisor-summatory quantity that the implementations actually evaluate. Because \(F(n)\) depends only on the exponents in the prime factorization of \(n\), it is convenient to work directly with those exponents.

Step 1: Keep Only the Prime-Exponent Profile

If \(n=\prod p_i^{e_i}\), then every divisor \(d\mid n\) has the form

$d=\prod_{i=1}^{r} p_i^{a_i},\qquad 0\le a_i\le e_i.$

The reduced counting formula ignores the actual primes \(p_i\) and uses only the exponents. That means two integers with the same exponent profile give the same value of \(F\) and the same value of \(S\).

For one prime-power exponent \(e\), define the five local factors

$B_1(e)=e+1,$

$B_2(e)=(e+1)^2,$

$B_3(e)=\left\lfloor\frac{(e+1)^2+1}{2}\right\rfloor,$

$B_4(e)=(e+1)^3,$

$B_5(e)=\frac{(e+1)(2e^2+4e+3)}{3}.$

These are exactly the prime-power building blocks used by the reduced formula.

Step 2: Build the Exact-Count Function \(F(n)\)

Since the local contributions factor over primes, the exact count is multiplicative and can be written as

$F(n)=7\prod_{i=1}^{r} B_1(e_i)-14\prod_{i=1}^{r} B_2(e_i)-4\prod_{i=1}^{r} B_3(e_i)+8\prod_{i=1}^{r} B_4(e_i)+4\prod_{i=1}^{r} B_5(e_i).$

The first factor \(B_1(e)=e+1\) is the familiar local contribution from the divisor-count function. The higher-degree factors encode the remaining symmetry classes that survive the reduction from geometry to arithmetic.

Nothing in this formula depends on the magnitudes of the primes themselves. Only the exponent vector matters.

Step 3: Lift from an Exact Parameter to All Divisors

The problem asks for the divisor-sum lift of \(F\), namely

$S(n)=\sum_{d\mid n} F(d)=\sum_{0\le a_i\le e_i} F(a_1,\dots,a_r).$

Because \(F\) is a fixed linear combination of products, the divisor sum separates prime by prime:

$S(n)=7\prod_{i=1}^{r}\left(\sum_{a=0}^{e_i} B_1(a)\right)-14\prod_{i=1}^{r}\left(\sum_{a=0}^{e_i} B_2(a)\right)-4\prod_{i=1}^{r}\left(\sum_{a=0}^{e_i} B_3(a)\right)+8\prod_{i=1}^{r}\left(\sum_{a=0}^{e_i} B_4(a)\right)+4\prod_{i=1}^{r}\left(\sum_{a=0}^{e_i} B_5(a)\right).$

So the entire divisor summation is absorbed into five closed-form cumulative factors.

Step 4: Evaluate the Cumulative Local Sums

Define

$A_j(e)=\sum_{a=0}^{e} B_j(a).$

Then the implementations use the closed forms

$A_1(e)=\frac{(e+1)(e+2)}{2},$

$A_2(e)=\frac{(e+1)(e+2)(2e+3)}{6},$

$A_3(e)=\sum_{a=0}^{e}\left\lfloor\frac{(a+1)^2+1}{2}\right\rfloor=\frac{1}{2}\left(A_2(e)+\left\lfloor\frac{e}{2}\right\rfloor+1\right),$

$A_4(e)=\sum_{a=0}^{e}(a+1)^3=A_1(e)^2,$

$A_5(e)=\sum_{a=0}^{e}\frac{(a+1)(2a^2+4a+3)}{3}=\frac{(e+1)(e+2)(e^2+3e+3)}{6}.$

The parity correction inside \(A_3(e)\) comes from the fact that odd squares contribute the extra \(+1\) when the integer division in \(B_3\) is written as a floor.

Step 5: Final Closed Form for the Required Quantity

Substituting the cumulative factors into the separated divisor sum gives

$\boxed{S(n)=7\prod_{i=1}^{r} A_1(e_i)-14\prod_{i=1}^{r} A_2(e_i)-4\prod_{i=1}^{r} A_3(e_i)+8\prod_{i=1}^{r} A_4(e_i)+4\prod_{i=1}^{r} A_5(e_i).}$

This is the whole reason the computation is so small: all geometric counting has already been compressed into five multiplicative prime-power terms and their divisor-sum lifts.

Worked Example: Exponent Vector \((2,1)\)

For a quick exact-count checkpoint, a single prime square has exponent vector \((2)\). Then

$B_1(2)=3,\quad B_2(2)=9,\quad B_3(2)=5,\quad B_4(2)=27,\quad B_5(2)=19,$

so

$F=7\cdot 3-14\cdot 9-4\cdot 5+8\cdot 27+4\cdot 19=167.$

Now lift to the divisor sum for exponent vector \((2,1)\). The cumulative factors are

$A_1(2)=6,\quad A_2(2)=14,\quad A_3(2)=8,\quad A_4(2)=36,\quad A_5(2)=26,$

$A_1(1)=3,\quad A_2(1)=5,\quad A_3(1)=3,\quad A_4(1)=9,\quad A_5(1)=7.$

Therefore

$\begin{aligned} S&=7(6\cdot 3)-14(14\cdot 5)-4(8\cdot 3)+8(36\cdot 9)+4(26\cdot 7)\\ &=126-980-96+2592+728\\ &=2370. \end{aligned}$

This matches the checkpoint embedded in the implementations. Another checkpoint is \(S(1,1,1)=5535\).

How the Code Works

The C++, Python, and Java implementations store the target exponent vector \((6,3,2,1,1,1,1,1)\) directly. No runtime factorization is needed, because the reduced arithmetic formula already depends only on these exponents.

Each implementation evaluates the five cumulative product terms \(\prod A_j(e_i)\), combines them with the coefficients \(7,-14,-4,8,4\), and prints the resulting exact integer. The structure is intentionally minimal: one generic product loop, five local formulas, and one final linear combination.

All arithmetic is exact. The C++ implementation uses unsigned 128-bit arithmetic, Python uses built-in arbitrary-precision integers, and Java uses arbitrary-precision integers as well. No floating-point calculations or geometric searches are involved.

Complexity Analysis

If the exponent vector has length \(r\), each of the five product terms requires one pass over that vector. The total running time is therefore \(O(r)\), and the extra memory usage is \(O(1)\) beyond the fixed storage for the exponents and a few accumulator variables. In the target instance, \(r=8\), so the computation is effectively constant time.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=723
2. Divisor function: Wikipedia — Divisor function
3. Multiplicative function: Wikipedia — Multiplicative function
4. Dirichlet convolution: Wikipedia — Dirichlet convolution
5. Faulhaber's formula: Wikipedia — Faulhaber's formula