Open computational mathematics. AI-audited, not peer-reviewed. All code and data open for independent verification.

by cahlen complete

Hardware

1x NVIDIA B200 (183 GB VRAM) Intel Xeon 1.4 TB RAM
number-theorydiophantine-equationscomputational-mathematics b200 exhaustive-enumerationprime-sievesolution-counting

Key Results

Problem
Verify the Erdos-Straus conjecture (4/n = 1/x + 1/y + 1/z) and count solutions f(p) for all primes p up to N
Status
Kernel compiled but hangs on cudaDeviceSynchronize — needs batched progress reporting
Open Questions
Growth rate of f(p): does mean f(p) grow as c*log(p)?, Distribution of barely-solvable primes (f(p) = 1), Are there primes > 10⁸ with unusually low f(p)?, Connection between f(p) and p mod 4, p mod 24 residue classes

Erdos-Straus Conjecture: GPU Solution Counting

The Conjecture

For every integer n2n \geq 2, the equation

4n=1x+1y+1z\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}

has a solution in positive integers x,y,zx, y, z. This has been verified computationally for all n1014n \leq 10^{14} (Swett 1999, Elsholtz-Tao 2013 framework) but remains unproven.

What This Experiment Does

Rather than just verifying the conjecture (which is known to hold to enormous bounds), we count the number of solutions f(p)f(p) for each prime pp. Composites always have solutions inherited from their factors, so primes are the hard cases.

The solution count f(p)f(p) and its distribution contain structural information:

  • Barely-solvable primes (f(p)=1f(p) = 1): How rare are these? Do they thin out?
  • Growth rate: Does mean f(p)f(p) grow logarithmically, polynomially?
  • Residue class structure: How does f(p)f(p) depend on p(mod4)p \pmod{4}, p(mod24)p \pmod{24}?

Method

For each prime pp, the kernel enumerates all ordered triples (x,y,z)(x, y, z) with xyzx \leq y \leq z satisfying 4/p=1/x+1/y+1/z4/p = 1/x + 1/y + 1/z:

  1. For xx in [p/4+1,3p/4][\lfloor p/4 \rfloor + 1, \lfloor 3p/4 \rfloor]:
    • Compute remainder: num/den=(4xp)/(px)\text{num}/\text{den} = (4x - p) / (px)
    • For yy in [den/num,2den/num][\lceil \text{den}/\text{num} \rceil, \lfloor 2\text{den}/\text{num} \rfloor]:
      • Check if z=deny/(numyden)z = \text{den} \cdot y / (\text{num} \cdot y - \text{den}) is a positive integer with zyz \geq y

Each GPU thread handles one prime independently. Primes are sieved on CPU (Eratosthenes), then the full array is uploaded to GPU for parallel processing.

Reproduce

nvcc -O3 -arch=sm_90 -o erdos_straus scripts/experiments/erdos-straus/erdos_straus.cu -lm
./erdos_straus 100   # all primes to 10^8

References

  1. Erdos, P. (1950). “On a Diophantine equation.” Mat. Lapok, 1, pp. 192-210.
  2. Elsholtz, C. and Tao, T. (2013). “Counting the number of solutions to the Erdos-Straus equation on unit fractions.” Journal of the Australian Mathematical Society, 94(1), pp. 59-105.
  3. Swett, A. (1999). “Proof that 4/n = 1/x + 1/y + 1/z for n up to 10^14.”

Human-AI collaboration (Cahlen Humphreys + Claude). All code and data open.

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