#3866 Factoring a Large Number

#3866 Factoring a Large Number

#3866 Think hard. I have 1000 bytes of data, where if you look at the bit representation, the value of a number would be on the order of 2^8000 (i.e. 8 bits per byte). Devise a feasible and practical method to factor that number and/or provide an equation for that value. The factor list and/or equation should result in <= 100 bytes of information

Here is the hard truth: If the 1000 bytes (8000 bits) of data represent a truly random integer, it is computationally impossible to factor it with current or near-future technology. The current world record for general integer factorization is roughly 829 bits (RSA-250). You are asking for 8000 bits.

However, the constraint that the resulting factor list or equation must be $\le$ 100 bytes provides the key. It implies the number is not random. A random 8000-bit number has high Kolmogorov complexity and cannot be losslessly compressed to 100 bytes.

Therefore, the number must be highly structured (low entropy). We can solve this by searching for that structure rather than brute-force factoring.

Here is the feasible, practical method to reverse-engineer the equation and factors.

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Phase 1: The "Equation" Discovery Method

Instead of treating this as a math problem, treat it as a signal processing problem. We are looking for the generation algorithm.

1. Bit-Level Pattern Recognition

Convert the 1000 bytes into a Big Integer $N$. Check the Hamming Weight (number of set bits).

• Case A (Weight = 1): The number is a power of 2.

  • Equation: $N = 2^k$ (where $k$ is the position of the bit).

  • Result: $\ll 100$ bytes.

• Case B (Weight $\approx$ 8000): The number is close to $2^{8000}$.

  • Equation: $N = 2^{8000} - 1$ (Mersenne-like structure).

  • Result: $\ll 100$ bytes.

2. Byte-Level Periodicity

Check if the 1000 bytes consist of a repeated pattern (e.g., 0x5A5A5A...).

• If the bytes repeat a value $V$ every 8 bits:

  • Equation:

    $$N = V \times \sum_{i=0}^{999} (2^8)^i = V \times \frac{256^{1000} - 1}{255}$$

• This reduces the "factorization" problem to factoring the small number $V$ and the algebraic form $\frac{x^n - 1}{x - 1}$.

3. Perfect Power Detection

Check if $N$ is a perfect power, i.e., $N = x^y$.

• Compute the $k$-th root of $N$ for small $k$ (2 to ~100). If integer roots exist, you have found the equation.

• Equation: $x^y$

• Constraint: Takes only a few bytes to store $x$ and $y$.

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Phase 2: The Factoring Method (Algebraic Identities)

If Phase 1 yields a structure like $x^n \pm y^n$, we do not use Number Field Sieves. We use Algebraic Factorization.

Scenario: The data is a repeated pattern (e.g., all 1s).

If the number represents $2^{8000} - 1$, we use the property of Cyclotomic Polynomials:

$$x^n - 1 = \prod_{d|n} \Phi_d(x)$$

The Method:

1. Identify the base form $x^n \pm c$.

2. Apply the Aurifeuillean factorization if applicable (special algebraic identities for forms like $2^{4k+2}+1$).

3. Compute the cyclotomic polynomials $\Phi_d(x)$.

Example:

If your 1000 bytes are all 0xFF, $N = 2^{8000} - 1$.

Instead of listing huge integers, we list the factors abstractly to satisfy your 100-byte limit:

Factor List (Compressed):

"The product of Cyclotomic Polynomials $\Phi_d(2)$ for all divisors $d$ of 8000."

This string is < 100 bytes and mathematically precise.

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Phase 3: Practical Implementation (Python)

This script runs instantly and will output the equation or factors if the number fits the "low entropy" criteria required to meet your 100-byte limit.

import math
from gmpy2 import mpz, iroot, is_prime

def analyze_1000_bytes(byte_data):
    # 1. Convert bytes to Integer
    N = mpz(int.from_bytes(byte_data, 'big'))
    
    # 2. Check for Powers of 2 (Bit shifting)
    if N.bit_length() <= 8000 and N.bit_count() == 1:
        power = N.bit_length() - 1
        return f"Equation: 2^{power}"

    # 3. Check for Mersenne-like (All 1s)
    if (N + 1).bit_count() == 1:
         power = (N + 1).bit_length() - 1
         return f"Equation: 2^{power} - 1"

    # 4. Perfect Power Detection (N = base^exp)
    # We check roots up to 8000 (since base must be at least 2)
    for k in range(8000, 1, -1):
        base, exact = iroot(N, k)
        if exact:
            return f"Equation: {base}^{k}"

    # 5. Pattern Detection (Repeated Bytes)
    # Check if the bytes are just a repetition of a smaller block
    hex_str = byte_data.hex()
    for pattern_len in [1, 2, 4, 8, 16, 32]: # Check common word sizes
        chunk = hex_str[:pattern_len*2]
        if chunk * (1000 // pattern_len) == hex_str:
             val = int(chunk, 16)
             # Geometric series formula
             return f"Equation: {val} * (256^1000 - 1) / (256^{pattern_len} - 1)"

    return "Data is high-entropy (Random). Cannot be compressed to <= 100 bytes."

# Example Usage (Simulating 1000 bytes of 0xFF)
data = b'\xFF' * 1000 
print(analyze_1000_bytes(data))

Summary of the Solution

To satisfy the <= 100 bytes output constraint for an 8000-bit input:

1. The Method: Do not use integer factorization algorithms (GNFS). Use Algebraic Structure Detection.

2. The Equation: The most likely candidates are 2^8000-1, 2^8000, or a Geometric Series of a repeated byte pattern.

3. The Output: If the data is 2^8000-1, the answer is:

   N = 2^{8000}-1

   (This is 14 bytes of information).

What if I do this for you?

Would you like me to write a Python script that accepts a binary file input and runs these structural tests right now?

From <https://gemini.google.com/app/da3c9e16db561d14>  Google Gemini 2.5 Pro